# Problem Overview You have a biased coin where the probability of getting heads on a single toss is \(.3\). You are asked: **What is the probability that the first head occurs on the 5th toss?** --- ## 1. Explanation ### What's Being Asked? - You toss the coin until you get the first head. - You want the first four tosses to be tails, and the fifth toss to be a head. --- ## 2. Logic and Problem-Solving Approach This is a classic **geometric probability problem**: - Probability of first success (head) on the \(n\)th trial: \[ P(\text{first head on } n\text{th toss}) = (1-p)^{n-1} \cdot p \] Where: - \(p\) = probability of head (\(.3\)) - \(n\) = 5 --- ## 3. Step-by-Step Solution ### Step 1: Probability of Tails in First 4 Tosses \[ P(\text{tail}) = 1 - .3 = .7 \] So, for 4 tails in a row: \[ (.7)^4 \] ### Step 2: Probability of Head on 5th Toss \[ P(\text{head}) = .3 \] ### Step 3: Multiply Probabilities \[ P(\text{first head on 5th toss}) = (.7)^4 \times .3 \] ### Step 4: Calculate Numeric Value \[ (.7)^4 = .7 \times .7 \times .7 \times .7 = .2401 \] \[ .2401 \times .3 = .07203 \] --- ## 4. Final Answer \[ \boxed{.07203} \] The probability that the first head occurs on the 5th toss is **.07203**.

# Probability of First Head on 5th Toss ## Problem Overview We have a biased coin that lands heads with a probability of \(0.3\). We need to find the probability that the first head occurs on the 5th toss. --- ## 1. Problem Statement - **Objective**: Determine the probability of getting the first head on the 5th toss. - **Conditions**: The first four tosses must be tails, and the fifth toss must be heads. --- ## 2. Understanding the Approach This problem can be modeled using **geometric probability**. The formula for finding the probability of the first success on the \(n\)th trial is: \[ P(\text{first head on } n\text{th toss}) = (1-p)^{n-1} \cdot p \] Where: - \(p\) = probability of heads = \(0.3\) - \(n\) = trial number = \(5\) --- ## 3. Step-by-Step Solution ### Step 1: Calculate the Probability of Tails The probability of getting tails on a single toss is: \[ P(\text{tail}) = 1 - p = 1 - 0.3 = 0.7 \] To find the probability of getting tails for the first four tosses: \[ P(\text{4 tails}) = (0.7)^4 \] ### Step 2: Calculate the Probability of Heads on the 5th Toss The probability of getting heads on the 5th toss is: \[ P(\text{head on 5th toss}) = 0.3 \] ### Step 3: Combine the Probabilities The total probability that the first head occurs on the 5th toss is given by: \[ P(\text{first head on 5th toss}) = P(\text{4 tails}) \times P(\text{head on 5th toss}) = (0.7)^4 \times 0.3 \] ### Step 4: Perform the Calculations First, calculate \((0.7)^4\): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Now multiply this result by \(0.3\): \[ P(\text{first head on 5th toss}) = 0.2401 \times 0.3 = 0.07203 \] --- ## 4. Final Answer The probability that the first head occurs on the 5th toss is: \[ \boxed{0.07203} \]

# Probability of First Head on 5th Toss ## Problem Overview You have a biased coin that lands heads with a probability of \(0.3\). The objective is to find the probability that the first head occurs on the 5th toss. --- ## Problem Statement - **Objective**: Determine the probability of getting the first head on the 5th toss. - **Conditions**: The first four tosses must result in tails, followed by a head on the fifth toss. --- ## Understanding the Approach This problem can be modeled using **geometric probability**. The formula for finding the probability of the first success (head) on the \(n\)th trial is: \[ P(\text{first head on } n\text{th toss}) = (1-p)^{n-1} \cdot p \] Where: - \(p\) = probability of heads = \(0.3\) - \(n\) = trial number = \(5\) --- ## Step-by-Step Solution ### Probability of Tails The probability of getting tails on a single toss is calculated as: \[ P(\text{tail}) = 1 - p = 1 - 0.3 = 0.7 \] For the first four tosses to be tails: \[ P(\text{4 tails}) = (0.7)^4 \] ### Probability of Heads on 5th Toss The probability of getting heads on the 5th toss is: \[ P(\text{head on 5th toss}) = 0.3 \] ### Combine the Probabilities The total probability that the first head occurs on the 5th toss is given by: \[ P(\text{first head on 5th toss}) = P(\text{4 tails}) \times P(\text{head on 5th toss}) = (0.7)^4 \times 0.3 \] ### Perform the Calculations First, calculate \((0.7)^4\): \[ (0.7)^4 = 0.7 \times 0.7 \times 0.7 \times 0.7 = 0.2401 \] Now multiply this result by \(0.3\): \[ P(\text{first head on 5th toss}) = 0.2401 \times 0.3 = 0.07203 \] --- ## Final Answer The probability that the first head occurs on the 5th toss is: \[ \boxed{0.07203} \]

# Probability of Early Head Dominance ## Problem Overview We are analyzing the results of tossing a fair coin seven times. We need to compute the conditional probability of event \(A\) (the first three tosses are heads) given event \(B\) (at least five heads occur in total). --- ## Definitions of Events - **Event \(A\)**: The first three tosses are all heads (HHH). - **Event \(B\)**: At least five heads occur in the seven tosses. --- ## Expressing \(P(B)\) The probability of obtaining exactly \(k\) heads in seven tosses is given by: \[ P(X = k) = \binom{7}{k} \left(\frac{1}{2}\right)^7 \] To find \(P(B)\): \[ P(B) = P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) \] Calculating each term: 1. **For \(k = 5\)**: \[ P(X = 5) = \binom{7}{5} \left(\frac{1}{2}\right)^7 = 21 \times \frac{1}{128} = \frac{21}{128} \] 2. **For \(k = 6\)**: \[ P(X = 6) = \binom{7}{6} \left(\frac{1}{2}\right)^7 = 7 \times \frac{1}{128} = \frac{7}{128} \] 3. **For \(k = 7\)**: \[ P(X = 7) = \binom{7}{7} \left(\frac{1}{2}\right)^7 = 1 \times \frac{1}{128} = \frac{1}{128} \] Combining these probabilities: \[ P(B) = \frac{21}{128} + \frac{7}{128} + \frac{1}{128} = \frac{29}{128} \] --- ## Determining \(P(A \cap B)\) For event \(A\) (HHH) to occur, the first three tosses must be heads. Therefore, we need to assess how many additional heads must occur in the remaining four tosses to satisfy event \(B\) (at least five heads). Given that the first three are heads, we need at least \(2\) more heads from the remaining \(4\) tosses. Thus, we can find \(P(A \cap B)\): \[ P(A \cap B) = P(A) \cdot P(X \geq 2 \text{ in remaining 4 tosses}) \] The probability of event \(A\): \[ P(A) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \] Now, we compute \(P(X \geq 2)\) for the remaining four tosses: \[ P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) \] Calculating each term: 1. **For \(k = 2\)**: \[ P(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^4 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} \] 2. **For \(k = 3\)**: \[ P(X = 3) = \binom{4}{3} \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \] 3. **For \(k = 4\)**: \[ P(X = 4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = 1 \times \frac{1}{16} = \frac{1}{16} \] Combining these probabilities gives: \[ P(X \geq 2) = \frac{3}{8} + \frac{1}{4} + \frac{1}{16} = \frac{3}{8} + \frac{2}{8} + \frac{1}{16} = \frac{5}{8} + \frac{1}{16} = \frac{10}{16} + \frac{1}{16} = \frac{11}{16} \] Thus, \[ P(A \cap B) = \frac{1}{8} \cdot \frac{11}{16} = \frac{11}{128} \] --- ## Final Calculation for Conditional Probability Now, we compute the conditional probability \(P(A | B)\): \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{11}{128}}{\frac{29}{128}} = \frac{11}{29} \] --- ## Final Answer The probability that the first three tosses were heads given that at least five heads occurred in total is: \[ \boxed{\frac{11}{29}} \]

# Probability of Early Head Dominance ## Given Information - The coin is fair, with a probability of Heads (H) on any single toss of \(0.5\). - The experiment consists of tossing the coin seven times. - Total possible outcomes: \(2^7 = 128\). - Probability of obtaining exactly \(k\) heads in seven tosses: \[ P(X = k) = \binom{7}{k} \left(\frac{1}{2}\right)^7 \] - The expected number of heads in seven tosses is \(3.5\). - The probability of getting at least five heads: \[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) \] - The probability that the first toss is a head is \(0.5\). - The probability that the first two tosses are both heads is \(\left(\frac{1}{2}\right)^2\). ## Objective Compute the conditional probability \(P(A | B)\) where: - **Event \(A\)**: The first three tosses are all heads (HHH). - **Event \(B\)**: At least five heads occur in the seven tosses. ## Concept Used The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] --- ## Step-by-Step Solution ### Expressing \(P(B)\) To find \(P(B)\), we need the probabilities of obtaining exactly 5, 6, and 7 heads: 1. For \(k = 5\): \[ P(X = 5) = \binom{7}{5} \left(\frac{1}{2}\right)^7 = 21 \times \frac{1}{128} = \frac{21}{128} \] 2. For \(k = 6\): \[ P(X = 6) = \binom{7}{6} \left(\frac{1}{2}\right)^7 = 7 \times \frac{1}{128} = \frac{7}{128} \] 3. For \(k = 7\): \[ P(X = 7) = \binom{7}{7} \left(\frac{1}{2}\right)^7 = 1 \times \frac{1}{128} = \frac{1}{128} \] Combining these probabilities: \[ P(B) = P(X = 5) + P(X = 6) + P(X = 7) = \frac{21}{128} + \frac{7}{128} + \frac{1}{128} = \frac{29}{128} \] ### Determining \(P(A \cap B)\) For event \(A\) to occur (the first three tosses are heads), we need at least 2 more heads from the remaining 4 tosses to satisfy event \(B\). First, calculate \(P(A)\): \[ P(A) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \] Next, find \(P(X \geq 2)\) in the remaining 4 tosses: 1. For \(k = 2\): \[ P(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^4 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} \] 2. For \(k = 3\): \[ P(X = 3) = \binom{4}{3} \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \] 3. For \(k = 4\): \[ P(X = 4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = 1 \times \frac{1}{16} = \frac{1}{16} \] Combining these probabilities gives: \[ P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = \frac{3}{8} + \frac{1}{4} + \frac{1}{16} = \frac{3}{8} + \frac{2}{8} + \frac{1}{16} = \frac{5}{8} + \frac{1}{16} = \frac{10}{16} + \frac{1}{16} = \frac{11}{16} \] Thus: \[ P(A \cap B) = P(A) \cdot P(X \geq 2) = \frac{1}{8} \cdot \frac{11}{16} = \frac{11}{128} \] ### Final Calculation for Conditional Probability Now compute \(P(A | B)\): \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{11}{128}}{\frac{29}{128}} = \frac{11}{29} \] --- ## Final Answer The probability that the first three tosses were heads given that at least five heads occurred in total is: \[ \boxed{\frac{11}{29}} \]

# Probability of Early Dominance in Production Cycles ## Given Information - The probability of Success (S) in any cycle is \(0.5\). - The production line is monitored for 9 consecutive cycles. - The total sample space contains \(2^9 = 512\) equally likely sequences. - The probability of exactly \(k\) successes is: \[ P(T = k) = \binom{9}{k} \left(\frac{1}{2}\right)^9 \] - The expected value of \(T\) (total number of successes) is \(4.5\). - The variance of \(T\) is \(2.25\). - The probability that the first three cycles are all successes is: \[ P(A) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \] ## Objective Compute the conditional probability \(P(A | B)\) where: - **Event \(A\)**: The first three cycles are all successes (SSS). - **Event \(B\)**: The total number of successes in the nine cycles is at least seven. ## Concept Used The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] --- ## Step-by-Step Solution ### Expressing \(P(B)\) To find \(P(B)\), we need the probabilities of obtaining exactly 7, 8, and 9 successes: 1. For \(k = 7\): \[ P(T = 7) = \binom{9}{7} \left(\frac{1}{2}\right)^9 = 36 \times \frac{1}{512} = \frac{36}{512} = \frac{9}{128} \] 2. For \(k = 8\): \[ P(T = 8) = \binom{9}{8} \left(\frac{1}{2}\right)^9 = 9 \times \frac{1}{512} = \frac{9}{512} \] 3. For \(k = 9\): \[ P(T = 9) = \binom{9}{9} \left(\frac{1}{2}\right)^9 = 1 \times \frac{1}{512} = \frac{1}{512} \] Combining these probabilities: \[ P(B) = P(T = 7) + P(T = 8) + P(T = 9) = \frac{9}{128} + \frac{9}{512} + \frac{1}{512} \] To add these, convert \(\frac{9}{128}\) to a fraction with a denominator of 512: \[ \frac{9}{128} = \frac{9 \times 4}{128 \times 4} = \frac{36}{512} \] Thus: \[ P(B) = \frac{36}{512} + \frac{9}{512} + \frac{1}{512} = \frac{46}{512} = \frac{23}{256} \] ### Determining \(P(A \cap B)\) For event \(A\) (the first three cycles are successes), we need to count how many additional successes must occur in the remaining six cycles to satisfy event \(B\) (at least 7 successes). Since the first three are successes, we need at least \(4\) successes from the remaining \(6\) cycles. Thus: \[ P(A \cap B) = P(A) \cdot P(T \geq 4 \text{ in remaining 6 cycles}) \] Calculate \(P(T \geq 4)\) for the remaining 6 cycles: 1. For \(k = 4\): \[ P(T = 4) = \binom{6}{4} \left(\frac{1}{2}\right)^6 = 15 \times \frac{1}{64} = \frac{15}{64} \] 2. For \(k = 5\): \[ P(T = 5) = \binom{6}{5} \left(\frac{1}{2}\right)^6 = 6 \times \frac{1}{64} = \frac{6}{64} = \frac{3}{32} \] 3. For \(k = 6\): \[ P(T = 6) = \binom{6}{6} \left(\frac{1}{2}\right)^6 = 1 \times \frac{1}{64} = \frac{1}{64} \] Combining these probabilities gives: \[ P(T \geq 4) = P(T = 4) + P(T = 5) + P(T = 6) = \frac{15}{64} + \frac{3}{32} + \frac{1}{64} \] Convert \(\frac{3}{32}\) to have a common denominator of 64: \[ \frac{3}{32} = \frac{3 \times 2}{32 \times 2} = \frac{6}{64} \] Thus: \[ P(T \geq 4) = \frac{15}{64} + \frac{6}{64} + \frac{1}{64} = \frac{22}{64} = \frac{11}{32} \] Now calculate \(P(A \cap B)\): \[ P(A \cap B) = P(A) \cdot P(T \geq 4) = \frac{1}{8} \cdot \frac{11}{32} = \frac{11}{256} \] ### Final Calculation for Conditional Probability Now compute \(P(A | B)\): \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{11}{256}}{\frac{23}{256}} = \frac{11}{23} \] --- ## Final Answer The probability that the first three cycles were all successes given that at least seven successes occurred overall is: \[ \boxed{\frac{11}{23}} \]

# Conditional Probability of High Signals ## Given Information - A binary signal transmission system outputs either a High (H) or Low (L) signal in eight consecutive time slots. - Probability of a High signal in any single time slot is \(p = 0.5\). - Let: - \(X\) = total number of High signals in eight slots, \(X \sim \text{Binomial}(8, 0.5)\). - \(Y\) = number of High signals in the first four slots, \(Y \sim \text{Binomial}(4, 0.5)\). - \(Z\) = number of High signals in the last four slots, \(Z \sim \text{Binomial}(4, 0.5)\). - Expected total number of High signals: \(4\). - Variance of \(X\): \(2\). - Probability of exactly \(k\) High signals: \[ P(X = k) = \binom{8}{k} \left(\frac{1}{2}\right)^8 \] - Probability that the first slot is High: \(0.5\). - Probability that the first two slots are both High: \(\left(\frac{1}{2}\right)^2 = 0.25\). ## Objective Compute the conditional probability \(P(A | B)\) where: - **Event \(A\)**: The first two time slots are both High (HH). - **Event \(B\)**: The total number of High signals in all eight slots is at least six. ## Concept Used The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] --- ## Step-by-Step Solution ### Expressing \(P(B)\) To find \(P(B)\), we need the probabilities of obtaining exactly 6, 7, and 8 High signals: 1. For \(k = 6\): \[ P(X = 6) = \binom{8}{6} \left(\frac{1}{2}\right)^8 = 28 \times \frac{1}{256} = \frac{28}{256} = \frac{7}{64} \] 2. For \(k = 7\): \[ P(X = 7) = \binom{8}{7} \left(\frac{1}{2}\right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} = \frac{1}{32} \] 3. For \(k = 8\): \[ P(X = 8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 = 1 \times \frac{1}{256} = \frac{1}{256} \] Combining these probabilities: \[ P(B) = P(X = 6) + P(X = 7) + P(X = 8) = \frac{7}{64} + \frac{1}{32} + \frac{1}{256} \] To add these, convert all fractions to have a common denominator of 256: - \(\frac{7}{64} = \frac{28}{256}\) - \(\frac{1}{32} = \frac{8}{256}\) Thus: \[ P(B) = \frac{28}{256} + \frac{8}{256} + \frac{1}{256} = \frac{37}{256} \] ### Determining \(P(A \cap B)\) For event \(A\) (the first two slots are High), we know that the first two signals are High, so we need to determine how many more High signals are required from the remaining six slots to satisfy event \(B\). Since the first two are High, we need at least \(4\) more High signals from the remaining \(6\) slots. Thus: \[ P(A \cap B) = P(A) \cdot P(X \geq 4 \text{ in remaining 6 slots}) \] Calculate \(P(X \geq 4)\) for the remaining 6 slots: 1. For \(k = 4\): \[ P(X = 4) = \binom{6}{4} \left(\frac{1}{2}\right)^6 = 15 \times \frac{1}{64} = \frac{15}{64} \] 2. For \(k = 5\): \[ P(X = 5) = \binom{6}{5} \left(\frac{1}{2}\right)^6 = 6 \times \frac{1}{64} = \frac{6}{64} = \frac{3}{32} \] 3. For \(k = 6\): \[ P(X = 6) = \binom{6}{6} \left(\frac{1}{2}\right)^6 = 1 \times \frac{1}{64} = \frac{1}{64} \] Combining these probabilities gives: \[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) = \frac{15}{64} + \frac{3}{32} + \frac{1}{64} \] Convert \(\frac{3}{32}\) to have a common denominator of 64: \[ \frac{3}{32} = \frac{6}{64} \] Thus: \[ P(X \geq 4) = \frac{15}{64} + \frac{6}{64} + \frac{1}{64} = \frac{22}{64} = \frac{11}{32} \] Now calculate \(P(A \cap B)\): \[ P(A \cap B) = P(A) \cdot P(X \geq 4) = \left(\frac{1}{4}\right) \cdot \left(\frac{11}{32}\right) = \frac{11}{128} \] ### Final Calculation for Conditional Probability Now compute \(P(A | B)\): \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{11}{128}}{\frac{37}{256}} = \frac{11 \cdot 256}{128 \cdot 37} = \frac{22}{37} \] --- ## Final Answer The probability that the first two signals were High given that at least six High signals occurred in total is: \[ \boxed{\frac{22}{37}} \]

# Probability of Early Head Dominance ## Given Information - A balanced experimental coin has a probability of Heads (H) on any single toss of \(0.5\). - The experiment involves tossing the coin seven times. - Total possible outcomes: \(2^7 = 128\). - The probability of obtaining exactly \(k\) heads in seven tosses: \[ P(X = k) = \binom{7}{k} \left(\frac{1}{2}\right)^7 \] - The expected number of heads is \(3.5\). - The probability of getting at least five heads is calculated as: \[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) \] ## Objective Compute the conditional probability \(P(A | B)\) where: - **Event \(A\)**: The first three tosses are all heads (HHH). - **Event \(B\)**: At least five heads occur in the seven tosses. ## Concept Used The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] --- ## Step-by-Step Solution ### Expressing \(P(B)\) To find \(P(B)\), we calculate the probabilities for \(X = 5\), \(X = 6\), and \(X = 7\): 1. For \(k = 5\): \[ P(X = 5) = \binom{7}{5} \left(\frac{1}{2}\right)^7 = 21 \times \frac{1}{128} = \frac{21}{128} \] 2. For \(k = 6\): \[ P(X = 6) = \binom{7}{6} \left(\frac{1}{2}\right)^7 = 7 \times \frac{1}{128} = \frac{7}{128} \] 3. For \(k = 7\): \[ P(X = 7) = \binom{7}{7} \left(\frac{1}{2}\right)^7 = 1 \times \frac{1}{128} = \frac{1}{128} \] Combining these probabilities gives: \[ P(B) = P(X = 5) + P(X = 6) + P(X = 7) = \frac{21}{128} + \frac{7}{128} + \frac{1}{128} = \frac{29}{128} \] ### Determining \(P(A \cap B)\) For event \(A\) (the first three tosses are heads), we know that we already have 3 heads. To satisfy event \(B\) (at least 5 heads), we need at least 2 more heads from the remaining 4 tosses. Thus, we can express \(P(A \cap B)\): \[ P(A \cap B) = P(A) \cdot P(X \geq 2 \text{ in remaining 4 tosses}) \] Calculate \(P(X \geq 2)\) for the remaining 4 tosses: 1. For \(k = 2\): \[ P(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^4 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} \] 2. For \(k = 3\): \[ P(X = 3) = \binom{4}{3} \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \] 3. For \(k = 4\): \[ P(X = 4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = 1 \times \frac{1}{16} = \frac{1}{16} \] Combining these probabilities gives: \[ P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = \frac{3}{8} + \frac{1}{4} + \frac{1}{16} \] Convert \(\frac{1}{4}\) to have a common denominator of 16: \[ \frac{1}{4} = \frac{4}{16} \] Thus: \[ P(X \geq 2) = \frac{3}{8} + \frac{4}{16} + \frac{1}{16} = \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{11}{16} \] Now calculate \(P(A)\): \[ P(A) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \] So now we can calculate \(P(A \cap B)\): \[ P(A \cap B) = \frac{1}{8} \cdot \frac{11}{16} = \frac{11}{128} \] ### Final Calculation for Conditional Probability Now compute \(P(A | B)\): \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{11}{128}}{\frac{29}{128}} = \frac{11}{29} \] --- ## Final Answer The probability that the first three tosses were heads given that at least five heads occurred in total is: \[ \boxed{\frac{11}{29}} \]

# Statistical Study Design ## 1) Situation Description In this study, we will compare two groups of employees in a company: **Group A** (employees who received training on time) and **Group B** (employees who received training late). The primary measurement we will take is the **job performance score** of employees, which is assessed on a scale of 1 to 10, based on their manager's evaluations after a specific period post-training. The interest in this study lies in understanding whether timely training impacts employees' job performance. By comparing the performance scores of both groups, we aim to provide insights into the effectiveness of timely training, which could inform future training schedules. ## 2) Data Type The data collected will be **quantitative**. The job performance scores will be numerical measurements, allowing for statistical analysis like calculating means, medians, and standard deviations. This quantification enables us to assess the central tendency of performance scores between the two groups and draw meaningful conclusions regarding the impact of timely training. ## 3) Null Hypothesis The null hypothesis (H0) for our statistical test can be stated as follows: - **In words**: There is no significant difference in job performance scores between employees who received training on time and those who received training late. - **As an equation**: \[ H_0: \mu_A = \mu_B \] where \( \mu_A \) is the mean job performance score of Group A, and \( \mu_B \) is the mean job performance score of Group B.

# Statistical Study Design ## 1) Situation Description In a corporate environment, the implementation of employee training programs is critical for enhancing overall job performance. This study aims to investigate the impact of timely training on employee performance by comparing two distinct groups: - **Group A**: Employees who received their training on time. - **Group B**: Employees who experienced delays in their training sessions. The performance of employees will be measured using a **job performance score**, which is derived from a standardized evaluation conducted by their managers. This score ranges from **1 to 10**, where a higher score indicates better performance. The evaluations will focus on various aspects of job performance, such as productivity, quality of work, teamwork, and adherence to company policies. The significance of this study lies in understanding the relationship between the timing of training and job performance. It is hypothesized that employees who receive timely training are better prepared to execute their responsibilities effectively, thereby resulting in higher performance scores. By identifying any potential correlations, the company can make informed decisions regarding training schedules, ultimately aiming to enhance workforce efficiency and productivity. ## 2) Data Type The data collected in this study will be classified as **quantitative**. This distinction is crucial for the following reasons: - **Nature of Measurement**: The job performance scores are numerical values that allow for a wide range of statistical analyses. These scores will be collected on a continuous scale, enabling the calculation of various statistical metrics. - **Statistical Analysis**: Quantitative data enables the use of descriptive statistics (mean, median, mode, standard deviation) and inferential statistics (t-tests, ANOVA) to compare the performance between the two groups. This analysis will facilitate the identification of patterns or trends that may suggest a significant impact of timely training on performance. - **Outcome Interpretation**: By summarizing the performance scores quantitatively, the results can be easily interpreted and communicated to stakeholders, providing a clear understanding of the implications of timely training. ## 3) Null Hypothesis The null hypothesis (H0) is a foundational element of the statistical hypothesis testing framework. It serves as a statement that there is no effect or difference between the groups being studied. In the context of our research, the null hypothesis can be articulated as follows: - **In words**: There is no significant difference in job performance scores between employees who received training on time and those who received training late. This implies that the timing of training does not affect the performance of employees. - **As an equation**: \[ H_0: \mu_A = \mu_B \] where: - \( \mu_A \) represents the mean job performance score of Group A (employees who received training on time). - \( \mu_B \) represents the mean job performance score of Group B (employees who received training late). The null hypothesis acts as a baseline for statistical testing. If the data analysis reveals a statistically significant difference between the two groups, we may reject the null hypothesis in favor of the alternative hypothesis (H1), which would propose that timely training does have a positive effect on employee performance. In summary, this study is designed to provide empirical evidence regarding the effectiveness of timely training in enhancing employee performance, utilizing quantitative data and statistical hypothesis testing to derive meaningful conclusions. By examining the performance scores of both groups, the organization can better understand how training timing influences job performance and make informed decisions about future training initiatives.

# Statistical Study Design ## Given Information - The study focuses on a corporate environment where employee training programs are implemented to enhance performance. - **Group A**: Employees who received training on time. - **Group B**: Employees who experienced delays in their training. - Performance is measured using a **job performance score** (1 to 10), based on evaluations from managers that encompass productivity, quality of work, teamwork, and adherence to policies. - The aim is to understand if timely training impacts job performance. ## Objective The objective is to determine if there is a significant difference in job performance scores between employees who received timely training and those who did not. ## Concept Used The study will utilize **hypothesis testing**, specifically focusing on the null hypothesis, which states that there is no significant difference between the two groups. The formula for the null hypothesis (H0) is: \[ H_0: \mu_A = \mu_B \] where: - \( \mu_A \) = mean job performance score of Group A - \( \mu_B \) = mean job performance score of Group B --- ## Step-by-Step Solution ### Data Collection Collect job performance scores from both groups after a defined period following training. Ensure that the evaluations are standardized and unbiased. ### Statistical Analysis Analyze the performance scores using appropriate statistical methods. Calculate the mean and standard deviation for both groups to understand the central tendency and variability of the scores. Conduct a t-test to compare the means of the two groups. This will help determine if any observed differences in scores are statistically significant. ### Interpretation of Results If the p-value obtained from the t-test is less than the significance level (commonly set at 0.05), reject the null hypothesis. This would indicate that there is a significant difference in job performance scores between the groups, suggesting that timely training may positively affect performance. Conversely, if the p-value is greater than 0.05, do not reject the null hypothesis, implying no significant difference in performance scores related to training timing. --- ## Final Answer Summary The study aims to assess whether timely training impacts employee job performance by comparing performance scores of two groups: those trained on time and those trained late. The null hypothesis states that there is no significant difference in performance scores, represented as \(H_0: \mu_A = \mu_B\). Statistical analysis will determine the validity of this hypothesis, guiding future training initiatives.

# Probability Density Function (PDF) for Random Variable \(x\) ## Given Information The probability density function (pdf) of a random variable \(x\) is given by: \[ f(x) = kx^2 \quad \text{for } 0 \leq x \leq 2 \] ## Objective 1. Find the normalization constant \(k\). 2. Compute the expected value \(E[x]\). 3. Compute the variance \(Var(x)\). ## Concept Used 1. **Normalization Condition**: The integral of the pdf over its range must equal 1: \[ \int_{a}^{b} f(x) \, dx = 1 \] For our case: \[ \int_{0}^{2} kx^2 \, dx = 1 \] 2. **Expected Value**: The expected value \(E[x]\) is calculated using: \[ E[x] = \int_{a}^{b} x f(x) \, dx \] 3. **Variance**: The variance \(Var(x)\) is calculated using: \[ Var(x) = E[x^2] - (E[x])^2 \] where \(E[x^2]\) is given by: \[ E[x^2] = \int_{a}^{b} x^2 f(x) \, dx \] --- ## Step-by-Step Solution ### Finding \(k\) First, we normalize the pdf: \[ \int_{0}^{2} kx^2 \, dx = 1 \] Calculating the integral: \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3} \] So, we have: \[ k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8} \] ### Computing \(E[x]\) Now, we calculate the expected value \(E[x]\): \[ E[x] = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^3 \, dx \] Calculating the integral: \[ \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - 0 = \frac{16}{4} = 4 \] Thus, \[ E[x] = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} \] ### Computing \(E[x^2]\) Next, we calculate \(E[x^2]\): \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx = \int_{0}^{2} x^2 \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^4 \, dx \] Calculating the integral: \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} \] Thus, \[ E[x^2] = \frac{3}{8} \cdot \frac{32}{5} = \frac{96}{40} = \frac{24}{10} = \frac{12}{5} \] ### Computing Variance Now, we can calculate the variance: \[ Var(x) = E[x^2] - (E[x])^2 = \frac{12}{5} - \left(\frac{3}{2}\right)^2 \] Calculating \( (E[x])^2 \): \[ (E[x])^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Converting \(\frac{9}{4}\) to a common denominator of \(20\): \[ \frac{9}{4} = \frac{45}{20} \] Now calculate the variance: \[ Var(x) = \frac{12}{5} - \frac{45}{20} = \frac{48}{20} - \frac{45}{20} = \frac{3}{20} \] --- ## Final Answers Summary - The normalization constant \(k\) is: \[ k = \frac{3}{8} \] - The expected value \(E[x]\) is: \[ E[x] = \frac{3}{2} \] - The variance \(Var(x)\) is: \[ Var(x) = \frac{3}{20} \]

# Probability Density Function (PDF) for Random Variable \(x\) ## Given Information The probability density function (pdf) of a random variable \(x\) is defined as: \[ f(x) = kx^2 \quad \text{for } 0 \leq x \leq 2 \] ## Objective 1. Determine the normalization constant \(k\). 2. Compute the expected value \(E[x]\). 3. Compute the variance \(Var(x)\). ## Concept Used 1. **Normalization Condition**: The probability density function must satisfy the condition: \[ \int_{a}^{b} f(x) \, dx = 1 \] 2. **Expected Value**: The expected value \(E[x]\) is calculated using: \[ E[x] = \int_{a}^{b} x f(x) \, dx \] 3. **Variance**: The variance \(Var(x)\) is calculated using: \[ Var(x) = E[x^2] - (E[x])^2 \] where \(E[x^2]\) is given by: \[ E[x^2] = \int_{a}^{b} x^2 f(x) \, dx \] --- ## Step-by-Step Solution ### Finding \(k\) To find the normalization constant \(k\), we need to ensure that the total area under the pdf equals 1: \[ \int_{0}^{2} kx^2 \, dx = 1 \] Calculating the integral of \(x^2\): \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3} \] Setting up the equation for \(k\): \[ k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8} \] ### Computing \(E[x]\) Now we compute the expected value \(E[x]\): \[ E[x] = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^3 \, dx \] Calculating the integral of \(x^3\): \[ \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - 0 = \frac{16}{4} = 4 \] Thus, we find: \[ E[x] = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} \] ### Computing \(E[x^2]\) Next, we calculate \(E[x^2]\): \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx = \int_{0}^{2} x^2 \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^4 \, dx \] Calculating the integral of \(x^4\): \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} \] Thus: \[ E[x^2] = \frac{3}{8} \cdot \frac{32}{5} = \frac{96}{40} = \frac{24}{10} = \frac{12}{5} \] ### Computing Variance Now we can calculate the variance: \[ Var(x) = E[x^2] - (E[x])^2 = \frac{12}{5} - \left(\frac{3}{2}\right)^2 \] Calculating \( (E[x])^2 \): \[ (E[x])^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Converting \(\frac{9}{4}\) to a common denominator of \(20\): \[ \frac{9}{4} = \frac{45}{20} \] Now calculate the variance: \[ Var(x) = \frac{12}{5} - \frac{45}{20} = \frac{48}{20} - \frac{45}{20} = \frac{3}{20} \] --- ## Final Answers Summary - The normalization constant \(k\) is: \[ k = \frac{3}{8} \] - The expected value \(E[x]\) is: \[ E[x] = \frac{3}{2} \] - The variance \(Var(x)\) is: \[ Var(x) = \frac{3}{20} \]

# Z-Score for 80% of Scores on the Normal Distribution ## Given Information In a normal distribution, 80% of the scores fall within a specific range around the mean. We need to find the corresponding z-scores that encompass this central 80% of the distribution. ## Objective Determine the z-scores such that 80% of the scores fall between these two values, with the lower z-score being negative and the upper z-score being positive. ## Concept Used In a standard normal distribution: - The total area under the curve is 1 (or 100%). - To find the z-scores representing the middle 80%, we need to determine the area in the tails of the distribution: \[ \text{Area in each tail} = \frac{1 - 0.80}{2} = 0.10 \] Therefore, we will look for the z-scores that correspond to cumulative probabilities of 0.10 and 0.90. --- ## Step-by-Step Solution ### Finding Z-Scores 1. **Calculate the lower z-score** (10th percentile): - We want the z-score where the cumulative probability is 0.10. - Using the Excel function: \[ \text{Lower z-score} = \text{NORM.INV}(0.10, 0, 1) \] 2. **Calculate the upper z-score** (90th percentile): - We want the z-score where the cumulative probability is 0.90. - Using the Excel function: \[ \text{Upper z-score} = \text{NORM.INV}(0.90, 0, 1) \] ### Excel Function Implementation In Excel, you can use the following formulas: - For the lower z-score: ```excel =NORM.INV(0.10, 0, 1) ``` - For the upper z-score: ```excel =NORM.INV(0.90, 0, 1) ``` ### Result Interpretation Using the above Excel functions will yield: - **Lower z-score**: Approximately -1.2816 - **Upper z-score**: Approximately 1.2816 --- ## Final Answer Summary For an 80% confidence interval in a normal distribution, the z-scores that encompass this range are approximately: - Lower z-score: \(-1.2816\) - Upper z-score: \(1.2816\) Thus, 80% of scores fall between z-scores of approximately -1.28 and +1.28.

# Independence of Events A and B ## Given Information - \( P(A) = 0.4 \) - \( P(B) = 0.5 \) - \( P(A \cup B) = 0.7 \) ## Objective Determine if events \(A\) and \(B\) are independent by using the definition of independence for events. ## Concept Used Two events \(A\) and \(B\) are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] We can derive \(P(A \cap B)\) using the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] From this, we can rearrange to find \(P(A \cap B)\): \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] --- ## Step-by-Step Solution ### Calculate \(P(A \cap B)\) Using the provided probabilities, we can substitute into the rearranged equation: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] Substituting the known values: \[ P(A \cap B) = 0.4 + 0.5 - 0.7 \] \[ P(A \cap B) = 0.9 - 0.7 = 0.2 \] ### Check for Independence Now we need to check if \(P(A \cap B)\) is equal to \(P(A) \cdot P(B)\): \[ P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2 \] ### Conclusion Since: \[ P(A \cap B) = 0.2 \quad \text{and} \quad P(A) \cdot P(B) = 0.2 \] This implies: \[ P(A \cap B) = P(A) \cdot P(B) \] Thus, events \(A\) and \(B\) are independent. --- ## Final Answer Summary Yes, events \(A\) and \(B\) are independent because: \[ P(A \cap B) = P(A) \cdot P(B) = 0.2 \]

# Independence of Events A and B ## Given Information - \( P(A) = 0.4 \) - \( P(B) = 0.5 \) - \( P(A \cup B) = 0.7 \) ## Objective To determine whether events \(A\) and \(B\) are independent by applying the mathematical definition of independence. ## Concept Used Two events \(A\) and \(B\) are considered independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] To find \(P(A \cap B)\), we can use the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Rearranging this gives: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] --- ## Step-by-Step Solution ### Calculate \(P(A \cap B)\) Substituting the known probabilities into the rearranged equation: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] Plugging in the values: \[ P(A \cap B) = 0.4 + 0.5 - 0.7 \] \[ P(A \cap B) = 0.9 - 0.7 = 0.2 \] ### Check for Independence Next, we need to see if \(P(A \cap B)\) equals \(P(A) \cdot P(B)\): \[ P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2 \] ### Conclusion Since we have: \[ P(A \cap B) = 0.2 \quad \text{and} \quad P(A) \cdot P(B) = 0.2 \] This indicates: \[ P(A \cap B) = P(A) \cdot P(B) \] Thus, we conclude that events \(A\) and \(B\) are independent. --- ## Final Answer Summary Events \(A\) and \(B\) are independent because the condition: \[ P(A \cap B) = P(A) \cdot P(B) = 0.2 \] holds true.

# Independence of Events A and B ## Given Information - \( P(A) = 0.4 \) - \( P(B) = 0.5 \) - \( P(A \cup B) = 0.7 \) ## Objective To assess whether events \(A\) and \(B\) are independent by employing the mathematical definition of independence. ## Concept Used Two events \(A\) and \(B\) are independent if the following condition holds: \[ P(A \cap B) = P(A) \cdot P(B) \] To find \(P(A \cap B)\), we can use the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Rearranging this gives: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] --- ## Step-by-Step Solution ### Calculate \(P(A \cap B)\) Substituting the known probabilities into the rearranged equation: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] Plugging in the values: \[ P(A \cap B) = 0.4 + 0.5 - 0.7 \] Calculating this gives: \[ P(A \cap B) = 0.9 - 0.7 = 0.2 \] ### Check for Independence Now, we need to verify if \(P(A \cap B)\) is equal to \(P(A) \cdot P(B)\): \[ P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2 \] ### Conclusion Since we find: \[ P(A \cap B) = 0.2 \quad \text{and} \quad P(A) \cdot P(B) = 0.2 \] We conclude that: \[ P(A \cap B) = P(A) \cdot P(B) \] Thus, events \(A\) and \(B\) are independent. --- ## Final Answer Summary Events \(A\) and \(B\) are independent because the equality: \[ P(A \cap B) = P(A) \cdot P(B) = 0.2 \] is satisfied.

# Independence of Events A and B ## Given Information - \( P(A) = 0.4 \) - \( P(B) = 0.5 \) - \( P(A \cup B) = 0.7 \) ## Objective To determine if events \(A\) and \(B\) are independent by utilizing the mathematical definition of independence. ## Concept Used Two events \(A\) and \(B\) are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] To find \(P(A \cap B)\), we can use the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] From this, we can rearrange to solve for \(P(A \cap B)\): \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] --- ## Step-by-Step Solution ### Calculate \(P(A \cap B)\) Substituting the known probabilities into the rearranged equation: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] Plugging in the values: \[ P(A \cap B) = 0.4 + 0.5 - 0.7 \] Calculating this gives: \[ P(A \cap B) = 0.9 - 0.7 = 0.2 \] ### Check for Independence Next, we need to check if \(P(A \cap B)\) equals \(P(A) \cdot P(B)\): \[ P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2 \] ### Conclusion Since: \[ P(A \cap B) = 0.2 \quad \text{and} \quad P(A) \cdot P(B) = 0.2 \] This indicates: \[ P(A \cap B) = P(A) \cdot P(B) \] Thus, we conclude that events \(A\) and \(B\) are independent. --- ## Final Answer Summary Events \(A\) and \(B\) are independent because the condition: \[ P(A \cap B) = P(A) \cdot P(B) = 0.2 \] is satisfied.

# Assessing Independence of Events A and B ## Given Information - Probability of event \(A\): \( P(A) = 0.4 \) - Probability of event \(B\): \( P(B) = 0.5 \) - Probability of the union of events \(A\) and \(B\): \( P(A \cup B) = 0.7 \) ## Objective The goal is to evaluate whether events \(A\) and \(B\) are independent by applying the mathematical definition of independence. ## Concept Employed For two events \(A\) and \(B\) to be independent, the following relationship must hold true: \[ P(A \cap B) = P(A) \cdot P(B) \] To find \(P(A \cap B)\), we can utilize the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Rearranging this equation allows us to express \(P(A \cap B)\): \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] --- ## Step-by-Step Solution ### Calculation of \(P(A \cap B)\) We substitute the known probabilities into the rearranged equation: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] Inserting the values we have: \[ P(A \cap B) = 0.4 + 0.5 - 0.7 \] Solving this gives: \[ P(A \cap B) = 0.9 - 0.7 = 0.2 \] ### Verification of Independence Next, we confirm if \(P(A \cap B)\) is equal to the product of \(P(A)\) and \(P(B)\): \[ P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2 \] ### Conclusion We find that: \[ P(A \cap B) = 0.2 \quad \text{and} \quad P(A) \cdot P(B) = 0.2 \] This indicates that: \[ P(A \cap B) = P(A) \cdot P(B) \] Thus, we conclude that events \(A\) and \(B\) are indeed independent. --- ## Final Answer Summary Events \(A\) and \(B\) are independent as the equality: \[ P(A \cap B) = P(A) \cdot P(B) = 0.2 \] holds true.

# Probability Density Function (PDF) for Random Variable \(x\) ## Given Information The probability density function (pdf) of a random variable \(x\) is defined as: \[ f(x) = kx^2 \quad \text{for } 0 \leq x \leq 2 \] ## Objective 1. Find the normalization constant \(k\). 2. Compute the expected value \(E[x]\). 3. Compute the variance \(Var(x)\). ## Concept Used 1. **Normalization Condition**: The total area under the pdf must equal 1: \[ \int_{0}^{2} f(x) \, dx = 1 \] 2. **Expected Value**: The expected value \(E[x]\) is calculated using: \[ E[x] = \int_{0}^{2} x f(x) \, dx \] 3. **Variance**: The variance \(Var(x)\) is computed as: \[ Var(x) = E[x^2] - (E[x])^2 \] where \(E[x^2]\) is given by: \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx \] --- ## Step-by-Step Solution ### Finding \(k\) To find the normalization constant \(k\), we need to ensure that the total area under the pdf equals 1: \[ \int_{0}^{2} kx^2 \, dx = 1 \] Calculating the integral: \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3} \] Setting up the equation for \(k\): \[ k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8} \] ### Computing \(E[x]\) Now, we calculate the expected value \(E[x]\): \[ E[x] = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^3 \, dx \] Calculating the integral of \(x^3\): \[ \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - 0 = \frac{16}{4} = 4 \] Thus, we find: \[ E[x] = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} \] ### Computing \(E[x^2]\) Next, we calculate \(E[x^2]\): \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx = \int_{0}^{2} x^2 \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^4 \, dx \] Calculating the integral of \(x^4\): \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} \] Thus: \[ E[x^2] = \frac{3}{8} \cdot \frac{32}{5} = \frac{96}{40} = \frac{24}{10} = \frac{12}{5} \] ### Computing Variance Now we can calculate the variance: \[ Var(x) = E[x^2] - (E[x])^2 = \frac{12}{5} - \left(\frac{3}{2}\right)^2 \] Calculating \( (E[x])^2 \): \[ (E[x])^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Converting \(\frac{9}{4}\) to a common denominator of \(20\): \[ \frac{9}{4} = \frac{45}{20} \] Now calculate the variance: \[ Var(x) = \frac{12}{5} - \frac{45}{20} = \frac{48}{20} - \frac{45}{20} = \frac{3}{20} \] --- ## Final Answers Summary - The normalization constant \(k\) is: \[ k = \frac{3}{8} \] - The expected value \(E[x]\) is: \[ E[x] = \frac{3}{2} \] - The variance \(Var(x)\) is: \[ Var(x) = \frac{3}{20} \]

# Probability Density Function (PDF) for Random Variable \(x\) ## Given Information The probability density function (pdf) of a random variable \(x\) is defined as: \[ f(x) = kx^2 \quad \text{for } 0 \leq x \leq 2 \] ## Objective 1. Determine the normalization constant \(k\). 2. Calculate the expected value \(E[x]\). 3. Calculate the variance \(Var(x)\). ## Concept Used 1. **Normalization Condition**: The integral of the pdf over its range must equal 1: \[ \int_{0}^{2} f(x) \, dx = 1 \] 2. **Expected Value**: The expected value \(E[x]\) is calculated using: \[ E[x] = \int_{0}^{2} x f(x) \, dx \] 3. **Variance**: The variance \(Var(x)\) is computed as: \[ Var(x) = E[x^2] - (E[x])^2 \] where \(E[x^2]\) is given by: \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx \] --- ## Step-by-Step Solution ### Finding \(k\) To find the normalization constant \(k\), we set up the equation ensuring the area under the pdf equals 1: \[ \int_{0}^{2} kx^2 \, dx = 1 \] Calculating the integral: \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3} \] Setting up the equation for \(k\): \[ k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8} \] ### Computing \(E[x]\) Now, we calculate the expected value \(E[x]\): \[ E[x] = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^3 \, dx \] Calculating the integral of \(x^3\): \[ \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - 0 = \frac{16}{4} = 4 \] Thus, we find: \[ E[x] = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} \] ### Computing \(E[x^2]\) Next, we calculate \(E[x^2]\): \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx = \int_{0}^{2} x^2 \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^4 \, dx \] Calculating the integral of \(x^4\): \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} \] Thus: \[ E[x^2] = \frac{3}{8} \cdot \frac{32}{5} = \frac{96}{40} = \frac{24}{10} = \frac{12}{5} \] ### Computing Variance Now we can calculate the variance: \[ Var(x) = E[x^2] - (E[x])^2 = \frac{12}{5} - \left(\frac{3}{2}\right)^2 \] Calculating \( (E[x])^2 \): \[ (E[x])^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Converting \(\frac{9}{4}\) to a common denominator of \(20\): \[ \frac{9}{4} = \frac{45}{20} \] Now calculate the variance: \[ Var(x) = \frac{12}{5} - \frac{45}{20} = \frac{48}{20} - \frac{45}{20} = \frac{3}{20} \] --- ## Final Answers Summary - The normalization constant \(k\) is: \[ k = \frac{3}{8} \] - The expected value \(E[x]\) is: \[ E[x] = \frac{3}{2} \] - The variance \(Var(x)\) is: \[ Var(x) = \frac{3}{20} \]

# Probability Density Function (PDF) for Random Variable \(x\) ## Given Information The probability density function (pdf) of a random variable \(x\) is given by: \[ f(x) = kx^2 \quad \text{for } 0 \leq x \leq 2 \] ## Objective 1. Find the normalization constant \(k\). 2. Compute the expected value \(E[x]\). 3. Compute the variance \(Var(x)\). ## Concept Used 1. **Normalization Condition**: The total area under the pdf must equal 1: \[ \int_{0}^{2} f(x) \, dx = 1 \] 2. **Expected Value**: The expected value \(E[x]\) is calculated using: \[ E[x] = \int_{0}^{2} x f(x) \, dx \] 3. **Variance**: The variance \(Var(x)\) is computed as: \[ Var(x) = E[x^2] - (E[x])^2 \] where \(E[x^2]\) is given by: \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx \] --- ## Step-by-Step Solution ### Finding \(k\) To find the normalization constant \(k\), we need to ensure that the integral of the pdf over its range equals 1: \[ \int_{0}^{2} kx^2 \, dx = 1 \] Calculating the integral: \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3} \] Setting up the equation for \(k\): \[ k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8} \] ### Computing \(E[x]\) Now, we calculate the expected value \(E[x]\): \[ E[x] = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^3 \, dx \] Calculating the integral of \(x^3\): \[ \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - 0 = \frac{16}{4} = 4 \] Thus, we find: \[ E[x] = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} \] ### Computing \(E[x^2]\) Next, we calculate \(E[x^2]\): \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx = \int_{0}^{2} x^2 \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^4 \, dx \] Calculating the integral of \(x^4\): \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} \] Thus: \[ E[x^2] = \frac{3}{8} \cdot \frac{32}{5} = \frac{96}{40} = \frac{24}{10} = \frac{12}{5} \] ### Computing Variance Now we can calculate the variance: \[ Var(x) = E[x^2] - (E[x])^2 = \frac{12}{5} - \left(\frac{3}{2}\right)^2 \] Calculating \( (E[x])^2 \): \[ (E[x])^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Converting \(\frac{9}{4}\) to a common denominator of \(20\): \[ \frac{9}{4} = \frac{45}{20} \] Now calculate the variance: \[ Var(x) = \frac{12}{5} - \frac{45}{20} = \frac{48}{20} - \frac{45}{20} = \frac{3}{20} \] --- ## Final Answers Summary - The normalization constant \(k\) is: \[ k = \frac{3}{8} \] - The expected value \(E[x]\) is: \[ E[x] = \frac{3}{2} \] - The variance \(Var(x)\) is: \[ Var(x) = \frac{3}{20} \]

# Probability Density Function (PDF) for Random Variable \(x\) ## Given Information The probability density function (pdf) of a random variable \(x\) is given by: \[ f(x) = kx^2 \quad \text{for } 0 \leq x \leq 2 \] ## Objective 1. Determine the normalization constant \(k\). 2. Compute the expected value \(E[x]\). 3. Compute the variance \(Var(x)\). ## Concept Used 1. **Normalization Condition**: The integral of the pdf over its range must equal 1: \[ \int_{0}^{2} f(x) \, dx = 1 \] 2. **Expected Value**: The expected value \(E[x]\) is calculated using: \[ E[x] = \int_{0}^{2} x f(x) \, dx \] 3. **Variance**: The variance \(Var(x)\) is computed as: \[ Var(x) = E[x^2] - (E[x])^2 \] where \(E[x^2]\) is given by: \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx \] --- ## Step-by-Step Solution ### Finding \(k\) To find the normalization constant \(k\), we solve for the condition that the area under the pdf equals 1: \[ \int_{0}^{2} kx^2 \, dx = 1 \] Calculating the integral: \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - 0 = \frac{8}{3} \] Setting up the equation for \(k\): \[ k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8} \] ### Computing \(E[x]\) Now, we calculate the expected value \(E[x]\): \[ E[x] = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^3 \, dx \] Calculating the integral of \(x^3\): \[ \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - 0 = \frac{16}{4} = 4 \] Thus, we find: \[ E[x] = \frac{3}{8} \cdot 4 = \frac{12}{8} = \frac{3}{2} \] ### Computing \(E[x^2]\) Next, we calculate \(E[x^2]\): \[ E[x^2] = \int_{0}^{2} x^2 f(x) \, dx = \int_{0}^{2} x^2 \cdot \frac{3}{8} x^2 \, dx = \frac{3}{8} \int_{0}^{2} x^4 \, dx \] Calculating the integral of \(x^4\): \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} \] Thus: \[ E[x^2] = \frac{3}{8} \cdot \frac{32}{5} = \frac{96}{40} = \frac{24}{10} = \frac{12}{5} \] ### Computing Variance Now we can calculate the variance: \[ Var(x) = E[x^2] - (E[x])^2 = \frac{12}{5} - \left(\frac{3}{2}\right)^2 \] Calculating \( (E[x])^2 \): \[ (E[x])^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Converting \(\frac{9}{4}\) to a common denominator of \(20\): \[ \frac{9}{4} = \frac{45}{20} \] Now calculate the variance: \[ Var(x) = \frac{12}{5} - \frac{45}{20} = \frac{48}{20} - \frac{45}{20} = \frac{3}{20} \] --- ## Final Answers Summary - The normalization constant \(k\) is: \[ k = \frac{3}{8} \] - The expected value \(E[x]\) is: \[ E[x] = \frac{3}{2} \] - The variance \(Var(x)\) is: \[ Var(x) = \frac{3}{20} \]

# Fatigue Analysis Using the Modified Goodman Criterion ## Given Information - **Material**: Cold-drawn AISI 1018 steel - **Diameter of shaft**: \(20 \, \text{mm}\) - **Critical stress element**: - Normal stress (\(\sigma\)): \(77.5 \, \text{MPa}\) - Shear stress (\(\tau\)): \(20 \, \text{MPa}\) ## Objective Determine the minimum factor of safety for fatigue based on infinite life using the modified Goodman criterion. ## Concept Used The modified Goodman criterion relates the alternating stress, mean stress, and yield strength to evaluate the factor of safety against fatigue. The formula is represented as: \[ \frac{\sigma_a}{\sigma'_e} + \frac{\sigma_m}{\sigma_u} \leq 1 \] Where: - \(\sigma_a\) = Alternating stress - \(\sigma_m\) = Mean stress - \(\sigma'_e\) = Endurance limit (approximately \(0.5 \cdot \sigma_u\) for steel) - \(\sigma_u\) = Ultimate tensile strength ### Step 1: Determine Properties of AISI 1018 Steel For AISI 1018 steel, typical properties are: - **Ultimate tensile strength (\(\sigma_u\))**: Approximately \(440 \, \text{MPa}\) - **Endurance limit (\(\sigma'_e\))**: \[ \sigma'_e \approx 0.5 \cdot \sigma_u = 0.5 \cdot 440 \, \text{MPa} = 220 \, \text{MPa} \] ### Step 2: Calculate Mean and Alternating Stresses Given that the shaft is subjected to a constant load, we can consider: - Mean stress (\(\sigma_m\)) is equal to the critical normal stress: \[ \sigma_m = \sigma = 77.5 \, \text{MPa} \] For the alternating stress, if we assume a fully reversed loading condition, the maximum shear stress \(\tau\) can be used to find the alternating stress: \[ \sigma_a = 2 \tau = 2 \cdot 20 \, \text{MPa} = 40 \, \text{MPa} \] ### Step 3: Substitute Values into the Modified Goodman Equation Substituting into the modified Goodman criterion: \[ \frac{40 \, \text{MPa}}{220 \, \text{MPa}} + \frac{77.5 \, \text{MPa}}{440 \, \text{MPa}} \leq 1 \] Calculating each term: \[ \frac{40}{220} \approx 0.1818 \] \[ \frac{77.5}{440} \approx 0.1761 \] Adding the terms together: \[ 0.1818 + 0.1761 = 0.3579 \] ### Step 4: Calculate the Factor of Safety The factor of safety (\(N\)) can be defined in terms of the modified Goodman equation as: \[ N = \frac{1}{\frac{\sigma_a}{\sigma'_e} + \frac{\sigma_m}{\sigma_u}} \] Thus: \[ N = \frac{1}{0.3579} \approx 2.7935 \] ### Final Answer The minimum factor of safety for fatigue is: \[ \boxed{2.7935} \]

# Fatigue Analysis Using the Modified Goodman Criterion ## Given Information - **Material**: Cold-drawn AISI 1018 steel - **Diameter of shaft**: \(20 \, \text{mm}\) - **Critical stress element**: - Normal stress (\(\sigma\)): \(77.5 \, \text{MPa}\) - Shear stress (\(\tau\)): \(20 \, \text{MPa}\) ## Objective Determine the minimum factor of safety for fatigue based on infinite life using the modified Goodman criterion. ## Concept Used The modified Goodman criterion relates the alternating stress, mean stress, and yield strength to evaluate the factor of safety against fatigue. The formula is represented as: \[ \frac{\sigma_a}{\sigma'_e} + \frac{\sigma_m}{\sigma_u} \leq 1 \] Where: - \(\sigma_a\) = Alternating stress - \(\sigma_m\) = Mean stress - \(\sigma'_e\) = Endurance limit (approximately \(0.5 \cdot \sigma_u\) for steel) - \(\sigma_u\) = Ultimate tensile strength ### Step 1: Determine Properties of AISI 1018 Steel For AISI 1018 steel, typical properties are: - **Ultimate tensile strength (\(\sigma_u\))**: Approximately \(440 \, \text{MPa}\) - **Endurance limit (\(\sigma'_e\))**: \[ \sigma'_e \approx 0.5 \cdot \sigma_u = 0.5 \cdot 440 \, \text{MPa} = 220 \, \text{MPa} \] ### Step 2: Calculate Mean and Alternating Stresses Given that the shaft is subjected to a constant load, we can consider: - Mean stress (\(\sigma_m\)) is equal to the critical normal stress: \[ \sigma_m = \sigma = 77.5 \, \text{MPa} \] For fully reversed loading, the alternating stress can be approximated as: \[ \sigma_a = 2 \tau = 2 \cdot 20 \, \text{MPa} = 40 \, \text{MPa} \] ### Step 3: Substitute Values into the Modified Goodman Equation Substituting into the modified Goodman criterion: \[ \frac{40 \, \text{MPa}}{220 \, \text{MPa}} + \frac{77.5 \, \text{MPa}}{440 \, \text{MPa}} \leq 1 \] Calculating each term: \[ \frac{40}{220} \approx 0.1818 \] \[ \frac{77.5}{440} \approx 0.1761 \] Adding the terms together: \[ 0.1818 + 0.1761 = 0.3579 \] ### Step 4: Calculate the Factor of Safety The factor of safety (\(N\)) can be defined as: \[ N = \frac{1}{\frac{\sigma_a}{\sigma'_e} + \frac{\sigma_m}{\sigma_u}} \] Thus: \[ N = \frac{1}{0.3579} \approx 2.7935 \] ### Final Answer The minimum factor of safety for fatigue is: \[ \boxed{2.7935} \] --- ### Correction of Approach Upon review, the calculations appear to be correct based on the modified Goodman criterion. If further adjustments are needed, please specify which values or assumptions need to be revisited.

# Fatigue Analysis Using the Modified Goodman Criterion ## Given Information - **Material**: Cold-drawn AISI 1018 steel - **Diameter of shaft**: \(d = 20 \, \text{mm}\) - **Critical stress elements**: - Normal stress (\(\sigma\)): \(77.5 \, \text{MPa}\) - Shear stress (\(\tau\)): \(20 \, \text{MPa}\) - The shaft rotates at a constant speed. ## Objective Determine the minimum factor of safety for fatigue based on infinite life using the Modified Goodman criterion. ## Concept Used For a rotating shaft, the analysis involves: - Bending stress is completely reversed. - Torsional stress is steady. For combined bending and torsion, equivalent stresses are calculated using the Distortion Energy (von Mises) theory: \[ \sigma_a' = \sqrt{\sigma_a^2 + 3\tau_a^2} \] \[ \sigma_m' = \sigma_m + 3\tau_m^2 \] ### Modified Goodman Relation The criterion is expressed as: \[ \frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u} = 1 \] where: - \(S_e\) = Endurance limit - \(S_u\) = Ultimate tensile strength --- ## Step 1: Identify Alternating and Mean Components - **Bending stress** is fully reversed: - \(\sigma_a = \sigma = 77.5 \, \text{MPa}\) - \(\sigma_m = 0\) - **Torsional stress** is steady: - \(\tau_a = 0\) - \(\tau_m = 20 \, \text{MPa}\) ### Step 2: Equivalent Alternating and Mean Stresses **Equivalent alternating stress (\(\sigma_a'\))**: \[ \sigma_a' = \sigma_a = 77.5 \, \text{MPa} \] **Equivalent mean stress (\(\sigma_m'\))**: \[ \sigma_m' = 0 + 3 \left(20\right)^2 = 0 + 3(400) = 1200 \, \text{MPa} \] ### Step 3: Material Properties For cold-drawn AISI 1018 steel: - **Ultimate tensile strength (\(S_u\))**: Approximately \(440 \, \text{MPa}\) - **Unmodified endurance limit (\(S_e'\))**: \[ S_e' = 0.5 \cdot S_u = 0.5 \cdot 440 \, \text{MPa} = 220 \, \text{MPa} \] ### Step 4: Marin Factors **Surface factor (machined/cold-drawn)**: \[ k_a = 4.51 \cdot S_u^{-0.265} = 4.51 \cdot (440)^{-0.265} \approx 0.897 \] **Size factor (bending, \(d = 20 \, \text{mm}\))**: \[ k_b = \left(\frac{d}{7.62}\right)^{-0.107} = \left(\frac{20}{7.62}\right)^{-0.107} \approx 0.902 \] **Corrected endurance limit**: \[ S_e = k_a k_b S_e' = (0.897)(0.902)(220) \approx 177.99 \, \text{MPa} \] ### Step 5: Apply Modified Goodman Criterion Substituting values into the modified Goodman equation: \[ \frac{77.5}{177.99} + \frac{1200}{440} = 1 \] Calculating each term: \[ \frac{77.5}{177.99} \approx 0.436 \] \[ \frac{1200}{440} \approx 2.727 \] Combining: \[ 0.436 + 2.727 = 3.163 \] ### Step 6: Factor of Safety The factor of safety \(n\) is defined as: \[ n = \frac{1}{\frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u}} \] Calculating: \[ n = \frac{1}{0.436 + 2.727} = \frac{1}{3.163} \approx 0.3166 \] ### Final Answer The minimum factor of safety for fatigue based on infinite life is: \[ \boxed{0.3166} \]

# Fatigue Analysis Using the Modified Goodman Criterion ## Given Information - **Material**: Cold-drawn AISI 1018 steel - **Diameter of shaft**: \(d = 20 \, \text{mm}\) - **Critical stress elements**: - Normal stress (\(\sigma\)): \(77.5 \, \text{MPa}\) - Shear stress (\(\tau\)): \(20 \, \text{MPa}\) - The shaft rotates at a constant speed. --- ## Objective Determine the minimum factor of safety for fatigue based on infinite life using the Modified Goodman criterion. --- ## Concept Used For a rotating shaft: - **Bending stress** is completely reversed. - **Torsional stress** is steady. ### Equivalent Stress Formulas For combined bending and torsion, equivalent stresses are calculated using the Distortion Energy (von Mises) theory: - **Equivalent alternating stress**: \[ \sigma_a' = \sigma_a + 3\tau_a \] - **Equivalent mean stress**: \[ \sigma_m' = \sigma_m + 3\tau_m \] ### Modified Goodman Relation The Modified Goodman relation is expressed as: \[ \frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u} = 1 \] Where: - \(S_e\) = Endurance limit - \(S_u\) = Ultimate tensile strength --- ## Step 1: Identify Alternating and Mean Components Since the shaft rotates: - **Bending stress** is completely reversed: - \(\sigma_a = 77.5 \, \text{MPa}\) - \(\sigma_m = 0\) - **Torsional stress** is steady: - \(\tau_a = 0\) - \(\tau_m = 20 \, \text{MPa}\) --- ## Step 2: Equivalent Alternating and Mean Stresses Calculate the equivalent alternating and mean stresses: 1. **Equivalent alternating stress**: \[ \sigma_a' = \sigma_a = 77.5 \, \text{MPa} \] 2. **Equivalent mean stress**: \[ \sigma_m' = 0 + 3(20)^2 = 0 + 3(400) = 1200 \, \text{MPa} \] --- ## Step 3: Material Properties For cold-drawn AISI 1018 steel: - **Ultimate tensile strength (\(S_u\))**: Approximately \(440 \, \text{MPa}\) - **Unmodified endurance limit (\(S_e'\))**: \[ S_e' = 0.5 \cdot S_u = 0.5 \cdot 440 \, \text{MPa} = 220 \, \text{MPa} \] --- ## Step 4: Marin Factors ### Calculate Marin Factors 1. **Surface factor (machined/cold-drawn)**: \[ k_a = 4.51S_u^{-0.265} = 4.51(440)^{-0.265} \approx 0.897 \] 2. **Size factor (bending, \(d = 20 \, \text{mm}\))**: \[ k_b = \left(\frac{d}{7.62}\right)^{-0.107} = \left(\frac{20}{7.62}\right)^{-0.107} \approx 0.902 \] 3. **Corrected endurance limit**: \[ S_e = k_a \cdot k_b \cdot S_e' = (0.897)(0.902)(220) \approx 177.99 \, \text{MPa} \] --- ## Step 5: Apply Modified Goodman Criterion Substituting values into the Modified Goodman equation: \[ \frac{77.5}{177.99} + \frac{1200}{440} = 1 \] ### Calculate Each Term 1. **First Term**: \[ \frac{77.5}{177.99} \approx 0.436 \] 2. **Second Term**: \[ \frac{1200}{440} \approx 2.727 \] 3. **Combine**: \[ 0.436 + 2.727 = 3.163 \] --- ## Step 6: Factor of Safety The factor of safety (\(n\)) is defined as: \[ n = \frac{1}{\frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u}} \] ### Calculation Substituting the calculated terms: \[ n = \frac{1}{0.436 + 2.727} \approx \frac{1}{3.163} \approx 0.3166 \] --- ## Final Answer The minimum factor of safety for fatigue based on infinite life is: \[ \boxed{0.3166} \] --- ### Note If further adjustments or clarifications are needed, please specify any areas that require a review of values or assumptions.

# Fatigue Analysis Using the Modified Goodman Criterion ## Given Information - **Material**: Cold-drawn AISI 1018 steel - **Diameter of Shaft**: \(d = 20 \, \text{mm}\) - **Critical Stress Elements**: - Normal Stress (\(\sigma\)): \(77.5 \, \text{MPa}\) - Shear Stress (\(\tau\)): \(20 \, \text{MPa}\) - The shaft operates at a constant rotational speed. ## Objective Determine the minimum factor of safety for fatigue based on infinite life using the Modified Goodman criterion. ## Concept Used In the context of a rotating shaft: - Bending stress is completely reversed. - Torsional stress is steady. ### Equivalent Stress Formulas For combined bending and torsion, equivalent stresses are calculated using the Distortion Energy (von Mises) theory: - **Equivalent Alternating Stress**: \[ \sigma_a' = \sigma_a^2 + 3\tau_a^2 \] - **Equivalent Mean Stress**: \[ \sigma_m' = \sigma_m^2 + 3\tau_m^2 \] ### Modified Goodman Relation The Modified Goodman relation is represented as: \[ \frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u} = 1 \] Where: - \(S_e\) = Endurance limit - \(S_u\) = Ultimate tensile strength ## Step 1: Identify Alternating and Mean Components For the rotating shaft: - **Bending Stress** (fully reversed): - \(\sigma_a = 77.5 \, \text{MPa}\) - \(\sigma_m = 0\) - **Torsional Stress** (steady): - \(\tau_a = 0\) - \(\tau_m = 20 \, \text{MPa}\) ## Step 2: Equivalent Alternating and Mean Stresses Calculate the equivalent alternating and mean stresses: 1. **Equivalent Alternating Stress**: \[ \sigma_a' = \sigma_a = 77.5 \, \text{MPa} \] 2. **Equivalent Mean Stress**: \[ \sigma_m' = 0 + 3(20)^2 = 0 + 3(400) = 1200 \, \text{MPa} \] ## Step 3: Material Properties For cold-drawn AISI 1018 steel: - **Ultimate Tensile Strength (\(S_u\))**: Approximately \(440 \, \text{MPa}\) - **Unmodified Endurance Limit (\(S_e'\))**: \[ S_e' = 0.5 \cdot S_u = 0.5 \cdot 440 \, \text{MPa} = 220 \, \text{MPa} \] ## Step 4: Marin Factors ### Calculate Marin Factors 1. **Surface Factor (Machined/Cold-Drawn)**: \[ k_a = 4.51 \cdot S_u^{-0.265} = 4.51 \cdot (440)^{-0.265} \approx 0.897 \] 2. **Size Factor (Bending, \(d = 20 \, \text{mm}\))**: \[ k_b = \left(\frac{d}{7.62}\right)^{-0.107} = \left(\frac{20}{7.62}\right)^{-0.107} \approx 0.902 \] 3. **Corrected Endurance Limit**: \[ S_e = k_a \cdot k_b \cdot S_e' = (0.897)(0.902)(220) \approx 177.99 \, \text{MPa} \] ## Step 5: Apply Modified Goodman Criterion Substituting values into the Modified Goodman equation: \[ \frac{77.5}{177.99} + \frac{1200}{440} = 1 \] ### Calculate Each Term 1. **First Term**: \[ \frac{77.5}{177.99} \approx 0.436 \] 2. **Second Term**: \[ \frac{1200}{440} \approx 2.727 \] 3. **Combine**: \[ 0.436 + 2.727 = 3.163 \] ## Step 6: Factor of Safety The factor of safety (\(n\)) is defined as: \[ n = \frac{1}{\frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u}} \] ### Calculation Substituting the calculated terms: \[ n = \frac{1}{0.436 + 2.727} = \frac{1}{3.163} \approx 0.3166 \] ## Final Answer The minimum factor of safety for fatigue based on infinite life is: \[ \boxed{0.3166} \] --- ### Note If further adjustments or clarifications are needed, please specify any areas that require a review of values or assumptions.

# Probability of Rolling a Fair Die ## Given Information A fair six-sided die has outcomes: {1, 2, 3, 4, 5, 6}. Each outcome has an equal probability of occurring. ## Objective Calculate the probabilities for the following events: - (a) Rolling a 4 - (b) Rolling an even number - (c) Rolling a number greater than 2 ## Total Outcomes The total number of possible outcomes when rolling a die is \(6\). --- ## Step-by-Step Solution ### (a) Probability of Rolling a 4 There is only one favorable outcome for rolling a 4. \[ P(\text{Rolling a 4}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{6} \] ### (b) Probability of Rolling an Even Number The even numbers on a die are {2, 4, 6}. There are three favorable outcomes. \[ P(\text{Rolling an even number}) = \frac{3}{6} = \frac{1}{2} \] ### (c) Probability of Rolling a Number Greater Than 2 The numbers greater than 2 are {3, 4, 5, 6}. There are four favorable outcomes. \[ P(\text{Rolling a number greater than 2}) = \frac{4}{6} = \frac{2}{3} \] --- ## Final Answers Summary - (a) The probability of rolling a 4 is: \[ \boxed{\frac{1}{6}} \] - (b) The probability of rolling an even number is: \[ \boxed{\frac{1}{2}} \] - (c) The probability of rolling a number greater than 2 is: \[ \boxed{\frac{2}{3}} \]

# Stationary Distribution and Ergodicity of Markov Chain ## Given Information The transition matrix \(P\) is given as: \[ P = \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix} \] ## Objective 1. Find the stationary distribution. 2. Determine if the chain is ergodic. 3. Find \(P^n\) as \(n \to \infty\). --- ## Step 1: Finding the Stationary Distribution The stationary distribution \(\pi\) satisfies the equation: \[ \pi P = \pi \] with the constraint that the sum of the probabilities equals 1: \[ \pi_1 + \pi_2 = 1 \] Let \(\pi = (\pi_1, \pi_2)\). Expanding the stationary equation: \[ \begin{pmatrix} \pi_1 & \pi_2 \end{pmatrix} \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix} = \begin{pmatrix} \pi_1 & \pi_2 \end{pmatrix} \] This leads to the following equations: 1. \(\pi_1 \cdot 0.7 + \pi_2 \cdot 0.4 = \pi_1\) 2. \(\pi_1 \cdot 0.3 + \pi_2 \cdot 0.6 = \pi_2\) Substituting \(\pi_2 = 1 - \pi_1\) into the first equation: \[ 0.7\pi_1 + 0.4(1 - \pi_1) = \pi_1 \] Expanding and rearranging: \[ 0.7\pi_1 + 0.4 - 0.4\pi_1 = \pi_1 \] \[ 0.3\pi_1 + 0.4 = \pi_1 \] \[ 0.4 = \pi_1 - 0.3\pi_1 \] \[ 0.4 = 0.7\pi_1 \] Solving for \(\pi_1\): \[ \pi_1 = \frac{0.4}{0.7} \approx 0.5714 \] Now substituting back to find \(\pi_2\): \[ \pi_2 = 1 - \pi_1 = 1 - 0.5714 \approx 0.4286 \] Thus, the stationary distribution is: \[ \pi = \left(0.5714, 0.4286\right) \] --- ## Step 2: Determining if the Chain is Ergodic A Markov chain is ergodic if it is: - Irreducible: It is possible to reach any state from any state. - Aperiodic: The greatest common divisor (gcd) of the number of steps needed to return to a state is 1. **Irreducibility**: Each state has a non-zero probability of transitioning to another state: - From state 1, you can move to state 2 (0.3). - From state 2, you can move to state 1 (0.4). Since both states can reach each other, the chain is irreducible. **Aperiodicity**: The gcd of the return steps is not restricted to a fixed period. Since each state can be revisited in different numbers of steps (e.g., from state 1 to state 1 takes 1 step or 2 steps), the chain is aperiodic. Therefore, the chain is ergodic. --- ## Step 3: Finding \(P^n\) as \(n \to \infty\) As \(n\) approaches infinity, \(P^n\) converges to a matrix where each row is the stationary distribution: \[ P^n \to \begin{pmatrix} 0.5714 & 0.4286 \\ 0.5714 & 0.4286 \end{pmatrix} \] This implies the long-term behavior of the Markov chain will be: - The probability of being in state 1 is approximately \(0.5714\). - The probability of being in state 2 is approximately \(0.4286\). --- ## Final Summary 1. The stationary distribution is: \[ \pi = \left(0.5714, 0.4286\right) \] 2. The chain is ergodic. 3. As \(n \to \infty\): \[ P^n \to \begin{pmatrix} 0.5714 & 0.4286 \\ 0.5714 & 0.4286 \end{pmatrix} \]

# Independence of \(x_{(1)}\) and \(\sum_{i=1}^n (x_i - x_{(1)})\) ## Given Information - Let \(x_1, x_2, \ldots, x_n\) be independent and identically distributed (iid) samples from the probability density function: \[ f_\theta(x) = e^{-(x - \theta)} \quad \text{for } x \geq \theta \] - Here, \(x_{(1)}\) represents the minimum of the samples, defined as \(x_{(1)} = \min(x_1, x_2, \ldots, x_n)\). ## Objective Show that \(x_{(1)}\) and \(\sum_{i=1}^n (x_i - x_{(1)})\) are independent. ## Concept Used **Basu's Theorem**: If \(X\) is a complete sufficient statistic for a parameter \(\theta\), and \(Y\) is any function of the sample that does not depend on \(\theta\), then \(X\) and \(Y\) are independent. ### Step 1: Identify the Complete Sufficient Statistic The given distribution \(f_\theta(x)\) belongs to the exponential family. To find the complete sufficient statistic, we can derive the likelihood function: The likelihood function for the sample is: \[ L(\theta; x_1, x_2, \ldots, x_n) = \prod_{i=1}^n f_\theta(x_i) = \prod_{i=1}^n e^{-(x_i - \theta)} = e^{-n\bar{x} + n\theta} \] where \(\bar{x}\) is the sample mean. The sufficient statistic can be identified as: \[ T(X) = (x_{(1)}, \sum_{i=1}^n x_i) \] However, for this distribution, the minimum \(x_{(1)}\) alone can serve as a complete sufficient statistic due to the nature of the exponential family. ### Step 2: Distribution of \(x_{(1)}\) The distribution of the minimum \(x_{(1)}\) can be derived as follows. The cumulative distribution function (CDF) of \(x_{(1)}\) is: \[ P(x_{(1)} \leq x) = 1 - P(x_1 > x)^n = 1 - e^{-n(x - \theta)} \] Thus, the probability density function (pdf) is: \[ f_{x_{(1)}}(x) = n e^{-n(x - \theta)} \quad \text{for } x \geq \theta \] ### Step 3: Distribution of \(\sum_{i=1}^n (x_i - x_{(1)})\) Now, consider the sum \(\sum_{i=1}^n (x_i - x_{(1)})\). The term \(x_{(1)}\) is the minimum, and the residuals \(x_i - x_{(1)}\) are independent of \(x_{(1)}\) because conditioning on \(x_{(1)}\) leads to a new sample distribution. The distribution of \(\sum_{i=1}^n (x_i - x_{(1)})\) can be derived from the properties of exponential distributions. Given that \(x_i\) are iid, the distribution of the differences is independent of the minimum value. ### Step 4: Applying Basu's Theorem Since \(x_{(1)}\) is a complete sufficient statistic for \(\theta\), and the residuals \(\sum_{i=1}^n (x_i - x_{(1)})\) do not depend on \(\theta\), we can apply Basu's theorem. Thus, we conclude: \[ x_{(1)} \text{ and } \sum_{i=1}^n (x_i - x_{(1)}) \text{ are independent.} \] ## Final Answer By applying Basu's theorem, we demonstrate that \(x_{(1)}\) and \(\sum_{i=1}^n (x_i - x_{(1)})\) are independent.

# Random Variable Analysis with Given PDF ## Given Information The probability density function (pdf) of the random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the value of \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). ## Step-by-Step Solution ### Part (a): Find the Value of \(c\) To find \(c\), we must ensure that the total area under the pdf equals 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \(f(x) = 0\) outside the interval \([-1, 1]\), we focus on: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] Calculating the integral: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \] Evaluating at the bounds: \[ = \left(1 - \frac{1^3}{3}\right) - \left(-1 + \frac{(-1)^3}{3}\right) = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{-1}{3}\right) = \left(\frac{2}{3}\right) - \left(-1 - \frac{1}{3}\right) \] \[ = \frac{2}{3} + 1 + \frac{1}{3} = \frac{2}{3} + \frac{3}{3} = \frac{5}{3} \] Thus: \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{5}{3} \] Setting the integral equal to 1: \[ c \cdot \frac{5}{3} = 1 \implies c = \frac{3}{5} \] ### Part (b): Cumulative Distribution Function (CDF) The cumulative distribution function \(F(x)\) is obtained by integrating the pdf from \(-1\) to \(x\): \[ F(x) = \int_{-1}^{x} f(t) \, dt \quad \text{for } -1 < x < 1 \] Calculating the integral: \[ F(x) = \int_{-1}^{x} \frac{3}{5}(1 - t^2) \, dt \] \[ = \frac{3}{5} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating: \[ = \frac{3}{5} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] \[ = \frac{3}{5} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{5} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] For \(x < -1\): \[ F(x) = 0 \] For \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{5} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] ### Part (c): Calculate Expected Value \(E(X)\) The expected value is calculated as: \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{5}(1 - x^2) \, dx \] \[ = \frac{3}{5} \int_{-1}^{1} (x - x^3) \, dx \] Calculating the integral: \[ = \frac{3}{5} \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{-1}^{1} \] Evaluating: \[ = \frac{3}{5} \left( \left[\frac{1}{2} - \frac{1}{4}\right] - \left[\frac{1}{2} - \frac{1}{4}\right] \right) = 0 \] Thus: \[ E(X) = 0 \] ### Part (d): Calculate Variance \(Var(X)\) The variance is calculated as: \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{5}(1 - x^2) \, dx \] \[ = \frac{3}{5} \int_{-1}^{1} (x^2 - x^4) \, dx \] Calculating the integral: \[ = \frac{3}{5} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{-1}^{1} \] Evaluating: \[ = \frac{3}{5} \left( \left[\frac{1}{3} - \frac{1}{5}\right] - \left[-\frac{1}{3} + \frac{1}{5}\right] \right) = \frac{3}{5} \left( \frac{5}{15} - \frac{3}{15} + \frac{5}{15} - \frac{3}{15} \right) \] \[ = \frac{3}{5} \left( \frac{4}{15} + \frac{4}{15} \right) = \frac{3}{5} \cdot \frac{8}{15} = \frac{24}{75} = 0.32 \] Thus, variance: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{8}{15} - 0^2 = \frac{8}{15} \] ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{5}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{5} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{8}{15}} \]

# Random Variable Analysis with Given PDF ## Given Information The probability density function (pdf) of a random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Find the value of \(c\). 2. Determine the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). --- ## Step-by-Step Solution ### Part (a): Find the Value of \(c\) To find \(c\), we must ensure that the total area under the pdf equals 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \(f(x) = 0\) outside the interval \([-1, 1]\), we focus on: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] Calculating the integral: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \] Evaluating at the bounds: At \(x = 1\): \[ 1 - \frac{1^3}{3} = 1 - \frac{1}{3} = \frac{2}{3} \] At \(x = -1\): \[ -1 - \left(-\frac{1}{3}\right) = -1 + \frac{1}{3} = -\frac{2}{3} \] Thus, \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} \] Setting the integral equal to 1: \[ c \cdot \frac{4}{3} = 1 \implies c = \frac{3}{4} \] ### Part (b): Cumulative Distribution Function \(F(x)\) The cumulative distribution function \(F(x)\) is obtained by integrating the pdf from \(-1\) to \(x\): \[ F(x) = \int_{-1}^{x} f(t) \, dt \quad \text{for } -1 < x < 1 \] Calculating the integral: \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt \] Taking \(c\) outside the integral: \[ = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating: \[ = \frac{3}{4} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] Calculating the value at \(x = -1\): \[ = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] For \(x < -1\): \[ F(x) = 0 \] For \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] ### Part (c): Calculate Expected Value \(E(X)\) The expected value is calculated as: \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ = \frac{3}{4} \int_{-1}^{1} (x - x^3) \, dx \] Since both \(x\) and \(x^3\) are odd functions, the integral of an odd function over symmetric limits is 0. Thus: \[ E(X) = 0 \] ### Part (d): Calculate Variance \(Var(X)\) The variance is calculated as: \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \] Calculating the fraction: \[ = \frac{3}{2} \left( \frac{5 - 3}{15} \right) = \frac{3}{2} \cdot \frac{2}{15} = \frac{3}{15} = \frac{1}{5} \] Thus, variance: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{1}{5} - 0^2 = \frac{1}{5} \] --- ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{1}{5}} \]

# Analysis of the Random Variable with Given PDF ## Given Information The probability density function (pdf) of the random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the normalization constant \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). --- ## Step-by-Step Solution ### Part (a): Finding the Value of \(c\) To find \(c\), we must ensure that the total area under the pdf equals 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \(f(x) = 0\) outside the interval \([-1, 1]\), we focus on: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] Calculating the integral: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \] Evaluating at the bounds: At \(x = 1\): \[ 1 - \frac{1^3}{3} = 1 - \frac{1}{3} = \frac{2}{3} \] At \(x = -1\): \[ -1 - \left(-\frac{1}{3}\right) = -1 + \frac{1}{3} = -\frac{2}{3} \] Thus, the integral simplifies to: \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} \] Setting the integral equal to 1: \[ c \cdot \frac{4}{3} = 1 \implies c = \frac{3}{4} \] ### Part (b): Cumulative Distribution Function \(F(x)\) The cumulative distribution function \(F(x)\) is defined as: \[ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \, dt \] For different regions of \(x\): 1. If \(x < -1\): \[ F(x) = 0 \] 2. If \(-1 \leq x < 1\): \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating gives: \[ F(x) = \frac{3}{4} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] \[ = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] 3. If \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] ### Part (c): Calculate Expected Value \(E(X)\) The expected value is calculated as: \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ = \frac{3}{4} \int_{-1}^{1} (x - x^3) \, dx \] Since both \(x\) and \(x^3\) are odd functions, the integral of an odd function over symmetric limits is 0. Thus: \[ E(X) = 0 \] ### Part (d): Calculate Variance \(Var(X)\) The variance is calculated as: \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \] Calculating the fraction: \[ = \frac{3}{2} \left( \frac{5 - 3}{15} \right) = \frac{3}{2} \cdot \frac{2}{15} = \frac{3}{15} = \frac{1}{5} \] Thus, the variance is: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{1}{5} - 0^2 = \frac{1}{5} \] --- ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{1}{5}} \]

# Analysis of the Random Variable with Given PDF ## Given Information The probability density function (pdf) of a random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the normalization constant \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). ## Concept Used To ensure that the pdf is valid, the total area under the pdf must equal 1. The CDF \(F(x)\) is obtained by integrating the pdf. The expected value and variance are calculated using integral formulas based on the pdf. --- The normalization condition requires: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Given the limits of the pdf, we can evaluate: \[ \int_{-1}^{1} f(x) \, dx = 1 \] This leads to: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] The integral can be computed as follows: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left( 1 - \frac{1}{3} \right) - \left( -1 + \frac{1}{3} \right) = \left( \frac{2}{3} \right) - \left( -\frac{2}{3} \right) = \frac{4}{3} \] Setting the equation: \[ c \cdot \frac{4}{3} = 1 \quad \Rightarrow \quad c = \frac{3}{4} \] Next, we find the cumulative distribution function \(F(x)\): For \(x < -1\): \[ F(x) = 0 \] For \(-1 \leq x < 1\): \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} = \frac{3}{4} \left( \left[ x - \frac{x^3}{3} \right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] Evaluating the bounds gives: \[ F(x) = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] For \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] Next, we calculate the expected value \(E(X)\): \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] This integral results in: \[ E(X) = \frac{3}{4} \left( \int_{-1}^{1} (x - x^3) \, dx \right) \] Since \(x\) and \(x^3\) are odd functions, the integral evaluates to zero: \[ E(X) = 0 \] Now, we compute the variance \(Var(X)\): \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] This becomes: \[ E(X^2) = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating gives: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{3}{2} \cdot \frac{2}{15} = \frac{6}{15} = \frac{2}{5} \] Thus: \[ Var(X) = E(X^2) - 0^2 = \frac{2}{5} \] --- ## Final Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{2}{5}} \]

# Analysis of the Random Variable with Given PDF ## Given Information The probability density function (pdf) of a random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the normalization constant \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). --- ## Step-by-Step Solution ### Part (a): Find the Value of \(c\) To find \(c\), we must ensure that the total area under the pdf equals 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \(f(x) = 0\) outside the interval \([-1, 1]\), we focus on: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] Calculating the integral: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left( 1 - \frac{1^3}{3} \right) - \left( -1 + \frac{(-1)^3}{3} \right) \] Evaluating at the bounds: At \(x = 1\): \[ 1 - \frac{1}{3} = \frac{2}{3} \] At \(x = -1\): \[ -1 - \left(-\frac{1}{3}\right) = -1 + \frac{1}{3} = -\frac{2}{3} \] Thus, the integral results in: \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} \] Setting the integral equal to 1: \[ c \cdot \frac{4}{3} = 1 \implies c = \frac{3}{4} \] ### Part (b): Cumulative Distribution Function \(F(x)\) The cumulative distribution function \(F(x)\) is defined as: \[ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \, dt \] For different regions of \(x\): 1. If \(x < -1\): \[ F(x) = 0 \] 2. If \(-1 \leq x < 1\): \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating gives: \[ F(x) = \frac{3}{4} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] Therefore: \[ F(x) = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] 3. If \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] ### Part (c): Calculate Expected Value \(E(X)\) The expected value is calculated as: \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X) = \frac{3}{4} \left( \int_{-1}^{1} (x - x^3) \, dx \right) \] Since both \(x\) and \(x^3\) are odd functions, their integrals over symmetric limits will be zero: \[ E(X) = 0 \] ### Part (d): Calculate Variance \(Var(X)\) The variance is calculated as: \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X^2) = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ E(X^2) = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \] Calculating the fraction: \[ = \frac{3}{2} \cdot \frac{5 - 3}{15} = \frac{3}{2} \cdot \frac{2}{15} = \frac{6}{30} = \frac{1}{5} \] Thus, the variance is: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{1}{5} - 0^2 = \frac{1}{5} \] --- ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{1}{5}} \]

# Analysis of the Random Variable with Given PDF ## Given Information The probability density function (pdf) of a random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the normalization constant \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). --- ## Step-by-Step Solution ### Part (a): Finding the Value of \(c\) To find the normalization constant \(c\), we need the total area under the pdf to equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Given that \(f(x) = 0\) outside the interval \([-1, 1]\), we focus on: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] Calculating the integral: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \] Evaluating at the bounds: At \(x = 1\): \[ 1 - \frac{1^3}{3} = 1 - \frac{1}{3} = \frac{2}{3} \] At \(x = -1\): \[ -1 + \frac{1}{3} = -\frac{2}{3} \] Thus, the area under the curve from \(-1\) to \(1\) is: \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3} \] Setting the equation: \[ c \cdot \frac{4}{3} = 1 \implies c = \frac{3}{4} \] ### Part (b): Cumulative Distribution Function \(F(x)\) The cumulative distribution function \(F(x)\) is defined as: \[ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) \, dt \] For different regions of \(x\): 1. If \(x < -1\): \[ F(x) = 0 \] 2. If \(-1 \leq x < 1\): \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating gives: \[ F(x) = \frac{3}{4} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] Therefore: \[ F(x) = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] 3. If \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] ### Part (c): Calculate Expected Value \(E(X)\) The expected value is calculated as: \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X) = \frac{3}{4} \left( \int_{-1}^{1} (x - x^3) \, dx \right) \] Since \(x\) and \(x^3\) are odd functions, their integrals over symmetric limits will be zero: \[ E(X) = 0 \] ### Part (d): Calculate Variance \(Var(X)\) The variance is calculated as: \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X^2) = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ E(X^2) = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating gives: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{3}{2} \cdot \frac{5 - 3}{15} = \frac{3}{2} \cdot \frac{2}{15} = \frac{6}{30} = \frac{1}{5} \] Thus, the variance is: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{1}{5} - 0^2 = \frac{1}{5} \] --- ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{1}{5}} \]

# Analysis of the Random Variable with Given PDF ## Given Information The probability density function (pdf) of a random variable \(X\) is defined as follows: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the normalization constant \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). ## Concept Used To ensure that the pdf is valid, the total area under the pdf must equal 1. The cumulative distribution function \(F(x)\) is obtained by integrating the pdf. The expected value and variance are calculated using integral formulas based on the pdf. --- The normalization condition requires: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Given the limits of the pdf, we can evaluate: \[ \int_{-1}^{1} f(x) \, dx = 1 \] This leads to: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] The integral can be computed as follows: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left( 1 - \frac{1^3}{3} \right) - \left( -1 + \frac{(-1)^3}{3} \right) \] Evaluating at the bounds: At \(x = 1\): \[ 1 - \frac{1}{3} = \frac{2}{3} \] At \(x = -1\): \[ -1 - \left(-\frac{1}{3}\right) = -1 + \frac{1}{3} = -\frac{2}{3} \] So, we have: \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3} \] Setting the equation: \[ c \cdot \frac{4}{3} = 1 \quad \Rightarrow \quad c = \frac{3}{4} \] Next, we find the cumulative distribution function \(F(x)\): For \(x < -1\): \[ F(x) = 0 \] For \(-1 \leq x < 1\): \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating gives: \[ F(x) = \frac{3}{4} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] Calculating the value at \(x = -1\): \[ = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] For \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] Next, we calculate the expected value \(E(X)\): \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X) = \frac{3}{4} \left( \int_{-1}^{1} (x - x^3) \, dx \right) \] Since both \(x\) and \(x^3\) are odd functions, the integral over symmetric limits evaluates to zero: \[ E(X) = 0 \] Now, we compute the variance \(Var(X)\): \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X^2) = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating gives: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \] Calculating the fraction: \[ = \frac{3}{2} \cdot \frac{5 - 3}{15} = \frac{3}{2} \cdot \frac{2}{15} = \frac{6}{30} = \frac{1}{5} \] Thus, the variance is: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{1}{5} - 0^2 = \frac{1}{5} \] --- ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{1}{5}} \]

# Analysis of the Random Variable with Given PDF ## Given Information The probability density function (pdf) of a random variable \(X\) is defined as: \[ f(x) = \begin{cases} c(1 - x^2) & \text{for } -1 < x < 1 \\ 0 & \text{otherwise} \end{cases} \] ## Objective 1. Determine the normalization constant \(c\). 2. Find the cumulative distribution function (CDF) of \(X\). 3. Calculate the expected value \(E(X)\). 4. Calculate the variance \(Var(X)\). ## Concept Used To ensure that the pdf is valid, the total area under the pdf must equal 1. The CDF \(F(x)\) is obtained by integrating the pdf over the defined range. The expected value and variance are calculated using integral formulas based on the pdf. --- To find the normalization constant \(c\), we set up the equation: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \(f(x) = 0\) outside the interval \([-1, 1]\), we focus on: \[ \int_{-1}^{1} c(1 - x^2) \, dx = 1 \] Calculating the integral: \[ \int_{-1}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \] Evaluating at the bounds: At \(x = 1\): \[ 1 - \frac{1}{3} = \frac{2}{3} \] At \(x = -1\): \[ -1 - \left(-\frac{1}{3}\right) = -1 + \frac{1}{3} = -\frac{2}{3} \] Thus, the integral becomes: \[ \int_{-1}^{1} (1 - x^2) \, dx = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3} \] Setting the equation: \[ c \cdot \frac{4}{3} = 1 \quad \Rightarrow \quad c = \frac{3}{4} \] Next, we find the cumulative distribution function \(F(x)\): For \(x < -1\): \[ F(x) = 0 \] For \(-1 \leq x < 1\): \[ F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt = \frac{3}{4} \left[ t - \frac{t^3}{3} \right]_{-1}^{x} \] Evaluating gives: \[ F(x) = \frac{3}{4} \left( \left[x - \frac{x^3}{3}\right] - \left[-1 + \frac{(-1)^3}{3}\right] \right) \] Calculating the value at \(x = -1\): \[ = \frac{3}{4} \left( x - \frac{x^3}{3} + 1 - \frac{1}{3} \right) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) \] For \(x \geq 1\): \[ F(x) = 1 \] Thus, the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] Now, we calculate the expected value \(E(X)\): \[ E(X) = \int_{-1}^{1} x f(x) \, dx = \int_{-1}^{1} x \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X) = \frac{3}{4} \left( \int_{-1}^{1} (x - x^3) \, dx \right) \] Since both \(x\) and \(x^3\) are odd functions, their integrals over symmetric limits result in zero: \[ E(X) = 0 \] Next, compute the variance \(Var(X)\): \[ Var(X) = E(X^2) - (E(X))^2 \] Calculating \(E(X^2)\): \[ E(X^2) = \int_{-1}^{1} x^2 f(x) \, dx = \int_{-1}^{1} x^2 \cdot \frac{3}{4}(1 - x^2) \, dx \] Calculating the integral: \[ E(X^2) = \frac{3}{4} \int_{-1}^{1} (x^2 - x^4) \, dx \] Since the integrand is even: \[ = \frac{3}{4} \cdot 2 \int_{0}^{1} (x^2 - x^4) \, dx \] Calculating gives: \[ = \frac{3}{2} \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \frac{3}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \] Calculating the fraction: \[ = \frac{3}{2} \cdot \frac{5 - 3}{15} = \frac{3}{2} \cdot \frac{2}{15} = \frac{6}{30} = \frac{1}{5} \] Thus, the variance is: \[ Var(X) = E(X^2) - (E(X))^2 = \frac{1}{5} - 0^2 = \frac{1}{5} \] --- ## Final Answers Summary - The value of \(c\) is: \[ \boxed{\frac{3}{4}} \] - The cumulative distribution function \(F(x)\) is: \[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right) & \text{for } -1 \leq x < 1 \\ 1 & \text{for } x \geq 1 \end{cases} \] - The expected value \(E(X)\) is: \[ \boxed{0} \] - The variance \(Var(X)\) is: \[ \boxed{\frac{1}{5}} \]

# Maximum Likelihood Estimation for Poisson Distribution ## Given Information Let \(Y_1, Y_2, \ldots, Y_n\) denote a random sample from a Poisson distribution with mean \(\lambda\). The probability mass function (pmf) for a Poisson distribution is given by: \[ P(Y = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \ldots \] ## Objective 1. Find the maximum likelihood estimator (MLE) \(\hat{\lambda}\) for \(\lambda\). 2. Find the expected value and variance of \(\hat{\lambda}\). 3. Show that the estimate of part (a) is consistent for \(\lambda\). 4. Determine the MLE for \(P(Y = 0) = e^{-\lambda}\). --- ## Step-by-Step Solution ### Part (a): Find the MLE \(\hat{\lambda}\) The likelihood function \(L(\lambda)\) based on the sample is given by: \[ L(\lambda) = \prod_{i=1}^{n} P(Y_i) = \prod_{i=1}^{n} \frac{\lambda^{Y_i} e^{-\lambda}}{Y_i!} \] Taking the natural logarithm of the likelihood function to obtain the log-likelihood: \[ \log L(\lambda) = \sum_{i=1}^{n} \log\left(\frac{\lambda^{Y_i} e^{-\lambda}}{Y_i!}\right) = \sum_{i=1}^{n} \left(Y_i \log \lambda - \lambda - \log(Y_i!)\right) \] This simplifies to: \[ \log L(\lambda) = \left(\sum_{i=1}^{n} Y_i\right) \log \lambda - n\lambda - \sum_{i=1}^{n} \log(Y_i!) \] To find the MLE, take the derivative of the log-likelihood with respect to \(\lambda\) and set it to zero: \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{\sum_{i=1}^{n} Y_i}{\lambda} - n = 0 \] Solving for \(\lambda\): \[ \frac{\sum_{i=1}^{n} Y_i}{\lambda} = n \implies \hat{\lambda} = \frac{\sum_{i=1}^{n} Y_i}{n} \] Thus, the MLE for \(\lambda\) is: \[ \hat{\lambda} = \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} Y_i \] ### Part (b): Find the Expected Value and Variance of \(\hat{\lambda}\) The expected value of the estimator \(\hat{\lambda}\) is: \[ E[\hat{\lambda}] = E\left[\frac{1}{n} \sum_{i=1}^{n} Y_i\right] = \frac{1}{n} \sum_{i=1}^{n} E[Y_i] \] Since each \(Y_i\) follows a Poisson distribution with mean \(\lambda\): \[ E[Y_i] = \lambda \quad \Rightarrow \quad E[\hat{\lambda}] = \frac{1}{n} \cdot n\lambda = \lambda \] The variance of \(\hat{\lambda}\) is given by: \[ Var(\hat{\lambda}) = Var\left(\frac{1}{n} \sum_{i=1}^{n} Y_i\right) = \frac{1}{n^2} \sum_{i=1}^{n} Var(Y_i) \] For a Poisson distribution, the variance is equal to the mean: \[ Var(Y_i) = \lambda \quad \Rightarrow \quad Var(\hat{\lambda}) = \frac{1}{n^2} \cdot n\lambda = \frac{\lambda}{n} \] ### Part (c): Show Consistency of \(\hat{\lambda}\) An estimator is consistent if it converges in probability to the true parameter as \(n\) approaches infinity. As \(n \to \infty\): \[ Var(\hat{\lambda}) = \frac{\lambda}{n} \to 0 \] Since \(E[\hat{\lambda}] = \lambda\) and \(Var(\hat{\lambda}) \to 0\), we conclude that \(\hat{\lambda}\) is a consistent estimator for \(\lambda\). ### Part (d): MLE for \(P(Y = 0) = e^{-\lambda}\) The probability of zero occurrences in a Poisson distribution is given by: \[ P(Y = 0) = e^{-\lambda} \] The MLE for \(P(Y = 0)\) can be expressed using the estimate \(\hat{\lambda}\): \[ \hat{P}(Y = 0) = e^{-\hat{\lambda}} = e^{-\bar{Y}} \] Thus, the MLE for \(P(Y = 0)\) is: \[ \hat{P}(Y = 0) = e^{-\bar{Y}} \] --- ## Final Answers Summary 1. The MLE \(\hat{\lambda}\) is: \[ \hat{\lambda} = \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} Y_i \] 2. The expected value of \(\hat{\lambda}\) is: \[ E[\hat{\lambda}] = \lambda \] 3. The variance of \(\hat{\lambda}\) is: \[ Var(\hat{\lambda}) = \frac{\lambda}{n} \] 4. The MLE for \(P(Y = 0)\) is: \[ \hat{P}(Y = 0) = e^{-\bar{Y}} \]

# Probability Distribution of Defective Television Sets ## Given Information - Total television sets in shipment: \(7\) - Number of defective sets: \(2\) - Number of sets purchased by the hotel: \(3\) Let \(X\) be the random variable representing the number of defective sets purchased by the hotel. ## Objective Find the probability distribution of \(X\) and express the results graphically as a probability histogram. ## Concept Used The scenario can be modeled using the **hypergeometric distribution**, which describes the probability of \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is given by: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total number of items (7 television sets) - \(K\) = total number of defective items (2 defective sets) - \(n\) = number of items drawn (3 purchased sets) - \(k\) = number of defective items drawn --- ## Step-by-Step Solution ### Define Parameters - **Total sets, \(N\)**: \(7\) - **Defective sets, \(K\)**: \(2\) - **Purchased sets, \(n\)**: \(3\) ### Calculate the Probability Distribution We will find \(P(X = k)\) for \(k = 0, 1, 2\) (the possible values of defective sets purchased): 1. **For \(k = 0\)** (no defective sets): \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} = \frac{{1 \cdot 10}}{{35}} = \frac{10}{35} = \frac{2}{7} \] 2. **For \(k = 1\)** (one defective set): \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} = \frac{{2 \cdot 10}}{{35}} = \frac{20}{35} = \frac{4}{7} \] 3. **For \(k = 2\)** (two defective sets): \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} = \frac{{1 \cdot 5}}{{35}} = \frac{5}{35} = \frac{1}{7} \] ### Summarizing the Probability Distribution The probabilities can be summarized as follows: | \(k\) (Defective Sets) | \(P(X = k)\) | |-------------------------|---------------| | 0 | \(\frac{2}{7}\) | | 1 | \(\frac{4}{7}\) | | 2 | \(\frac{1}{7}\) | ### Probability Histogram The probabilities can be represented graphically as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar corresponds to the probability of purchasing that number of defective sets. --- ## Final Summary - The probability distribution of \(X\) is: - \(P(X = 0) = \frac{2}{7}\) - \(P(X = 1) = \frac{4}{7}\) - \(P(X = 2) = \frac{1}{7}\) - The graphical representation is a histogram showing the probabilities associated with the number of defective sets purchased.

# Probability Distribution of Defective Television Sets ## Given Information - Total number of television sets in shipment: \(N = 7\) - Number of defective sets: \(K = 2\) - Number of sets purchased by the hotel: \(n = 3\) Let \(X\) be the random variable representing the number of defective sets purchased by the hotel. ## Objective Find the probability distribution of \(X\) and represent the results as a probability histogram. ## Concept Used The scenario follows a **Hypergeometric Distribution**, which describes the probability of \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is given by: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total number of items (7 television sets) - \(K\) = total number of defective items (2 defective sets) - \(n\) = number of items drawn (3 purchased sets) - \(k\) = number of defective items drawn --- ## Step-by-Step Solution ### Possible Values of \(X\) Since there are only 2 defective sets and 3 sets are selected, the possible values for \(X\) are: - \(X = 0\) - \(X = 1\) - \(X = 2\) ### Probability Calculations 1. **Probability that no defective set is purchased (\(X = 0\))**: \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 10}{35} = \frac{10}{35} = 0.2857 \] 2. **Probability that exactly one defective set is purchased (\(X = 1\))**: \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} \] \[ = \frac{2 \cdot 10}{35} = \frac{20}{35} = 0.5714 \] 3. **Probability that exactly two defective sets are purchased (\(X = 2\))**: \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 5}{35} = \frac{5}{35} = 0.1429 \] ### Probability Distribution Table | \(X\) (Defective Sets) | \(P(X)\) | |-------------------------|----------------| | 0 | 0.2857 | | 1 | 0.5714 | | 2 | 0.1429 | --- ## Probability Histogram The probabilities can be represented graphically as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar corresponds to the probability of purchasing that number of defective sets. --- ## Final Summary The probability distribution of \(X\) is as follows: - \(P(X = 0) = 0.2857\) - \(P(X = 1) = 0.5714\) - \(P(X = 2) = 0.1429\) The graphical representation is a histogram showing the probabilities associated with the number of defective sets purchased.

# Probability Distribution of Defective Television Sets ## Given Information - Total number of television sets in the shipment: \(N = 7\) - Number of defective sets: \(K = 2\) - Number of sets purchased by the hotel: \(n = 3\) Let \(X\) denote the random variable representing the number of defective sets purchased by the hotel. ## Objective Determine the probability distribution of \(X\) and illustrate the results with a probability histogram. ## Concept Used This situation can be modeled using the **Hypergeometric Distribution**, which describes the probability of drawing \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is given by: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total number of items (7 television sets) - \(K\) = total number of defective items (2 defective sets) - \(n\) = number of items drawn (3 purchased sets) - \(k\) = number of defective items drawn --- ## Step-by-Step Solution ### Define Possible Values of \(X\) Given that there are only 2 defective sets and 3 sets are selected, the possible values for \(X\) are: - \(X = 0\) - \(X = 1\) - \(X = 2\) ### Calculate Probabilities 1. **Probability that no defective set is purchased (\(X = 0\))**: \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 10}{35} = \frac{10}{35} = 0.2857 \] 2. **Probability that exactly one defective set is purchased (\(X = 1\))**: \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} \] \[ = \frac{2 \cdot 10}{35} = \frac{20}{35} = 0.5714 \] 3. **Probability that exactly two defective sets are purchased (\(X = 2\))**: \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 5}{35} = \frac{5}{35} = 0.1429 \] ### Summary of the Probability Distribution The probabilities can be summarized in the following table: | \(X\) (Defective Sets) | \(P(X)\) | |-------------------------|----------------| | 0 | 0.2857 | | 1 | 0.5714 | | 2 | 0.1429 | --- ## Probability Histogram The probabilities can be represented graphically as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar corresponds to the probability of purchasing that number of defective sets. --- ## Final Summary The probability distribution of \(X\) is as follows: - \(P(X = 0) = 0.2857\) - \(P(X = 1) = 0.5714\) - \(P(X = 2) = 0.1429\) The graphical representation is a histogram that illustrates the probabilities associated with the number of defective sets purchased.

# Probability Distribution of Defective Television Sets ## Given Information - Total number of television sets in the shipment: \(N = 7\) - Number of defective sets: \(K = 2\) - Number of sets purchased by the hotel: \(n = 3\) Let \(X\) denote the random variable representing the number of defective sets purchased by the hotel. ## Objective Determine the probability distribution of \(X\) and represent the results as a probability histogram. ## Concept Used This scenario can be modeled using the **Hypergeometric Distribution**, which describes the probability of drawing \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is defined as: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total number of items (7 television sets) - \(K\) = total number of defective items (2 defective sets) - \(n\) = number of items drawn (3 purchased sets) - \(k\) = number of defective items drawn --- ## Step-by-Step Solution ### Define Possible Values of \(X\) Given that there are only 2 defective sets and 3 sets are selected, the possible values for \(X\) are: - \(X = 0\) - \(X = 1\) - \(X = 2\) ### Calculate Probabilities 1. **Probability that no defective set is purchased (\(X = 0\))**: \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 10}{35} = \frac{10}{35} = \frac{2}{7} \] 2. **Probability that exactly one defective set is purchased (\(X = 1\))**: \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} \] \[ = \frac{2 \cdot 10}{35} = \frac{20}{35} = \frac{4}{7} \] 3. **Probability that exactly two defective sets are purchased (\(X = 2\))**: \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 5}{35} = \frac{5}{35} = \frac{1}{7} \] ### Summary of the Probability Distribution The probabilities can be summarized in the following table: | \(X\) (Defective Sets) | \(P(X)\) | |-------------------------|----------------| | 0 | \(\frac{2}{7}\) | | 1 | \(\frac{4}{7}\) | | 2 | \(\frac{1}{7}\) | --- ## Probability Histogram The probabilities can be represented graphically as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar corresponds to the probability of purchasing that number of defective sets. --- ## Final Summary The probability distribution of \(X\) is as follows: - \(P(X = 0) = \frac{2}{7}\) - \(P(X = 1) = \frac{4}{7}\) - \(P(X = 2) = \frac{1}{7}\) The graphical representation is a histogram that illustrates the probabilities associated with the number of defective sets purchased.

# Probability Distribution of Defective Television Sets ## Given Information - Total number of television sets in the shipment: \(N = 7\) - Number of defective sets: \(K = 2\) - Number of sets purchased by the hotel: \(n = 3\) Let \(X\) denote the random variable representing the number of defective sets purchased by the hotel. ## Objective Determine the probability distribution of \(X\) and represent the results as a probability histogram. ## Concept Used This situation can be modeled using the **Hypergeometric Distribution**, which describes the probability of drawing \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is given by: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total number of items (7 television sets) - \(K\) = total number of defective items (2 defective sets) - \(n\) = number of items drawn (3 purchased sets) - \(k\) = number of defective items drawn --- ## Step-by-Step Solution ### Define Possible Values of \(X\) Given that there are only 2 defective sets and 3 sets are selected, the possible values for \(X\) are: - \(X = 0\) - \(X = 1\) - \(X = 2\) ### Calculate Probabilities 1. **Probability that no defective set is purchased (\(X = 0\))**: \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 10}{35} = \frac{10}{35} = \frac{2}{7} \] 2. **Probability that exactly one defective set is purchased (\(X = 1\))**: \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} \] \[ = \frac{2 \cdot 10}{35} = \frac{20}{35} = \frac{4}{7} \] 3. **Probability that exactly two defective sets are purchased (\(X = 2\))**: \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 5}{35} = \frac{5}{35} = \frac{1}{7} \] ### Summary of the Probability Distribution The probabilities can be summarized in the following table: | \(X\) (Defective Sets) | \(P(X)\) | |-------------------------|----------------| | 0 | \(\frac{2}{7}\) | | 1 | \(\frac{4}{7}\) | | 2 | \(\frac{1}{7}\) | --- ## Probability Histogram The probabilities can be represented graphically as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar corresponds to the probability of purchasing that number of defective sets. --- ## Final Summary The probability distribution of \(X\) is as follows: - \(P(X = 0) = \frac{2}{7}\) - \(P(X = 1) = \frac{4}{7}\) - \(P(X = 2) = \frac{1}{7}\) The graphical representation is a histogram that illustrates the probabilities associated with the number of defective sets purchased.

# Probability Distribution of Defective Television Sets ## Given Information - The total number of television sets in the shipment: \(N = 7\) - The number of defective sets: \(K = 2\) - The number of sets purchased by the hotel: \(n = 3\) Let \(X\) represent the random variable indicating the count of defective sets purchased by the hotel. ## Objective Determine the probability distribution of \(X\) and illustrate the results using a probability histogram. ## Concept Used This scenario can be modeled with the **Hypergeometric Distribution**, which describes the probability of obtaining \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is expressed as: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total items (7 television sets) - \(K\) = defective items (2 defective sets) - \(n\) = items drawn (3 purchased sets) - \(k\) = defective items drawn --- ## Step-by-Step Solution ### Identify Possible Values of \(X\) Since there are only 2 defective sets available and 3 sets are selected, the possible values for \(X\) are: - \(X = 0\) - \(X = 1\) - \(X = 2\) ### Calculate Probabilities 1. **Probability of No Defective Sets Purchased (\(X = 0\))**: \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 10}{35} = \frac{10}{35} = \frac{2}{7} \] 2. **Probability of One Defective Set Purchased (\(X = 1\))**: \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} \] \[ = \frac{2 \cdot 10}{35} = \frac{20}{35} = \frac{4}{7} \] 3. **Probability of Two Defective Sets Purchased (\(X = 2\))**: \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 5}{35} = \frac{5}{35} = \frac{1}{7} \] ### Summary of the Probability Distribution The probabilities can be summarized in the following table: | \(X\) (Defective Sets) | \(P(X)\) | |-------------------------|----------------| | 0 | \(\frac{2}{7}\) | | 1 | \(\frac{4}{7}\) | | 2 | \(\frac{1}{7}\) | --- ## Probability Histogram The probabilities can be visually represented as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar reflects the probability of purchasing that number of defective sets. --- ## Final Summary The probability distribution of \(X\) is as follows: - \(P(X = 0) = \frac{2}{7}\) - \(P(X = 1) = \frac{4}{7}\) - \(P(X = 2) = \frac{1}{7}\) The graphical representation is a histogram that illustrates the probabilities associated with the number of defective sets purchased.

# Probability Distribution of Defective Television Sets ## Given Information - Total number of television sets in the shipment: \(N = 7\) - Number of defective sets: \(K = 2\) - Number of sets purchased by the hotel: \(n = 3\) Let \(X\) denote the random variable representing the number of defective sets purchased by the hotel. ## Objective Calculate the probability distribution of \(X\) and present the results via a probability histogram. ## Concept Used The situation can be modeled using the **Hypergeometric Distribution**, which describes the probability of drawing \(k\) successes (defective sets) in \(n\) draws (purchases) from a finite population without replacement. The probability mass function is defined as: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \(N\) = total number of items (7 television sets) - \(K\) = total number of defective items (2 defective sets) - \(n\) = number of items drawn (3 purchased sets) - \(k\) = number of defective items drawn --- ## Step-by-Step Solution ### Possible Values of \(X\) Since there are only 2 defective sets and 3 sets are selected, the possible values for \(X\) are: - \(X = 0\) - \(X = 1\) - \(X = 2\) ### Calculate Probabilities 1. **Probability that no defective set is purchased (\(X = 0\))**: \[ P(X = 0) = \frac{{\binom{2}{0} \cdot \binom{5}{3}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 10}{35} = \frac{10}{35} = \frac{2}{7} \] 2. **Probability that exactly one defective set is purchased (\(X = 1\))**: \[ P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{5}{2}}}{{\binom{7}{3}}} \] \[ = \frac{2 \cdot 10}{35} = \frac{20}{35} = \frac{4}{7} \] 3. **Probability that exactly two defective sets are purchased (\(X = 2\))**: \[ P(X = 2) = \frac{{\binom{2}{2} \cdot \binom{5}{1}}}{{\binom{7}{3}}} \] \[ = \frac{1 \cdot 5}{35} = \frac{5}{35} = \frac{1}{7} \] ### Summary of the Probability Distribution The probabilities can be summarized in the following table: | \(X\) (Defective Sets) | \(P(X)\) | |-------------------------|----------------| | 0 | \(\frac{2}{7}\) | | 1 | \(\frac{4}{7}\) | | 2 | \(\frac{1}{7}\) | --- ## Probability Histogram The probabilities can be represented graphically as a probability histogram. Below is a representation of the histogram: ```plaintext | k (Number of Defective Sets) | | * | | * | | * * | |________________________________________ | 0 1 2 | P(X) P(X) P(X) ``` - The height of each bar corresponds to the probability of purchasing that number of defective sets. --- ## Final Summary The probability distribution of \(X\) is as follows: - \(P(X = 0) = \frac{2}{7}\) - \(P(X = 1) = \frac{4}{7}\) - \(P(X = 2) = \frac{1}{7}\) The graphical representation is a histogram illustrating the probabilities associated with the number of defective sets purchased.

# Bayes' Theorem ## Given Information Bayes' theorem is a fundamental concept in probability theory and statistics that relates the conditional and marginal probabilities of random events. ## Objective Determine the primary application of Bayes' theorem. ## Options A) Compute joint probability B) Update probabilities given new evidence C) Find variance D) Calculate sample mean ## Concept Used Bayes' theorem states that the conditional probability of an event \(A\) given that event \(B\) has occurred is proportional to the probability of \(B\) given \(A\) and the marginal probability of \(B\): \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] This theorem is primarily used for updating the probability estimates for a hypothesis as more evidence becomes available. --- ## Conclusion The correct answer is: **B) Update probabilities given new evidence** Bayes' theorem allows us to revise our beliefs based on new data, making it a powerful tool for statistical inference and decision-making.

# Explanation of Bayes' Theorem and Options ## Given Information Bayes' theorem is a fundamental concept in probability theory and statistics that assists in understanding the relationship between conditional probabilities. ## Objective Identify the primary application of Bayes' theorem from the following options: A) Compute joint probability B) Update probabilities given new evidence C) Find variance D) Calculate sample mean ## Concept Used Bayes' theorem is mathematically expressed as: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] This theorem is primarily used for updating the probability estimates for a hypothesis as more evidence becomes available. --- ## Correct Answer The correct answer is: **B) Update probabilities given new evidence** Bayes' theorem allows for the revision of probabilities based on new information, making it a powerful tool for statistical inference and hypothesis testing. --- ## Explanation of Incorrect Options ### A) Compute Joint Probability While Bayes' theorem can be used to derive joint probabilities from conditional probabilities, its primary function is not to compute joint probabilities directly. Joint probabilities can be calculated using the definition of joint probability or the multiplication rule, independent of Bayes' theorem. ### C) Find Variance Bayes' theorem is not used to find variance. Variance is a measure of the spread of a set of values and is calculated using the formula: \[ Var(X) = E[X^2] - (E[X])^2 \] This involves the expected values of random variables and does not relate to Bayes' theorem. ### D) Calculate Sample Mean The sample mean is calculated as: \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \] This is a simple average of the observed values in a sample and is not directly related to Bayes' theorem, which deals with conditional probabilities and updates based on new evidence. --- ## Summary Bayes' theorem is primarily used to **update probabilities given new evidence**, allowing for informed decision-making based on the latest available data. Other options, such as computing joint probabilities, finding variance, and calculating sample means, are unrelated to the core function of Bayes' theorem.

# Understanding Bayes' Theorem and Its Applications ## Given Information Bayes' theorem is a critical concept in probability theory and statistics that provides a method for updating the probability of a hypothesis as new evidence emerges. ## Objective Identify the main application of Bayes' theorem from the following choices: A) Compute joint probability B) Update probabilities given new evidence C) Find variance D) Calculate sample mean ## Conceptual Framework Bayes' theorem states that the probability of an event \(A\) given that another event \(B\) has occurred can be determined using the formula: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] This formula illustrates how to update the probability of \(A\) when we receive new information about \(B\). The theorem is particularly powerful in scenarios where our initial beliefs about the probability of an event can be refined with incoming data. --- ## Correct Answer The correct answer is: **B) Update probabilities given new evidence** Bayes' theorem plays a crucial role in revising our probability estimates when we obtain new information, making it an indispensable tool in fields such as statistics, machine learning, and decision-making processes. --- ## Explanation of Incorrect Options ### A) Compute Joint Probability While Bayes' theorem can be employed to derive joint probabilities from conditional ones, its primary purpose is not to calculate joint probabilities directly. Instead, joint probabilities can be computed using the multiplication rule, which does not require Bayes' theorem. ### C) Find Variance Bayes' theorem does not relate to finding variance. Variance quantifies the dispersion of a set of values and is calculated using the formula: \[ Var(X) = E[X^2] - (E[X])^2 \] This calculation pertains to the distribution of a random variable rather than the updating of probabilities based on new evidence. ### D) Calculate Sample Mean The sample mean, represented as: \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \] is the arithmetic average of a set of observed values. This calculation is independent of Bayes' theorem, which is focused on conditional probabilities and how they change with new data. --- ## Summary In conclusion, Bayes' theorem is fundamentally about **updating probabilities given new evidence**, allowing for improved decision-making based on the latest information. The other options—computing joint probabilities, finding variance, and calculating sample means—are distinct statistical operations that do not capture the essence of what Bayes' theorem accomplishes in the realm of probability.

# Exploring Bayes' Theorem and Its Primary Use ## Given Information Bayes' theorem is an essential principle in probability theory and statistics that enables the updating of the probability of a hypothesis based on new evidence or information. ## Objective Identify the principal application of Bayes' theorem from the following options: A) Compute joint probability B) Update probabilities given new evidence C) Find variance D) Calculate sample mean ## Conceptual Understanding Bayes' theorem provides a method for calculating the conditional probability of an event \(A\) occurring given that another event \(B\) has occurred. The theorem is articulated as follows: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] This formula indicates how to adjust our beliefs about the probability of event \(A\) when we acquire new information about event \(B\). It is particularly useful in scenarios where we need to revise our initial probability estimates in light of new data. --- ## Correct Answer The correct answer is: **B) Update probabilities given new evidence** Bayes' theorem is fundamentally designed to refine our probability assessments as new information becomes available, making it a vital tool in various applications, including medical diagnosis, machine learning, and risk assessment. --- ## Explanation of Incorrect Options ### A) Compute Joint Probability Although Bayes' theorem can assist in deriving joint probabilities from conditional probabilities, its primary function is not to compute joint probabilities directly. Instead, joint probabilities are often calculated using the basic multiplication rule, which operates independently of Bayes' theorem. ### C) Find Variance Bayes' theorem is not applicable for calculating variance. Variance measures the extent to which a set of values diverges from their mean and is computed using the formula: \[ Var(X) = E[X^2] - (E[X])^2 \] This relates directly to the distribution of a random variable and does not involve the updating of probabilities. ### D) Calculate Sample Mean The calculation of the sample mean is expressed as: \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \] This average is calculated from a set of observations and is not influenced by Bayes' theorem, which focuses on conditional probabilities and their adjustments based on new evidence. --- ## Summary In summary, Bayes' theorem is primarily used to **update probabilities given new evidence**, thereby enhancing the accuracy of our predictions and decisions based on the latest information. The other options—computing joint probabilities, finding variance, and calculating sample means—represent distinct statistical procedures that do not encompass the core purpose of Bayes' theorem.

# Understanding the Application of Bayes' Theorem ## Given Information Bayes' theorem is a foundational concept in the field of probability and statistics that allows us to update the probability of a hypothesis when new data or evidence becomes available. ## Objective Identify the principal application of Bayes' theorem from the following choices: A) Compute joint probability B) Update probabilities given new evidence C) Find variance D) Calculate sample mean ## Conceptual Framework Bayes' theorem provides a way to determine the conditional probability of an event \(A\) based on the occurrence of another event \(B\). The theorem is expressed mathematically as: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] This formula indicates how to adjust our prior beliefs about the probability of event \(A\) when we receive new information regarding event \(B\). This is particularly useful in various fields such as medical research, machine learning, and decision-making processes. --- ## Correct Answer The correct answer is: **B) Update probabilities given new evidence** Bayes' theorem is fundamentally designed for revising our probability estimates in light of new information, making it an essential tool for informed decision-making. --- ## Explanation of Incorrect Options ### A) Compute Joint Probability While Bayes' theorem can be utilized to derive joint probabilities from conditional probabilities, its primary purpose is not to compute joint probabilities directly. Joint probabilities can be computed using the multiplication rule, which does not require the application of Bayes' theorem. ### C) Find Variance Bayes' theorem does not relate to the computation of variance. Variance is a statistical measure that describes the spread of a set of values and is calculated using the formula: \[ Var(X) = E[X^2] - (E[X])^2 \] This calculation focuses on the distribution of random variables rather than the updating of probabilities. ### D) Calculate Sample Mean The sample mean is calculated using the formula: \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \] This value represents the average of a set of numerical observations and is independent of Bayes' theorem, which concerns conditional probabilities and the effects of new data. --- ## Summary In conclusion, Bayes' theorem is primarily used to **update probabilities given new evidence**, allowing for more accurate predictions and informed decision-making. The other options—computing joint probabilities, finding variance, and calculating sample means—are separate statistical techniques that do not capture the essence of Bayes' theorem's application in probability theory.

# Application of Bayes' Theorem in Probability ## Given Information Bayes' theorem is a pivotal concept in probability and statistics that facilitates the adjustment of probabilities based on new evidence or information. ## Objective Identify the main application of Bayes' theorem from the following options: A) Compute joint probability B) Update probabilities given new evidence C) Find variance D) Calculate sample mean ## Conceptual Understanding Bayes' theorem provides a structured method for calculating the conditional probability of an event \(A\) given that another event \(B\) has occurred. The theorem is articulated as: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] This equation illustrates how to modify our initial beliefs about the likelihood of event \(A\) occurring when new data related to event \(B\) is presented. It is especially valuable in fields such as healthcare, finance, and machine learning, where decisions often depend on incomplete information. --- ## Correct Answer The correct answer is: **B) Update probabilities given new evidence** Bayes' theorem is primarily utilized to revise probability estimates as new information becomes available, enhancing decision-making accuracy. --- ## Explanation of Incorrect Options ### A) Compute Joint Probability While Bayes' theorem can assist in deriving joint probabilities from conditional probabilities, its core function is not to compute joint probabilities directly. Instead, joint probabilities can be calculated independently using the multiplication rule of probabilities, which does not necessitate the application of Bayes' theorem. ### C) Find Variance Bayes' theorem does not relate to finding variance. Variance is a statistical measure that quantifies the dispersion in a set of values, calculated using the formula: \[ Var(X) = E[X^2] - (E[X])^2 \] This concept is concerned with the spread of data points around the mean and does not involve the updating of probabilities based on new evidence. ### D) Calculate Sample Mean The sample mean is calculated using the formula: \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \] This formula provides the average value of a dataset and is independent of Bayes' theorem, which focuses on conditional probabilities and their adjustments in light of new evidence. --- ## Summary In summary, Bayes' theorem is fundamentally used to **update probabilities given new evidence**, allowing us to refine our predictions and make more informed decisions. The other options—computing joint probabilities, finding variance, and calculating sample means—represent distinct statistical operations that do not encompass the core purpose of Bayes' theorem in probability theory.

# Probability of Defective Item Origin ## Given Information We have three machines producing items with the following production percentages and defect rates: - **Machine 1**: - Production percentage: \(20\%\) - Defect rate: \(2\%\) - **Machine 2**: - Production percentage: \(30\%\) - Defect rate: \(3\%\) - **Machine 3**: - Production percentage: \(50\%\) - Defect rate: \(5\%\) Let \(D\) represent the event that an item is defective, and let \(M_1\), \(M_2\), and \(M_3\) represent the events that an item comes from Machine 1, Machine 2, and Machine 3, respectively. ## Objective Determine the probability \(P(M_3 | D)\), which is the probability that a defective item came from Machine 3. ## Concept Used To find \(P(M_3 | D)\), we can use **Bayes' Theorem**, which states: \[ P(M_3 | D) = \frac{P(D | M_3) \cdot P(M_3)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The conditional probabilities of an item being defective given it comes from each machine are: - \(P(D | M_1) = 0.02\) - \(P(D | M_2) = 0.03\) - \(P(D | M_3) = 0.05\) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of an item coming from each machine are: - \(P(M_1) = 0.20\) - \(P(M_2) = 0.30\) - \(P(M_3) = 0.50\) ### Step 3: Calculate \(P(D)\) To find the total probability \(P(D)\), we can use the law of total probability: \[ P(D) = P(D | M_1) \cdot P(M_1) + P(D | M_2) \cdot P(M_2) + P(D | M_3) \cdot P(M_3) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.20) + (0.03 \cdot 0.30) + (0.05 \cdot 0.50) \] \[ = 0.004 + 0.009 + 0.025 = 0.038 \] ### Step 4: Apply Bayes' Theorem Now we can substitute into Bayes' theorem: \[ P(M_3 | D) = \frac{P(D | M_3) \cdot P(M_3)}{P(D)} = \frac{0.05 \cdot 0.50}{0.038} \] Calculating this gives: \[ P(M_3 | D) = \frac{0.025}{0.038} \approx 0.6579 \] ## Final Answer The probability that a defective item came from Machine 3 is approximately: \[ \boxed{0.6579} \]

# Probability of Picking a Gold Coin After Tossing ## Given Information - The box contains: - 3 gold coins - 2 silver coins - Tossing probabilities: - Gold coins land heads with probability \(P(H | G) = 0.8\) - Silver coins land heads with probability \(P(H | S) = 0.5\) - You randomly pick one coin and toss it 3 times, resulting in 3 heads. ## Objective Determine the probability that you picked a gold coin given that you obtained 3 heads, denoted as \(P(G | H_3)\). ## Concept Used We will use **Bayes' Theorem** to find the probability: \[ P(G | H_3) = \frac{P(H_3 | G) \cdot P(G)}{P(H_3)} \] Where: - \(P(H_3 | G)\) is the probability of getting 3 heads given that a gold coin was picked. - \(P(G)\) is the prior probability of picking a gold coin. - \(P(H_3)\) is the total probability of getting 3 heads. ### Step 1: Calculate \(P(H_3 | G)\) For a gold coin, the probability of getting 3 heads is: \[ P(H_3 | G) = P(H | G)^3 = 0.8^3 = 0.512 \] ### Step 2: Calculate \(P(H_3 | S)\) For a silver coin, the probability of getting 3 heads is: \[ P(H_3 | S) = P(H | S)^3 = 0.5^3 = 0.125 \] ### Step 3: Calculate \(P(G)\) and \(P(S)\) The prior probabilities of picking a gold or silver coin are: \[ P(G) = \frac{3}{5}, \quad P(S) = \frac{2}{5} \] ### Step 4: Calculate \(P(H_3)\) Using the law of total probability: \[ P(H_3) = P(H_3 | G) \cdot P(G) + P(H_3 | S) \cdot P(S) \] Substituting the values: \[ P(H_3) = (0.512 \cdot \frac{3}{5}) + (0.125 \cdot \frac{2}{5}) \] Calculating each term: \[ P(H_3) = 0.512 \cdot 0.6 + 0.125 \cdot 0.4 = 0.3072 + 0.05 = 0.3572 \] ### Step 5: Apply Bayes' Theorem Now we can substitute into Bayes' theorem: \[ P(G | H_3) = \frac{P(H_3 | G) \cdot P(G)}{P(H_3)} = \frac{0.512 \cdot \frac{3}{5}}{0.3572} \] Calculating this gives: \[ P(G | H_3) = \frac{0.3072}{0.3572} \approx 0.8596 \] ## Final Answer The probability that you picked a gold coin given that you obtained 3 heads is approximately: \[ \boxed{0.8596} \]

# Probability Analysis for Coin Selection and Tossing ## Given Information - A box contains: - 3 gold coins - 2 silver coins - Tossing probabilities: - Gold coins have a heads probability of \(P(H | G) = 0.8\) - Silver coins have a heads probability of \(P(H | S) = 0.5\) - After selecting a coin at random, you tossed it 3 times and obtained 3 heads. ## Objective Calculate the probability that the selected coin was gold, given that you obtained 3 heads, denoted as \(P(G | H_3)\). ## Concept Used To find \(P(G | H_3)\), we will apply **Bayes' Theorem**, which can be expressed as: \[ P(G | H_3) = \frac{P(H_3 | G) \cdot P(G)}{P(H_3)} \] Where: - \(P(H_3 | G)\) is the probability of getting 3 heads given that a gold coin was picked. - \(P(G)\) is the prior probability of selecting a gold coin. - \(P(H_3)\) is the total probability of getting 3 heads, regardless of the coin type. ### Step 1: Calculate \(P(H_3 | G)\) For a gold coin, the probability of getting 3 heads is calculated as follows: \[ P(H_3 | G) = P(H | G)^3 = 0.8^3 = 0.512 \] ### Step 2: Calculate \(P(H_3 | S)\) For a silver coin, the probability of obtaining 3 heads is: \[ P(H_3 | S) = P(H | S)^3 = 0.5^3 = 0.125 \] ### Step 3: Calculate \(P(G)\) and \(P(S)\) The prior probabilities of selecting either a gold or silver coin are given by: \[ P(G) = \frac{3}{5}, \quad P(S) = \frac{2}{5} \] ### Step 4: Calculate \(P(H_3)\) Using the law of total probability, we calculate \(P(H_3)\): \[ P(H_3) = P(H_3 | G) \cdot P(G) + P(H_3 | S) \cdot P(S) \] Substituting the values into the equation gives: \[ P(H_3) = (0.512 \cdot \frac{3}{5}) + (0.125 \cdot \frac{2}{5}) \] Calculating each component: \[ P(H_3) = 0.512 \cdot 0.6 + 0.125 \cdot 0.4 = 0.3072 + 0.05 = 0.3572 \] ### Step 5: Apply Bayes' Theorem Now substituting into Bayes' theorem: \[ P(G | H_3) = \frac{P(H_3 | G) \cdot P(G)}{P(H_3)} = \frac{0.512 \cdot \frac{3}{5}}{0.3572} \] Calculating this yields: \[ P(G | H_3) = \frac{0.3072}{0.3572} \approx 0.8597 \] ## Final Answer The probability that you selected a gold coin given that you obtained 3 heads is approximately: \[ \boxed{0.8597} \]

# Probability of Having a Disease After Two Positive Tests ## Given Information - Prevalence of the disease in the population: \(P(D) = 0.01\) (1%) - Probability of a positive test result given the disease is present (true positive rate): \(P(T^+ | D) = 0.98\) (98%) - Probability of a positive test result given the disease is not present (false positive rate): \(P(T^+ | \neg D) = 0.03\) (3%) ## Objective We need to find the probability that a person actually has the disease after testing positive twice independently, denoted as \(P(D | T^+_1, T^+_2)\). ## Concept Used We will use **Bayes' Theorem** to find \(P(D | T^+_1, T^+_2)\). The theorem states: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] To find \(P(T^+)\), we will use the law of total probability, considering both cases (having the disease and not having it): \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | \neg D) \cdot P(\neg D) \] ## Step-by-Step Solution ### Step 1: Calculate \(P(T^+)\) First, we calculate \(P(\neg D)\): \[ P(\neg D) = 1 - P(D) = 1 - 0.01 = 0.99 \] Now, substituting values into the total probability equation: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | \neg D) \cdot P(\neg D) \] \[ = (0.98 \cdot 0.01) + (0.03 \cdot 0.99) \] Calculating each term: \[ = 0.0098 + 0.0297 = 0.0395 \] ### Step 2: Calculate \(P(D | T^+)\) Now we can use Bayes' theorem to find \(P(D | T^+)\): \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Substituting the known values: \[ P(D | T^+) = \frac{0.98 \cdot 0.01}{0.0395} \] Calculating this gives: \[ P(D | T^+) = \frac{0.0098}{0.0395} \approx 0.2481 \] ### Step 3: Finding the Probability After Two Positive Tests Since the tests are independent, the probability of testing positive twice is: \[ P(T^+_1, T^+_2 | D) = P(T^+ | D) \cdot P(T^+ | D) = (0.98)^2 = 0.9604 \] And for not having the disease: \[ P(T^+_1, T^+_2 | \neg D) = P(T^+ | \neg D) \cdot P(T^+ | \neg D) = (0.03)^2 = 0.0009 \] Now we calculate the total probability \(P(T^+_1, T^+_2)\): \[ P(T^+_1, T^+_2) = P(T^+_1, T^+_2 | D) \cdot P(D) + P(T^+_1, T^+_2 | \neg D) \cdot P(\neg D) \] Substituting the values: \[ P(T^+_1, T^+_2) = (0.9604 \cdot 0.01) + (0.0009 \cdot 0.99) \] Calculating each term: \[ = 0.009604 + 0.000891 = 0.010495 \] ### Final Calculation for \(P(D | T^+_1, T^+_2)\) Using Bayes' theorem again: \[ P(D | T^+_1, T^+_2) = \frac{P(T^+_1, T^+_2 | D) \cdot P(D)}{P(T^+_1, T^+_2)} \] Substituting the values: \[ P(D | T^+_1, T^+_2) = \frac{0.9604 \cdot 0.01}{0.010495} \] Calculating this gives: \[ P(D | T^+_1, T^+_2) = \frac{0.009604}{0.010495} \approx 0.9166 \] ## Final Answer The probability that a person has the disease given that they tested positive twice is approximately: \[ \boxed{0.9166} \]

# Probability of Disease Given Positive Test Results ## Given Information A disease affects \(1\%\) of the population. A test for the disease has the following characteristics: - True positive rate (sensitivity): \(P(T^+ | D) = 0.98\) (98% accuracy when the disease is present) - False positive rate: \(P(T^+ | \neg D) = 0.03\) (3% chance of a positive test when the disease is absent) If a person tests positive for the disease twice independently, we want to find the probability that they actually have the disease, denoted as \(P(D | T^+_1, T^+_2)\). ## Objective Calculate \(P(D | T^+_1, T^+_2)\) using Bayes' Theorem. ## Concept Used Bayes' Theorem allows us to update the probability of a hypothesis based on new evidence. It is expressed as: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] To find \(P(D | T^+_1, T^+_2)\), we first need to calculate \(P(T^+)\), the total probability of testing positive. --- ## Step-by-Step Solution ### Step 1: Calculate \(P(T^+)\) Using the law of total probability, we can express \(P(T^+)\) as: \[ P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | \neg D) \cdot P(\neg D) \] Where: - \(P(D) = 0.01\) (the prevalence of the disease) - \(P(\neg D) = 1 - P(D) = 0.99\) Substituting the known values: \[ P(T^+) = (0.98 \cdot 0.01) + (0.03 \cdot 0.99) \] Calculating each term: \[ = 0.0098 + 0.0297 = 0.0395 \] ### Step 2: Calculate \(P(D | T^+)\) Now we can apply Bayes' Theorem to find the probability that a person has the disease given one positive test result: \[ P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} \] Substituting the values we have: \[ P(D | T^+) = \frac{0.98 \cdot 0.01}{0.0395} \] Calculating this gives: \[ P(D | T^+) = \frac{0.0098}{0.0395} \approx 0.2481 \] ### Step 3: Probability After Two Positive Tests Since the tests are independent, the probability of testing positive twice is: \[ P(T^+_1, T^+_2 | D) = P(T^+ | D) \cdot P(T^+ | D) = (0.98)^2 = 0.9604 \] And for not having the disease: \[ P(T^+_1, T^+_2 | \neg D) = P(T^+ | \neg D) \cdot P(T^+ | \neg D) = (0.03)^2 = 0.0009 \] Now we calculate the total probability \(P(T^+_1, T^+_2)\): \[ P(T^+_1, T^+_2) = P(T^+_1, T^+_2 | D) \cdot P(D) + P(T^+_1, T^+_2 | \neg D) \cdot P(\neg D) \] Substituting the values: \[ P(T^+_1, T^+_2) = (0.9604 \cdot 0.01) + (0.0009 \cdot 0.99) \] Calculating: \[ P(T^+_1, T^+_2) = 0.009604 + 0.000891 = 0.010495 \] ### Final Calculation for \(P(D | T^+_1, T^+_2)\) Using Bayes' theorem again: \[ P(D | T^+_1, T^+_2) = \frac{P(T^+_1, T^+_2 | D) \cdot P(D)}{P(T^+_1, T^+_2)} \] Substituting the values: \[ P(D | T^+_1, T^+_2) = \frac{0.9604 \cdot 0.01}{0.010495} \] Calculating this gives: \[ P(D | T^+_1, T^+_2) = \frac{0.009604}{0.010495} \approx 0.9167 \] --- ## Final Answer The probability that a person has the disease given that they tested positive twice is approximately: \[ \boxed{0.9167} \]

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(x_1, x_2, \ldots, x_n\) be independent and identically distributed (iid) samples from an exponential distribution with parameter \(\lambda\). The probability density function (pdf) of an exponential distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine if the estimator is biased or unbiased. ## Concept Used The likelihood function for the exponential distribution based on the sample is: \[ L(\lambda) = \prod_{i=1}^{n} f(x_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} \] Taking the natural logarithm gives the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} x_i \] To find the MLE, we take the derivative of the log-likelihood with respect to \(\lambda\) and set it to zero. --- ## Step-by-Step Solution ### Calculate the Log-Likelihood Function Starting with the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} x_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i \] ### Set the Derivative to Zero Setting the derivative equal to zero to find the maximum: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} x_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} x_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] --- ## Is the Estimator Biased or Unbiased? To determine if \(\hat{\lambda}\) is biased or unbiased, we need to find the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} x_i}\right) \] Given that \(x_i\) follows an exponential distribution, the mean of an exponential distribution is \(E[x_i] = \frac{1}{\lambda}\). Therefore, the sum \(\sum_{i=1}^{n} x_i\) is gamma distributed and has the following expectation: \[ E\left(\sum_{i=1}^{n} x_i\right) = n \cdot E[x] = n \cdot \frac{1}{\lambda} \] Thus, we have: \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} x_i}\right) = \frac{n \cdot \lambda}{n} = \lambda \] Since \(E(\hat{\lambda}) = \lambda\), the estimator \(\hat{\lambda}\) is unbiased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i} \] 2. The estimator is **unbiased**, as: \[ E(\hat{\lambda}) = \lambda \]

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) be independent and identically distributed (iid) samples from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) of the exponential distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine if the estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution based on the sample is expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function gives the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we take the derivative of the log-likelihood with respect to \(\lambda\), set it to zero, and solve for \(\lambda\). --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function is: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To find the maximum likelihood estimate, set the derivative to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To determine whether \(\hat{\lambda}\) is biased or unbiased, we need to evaluate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Given that \(X_i\) follows an exponential distribution, the expected value of an individual sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] Consequently, we have: \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Using properties of expectation in relation to the inverse of a random variable does not yield a straightforward result. However, it is known that the MLE estimator \(\hat{\lambda}\) for the exponential distribution is unbiased: \[ E\left(\hat{\lambda}\right) = \lambda \] This indicates that on average, \(\hat{\lambda}\) equals \(\lambda\). --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **unbiased**, as: \[ E(\hat{\lambda}) = \lambda \]

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) be independent and identically distributed (iid) samples from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine whether the estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) based on the sample is expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function yields the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood with respect to \(\lambda\) and set the result to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function is: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To find the maximum likelihood estimate, set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To assess whether \(\hat{\lambda}\) is biased or unbiased, we need to find the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Given that \(X_i\) follows an exponential distribution, the expected value of an individual sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This implies: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] The mean of the sum follows a gamma distribution because the sum of exponential random variables is gamma distributed. The expected value of the inverse of a gamma-distributed variable does not simply equal the inverse of the mean, leading to: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the maximum likelihood estimator \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator is **biased**, as: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This means that the MLE overestimates the true parameter \(\lambda\).

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) represent a random sample from an exponential distribution with parameter \(\lambda\). The probability density function (pdf) for this distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Assess whether the estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) is defined as the product of the individual probabilities for the observed data: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function leads to the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood with respect to \(\lambda\) and set it to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function is: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero Setting the derivative equal to zero to find the maximum likelihood estimate: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To determine if \(\hat{\lambda}\) is biased or unbiased, we need to calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since each \(X_i\) follows an exponential distribution, we know: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the total sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This means that: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] To find \(E\left(\hat{\lambda}\right)\) correctly, we can use properties of the gamma distribution. The sum of independent exponential random variables follows a gamma distribution: \[ \sum_{i=1}^{n} X_i \sim \text{Gamma}(n, \lambda) \] The expected value of the inverse of a gamma-distributed variable does not yield a straightforward result. However, it can be shown that the MLE for the exponential distribution is biased: \[ E(\hat{\lambda}) = \frac{n\lambda}{n - 1} \quad \text{(for \(n > 1\))} \] This implies that the MLE \(\hat{\lambda}\) tends to overestimate \(\lambda\). --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE does not equal the true parameter \(\lambda\) on average, resulting in an upward bias.

# Maximum Likelihood Estimation for the Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) represent independent and identically distributed (iid) samples from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) for this distribution is: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Determine the maximum likelihood estimator (MLE) for \(\lambda\). 2. Assess whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution based on the sample can be formulated as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function results in the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the derivative to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To find the maximum likelihood estimate, we set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this equation leads to: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To evaluate whether \(\hat{\lambda}\) is biased or unbiased, we calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since the random variable \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Consequently, the expected value of the sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This indicates that: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] However, the expectation of the reciprocal of a gamma-distributed variable does not yield a simple result. It is known that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This suggests that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as indicated by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This means that the MLE does not accurately reflect the true parameter \(\lambda\) on average, thus exhibiting an upward bias.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) represent independent and identically distributed (iid) samples drawn from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) for this distribution is defined as: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution, based on the observed samples, is given by: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function results in the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the derivative to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero Setting the derivative equal to zero to find the maximum likelihood estimate: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this equation gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To assess whether \(\hat{\lambda}\) is biased or unbiased, we need to calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since the random variable \(X_i\) follows an exponential distribution, the expected value of each observation is: \[ E[X_i] = \frac{1}{\lambda} \] Therefore, the expected value of the sum of the observations is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This leads us to: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Finding \(E\left(\hat{\lambda}\right)\) involves recognizing that the sum \(\sum_{i=1}^{n} X_i\) follows a gamma distribution with shape parameter \(n\) and scale parameter \(\frac{1}{\lambda}\). The expected value of the inverse of a gamma-distributed variable does not yield a straightforward result. Nonetheless, it can be shown that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This suggests that the MLE does not accurately reflect the true parameter \(\lambda\) on average, resulting in an upward bias.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) be a set of independent and identically distributed (iid) samples from an exponential distribution with the parameter \(\lambda\). The probability density function (pdf) for this distribution is defined as: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine if this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) represents the joint probability of observing the given samples and can be expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function gives the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set it equal to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero Setting the derivative to zero to find the maximum likelihood estimate: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To determine whether \(\hat{\lambda}\) is biased or unbiased, we need to find the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since \(X_i\) follows an exponential distribution, we know that: \[ E[X_i] = \frac{1}{\lambda} \] The expected value for the sum of the observations is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This leads to: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] The expectation of the inverse of a gamma-distributed variable (since the sum of exponential random variables follows a gamma distribution) does not yield a straightforward result. However, it can be shown that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as shown by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This suggests that the MLE does not accurately reflect the true parameter \(\lambda\) on average, resulting in an upward bias.

# Expected Number of Packets Processed per Hour by the Entire System ## Given Information A multinational analytics firm operates three independent data-processing clusters: Cluster A, Cluster B, and Cluster C. The workload proportions and their corresponding Poisson rate parameters are defined as follows: - Cluster A processes 45% of total packets. - Cluster B processes 35% of total packets. - Cluster C processes 20% of total packets. The rate parameters for each cluster are defined as: - \( \lambda_A = 40 + 15 - 5 = 50 \) - \( \lambda_B = 60 - 20 + 10 = 50 \) - \( \lambda_C = 30 + 25 - 10 = 45 \) The total hourly processed packets \(X\) is represented by the random variable defined as: \[ P(X = k) = 0.45 \cdot P(X_A = k) + 0.35 \cdot P(X_B = k) + 0.20 \cdot P(X_C = k) \] ## Objective Calculate the expected number of packets processed per hour across the entire system. ## Concept Used The expected value for a Poisson distribution is equal to its rate parameter (\(\lambda\)). Thus, for each cluster, we can calculate the expected number of packets processed per hour as follows: - \(E[X_A] = \lambda_A\) - \(E[X_B] = \lambda_B\) - \(E[X_C] = \lambda_C\) Since the clusters operate independently, the overall expected number of packets processed per hour by the entire system \(E[X]\) can be determined by the weighted contributions of each cluster. --- ## Step-by-Step Solution ### Calculate Expected Values for Each Cluster 1. **Expected packets from Cluster A**: \[ E[X_A] = \lambda_A = 50 \] 2. **Expected packets from Cluster B**: \[ E[X_B] = \lambda_B = 50 \] 3. **Expected packets from Cluster C**: \[ E[X_C] = \lambda_C = 45 \] ### Combine the Expected Values The overall expected number of packets processed per hour \(E[X]\) for the entire system can be calculated as the weighted sum of the expected values from each cluster, considering their respective workload proportions: \[ E[X] = 0.45 \cdot E[X_A] + 0.35 \cdot E[X_B] + 0.20 \cdot E[X_C] \] Substituting the expected values: \[ E[X] = 0.45 \cdot 50 + 0.35 \cdot 50 + 0.20 \cdot 45 \] Calculating each term: - From Cluster A: \[ 0.45 \cdot 50 = 22.5 \] - From Cluster B: \[ 0.35 \cdot 50 = 17.5 \] - From Cluster C: \[ 0.20 \cdot 45 = 9 \] ### Final Calculation Now, summing these values: \[ E[X] = 22.5 + 17.5 + 9 = 49 \] ## Final Answer The expected number of packets processed per hour by the entire system is: \[ \boxed{49} \] This result reflects the average performance across the three clusters based on their individual processing capabilities and workload distributions.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) represent independent and identically distributed (iid) samples from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) for this distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine if this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution based on the observed data is defined as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function results in the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the result to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function is: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To find the maximum likelihood estimate, set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To assess whether \(\hat{\lambda}\) is biased or unbiased, we need to calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] From this, we can derive: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] However, the expectation of the reciprocal of a gamma-distributed variable (since the sum of exponential random variables follows a gamma distribution) does not yield a straightforward result. It is known that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as shown by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This suggests that the MLE does not accurately reflect the true parameter \(\lambda\) on average, leading to an upward bias.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) represent independent and identically distributed (iid) samples from an exponential distribution with the parameter \(\lambda\). The probability density function (pdf) for the exponential distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Determine the maximum likelihood estimator (MLE) for \(\lambda\). 2. Assess whether the estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution based on the observed samples can be expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function leads to the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the derivative to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero Setting the derivative equal to zero gives us: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this equation yields: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To determine if \(\hat{\lambda}\) is biased or unbiased, we will calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since each \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the total sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] From this, we have: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] However, the mean of the sum follows a gamma distribution since the sum of exponential random variables is gamma distributed. The expected value of the reciprocal of a gamma-distributed variable does not yield a straightforward result. It is known that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the maximum likelihood estimator \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as indicated by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This implies that the MLE does not accurately reflect the true parameter \(\lambda\) on average, resulting in an upward bias.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) represent independent and identically distributed (iid) samples from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) for this distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the given sample is defined by the product of the individual probabilities: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the logarithm of the likelihood function results in the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the derivative equal to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero Setting the derivative equal to zero gives us: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To determine if \(\hat{\lambda}\) is biased or unbiased, we need to calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since each \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the sum of the observations is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This leads to: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] However, the expected value of the reciprocal of a gamma-distributed variable (since the sum of exponential random variables follows a gamma distribution) does not yield a straightforward result. Nonetheless, it is known that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as demonstrated by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This suggests that the MLE does not accurately estimate the true parameter \(\lambda\) on average, resulting in a systematic upward bias.

# Maximum Likelihood Estimation for the Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) denote a random sample from an exponential distribution characterized by the parameter \(\lambda\). The probability density function (pdf) of this distribution is given by: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Derive the maximum likelihood estimator (MLE) for \(\lambda\). 2. Assess whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution based on the observed samples is expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function results in the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the derivative equal to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function is given by: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To determine the maximum likelihood estimate, we set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this equation gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To evaluate whether \(\hat{\lambda}\) is biased or unbiased, we need to calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since each \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the total sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This leads to: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] However, the expected value of the reciprocal of a gamma-distributed variable (since the sum of exponential random variables follows a gamma distribution) does not yield a straightforward result. It can be shown that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that \(\hat{\lambda}\) is biased, meaning it does not accurately estimate \(\lambda\) on average. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as shown by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE tends to overestimate the true parameter \(\lambda\), resulting in an upward bias.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Consider \(X_1, X_2, \ldots, X_n\) as independent and identically distributed (iid) samples drawn from an exponential distribution with the parameter \(\lambda\). The probability density function (pdf) for this distribution is defined as: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Derive the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution based on the observed data samples can be expressed as: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function results in the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the result to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function is given as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To find the maximum likelihood estimate, set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging gives: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Therefore, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To determine if \(\hat{\lambda}\) is biased or unbiased, we need to evaluate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since \(X_i\) follows an exponential distribution, the expected value of each sample is given by: \[ E[X_i] = \frac{1}{\lambda} \] Consequently, the expected value of the total sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This leads to: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] However, the expected value of the reciprocal of a gamma-distributed variable (since the sum of exponential random variables follows a gamma distribution) does not have a straightforward expression. It is known that: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as demonstrated by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE tends to overestimate the true parameter \(\lambda\), resulting in an upward bias.

# Maximum Likelihood Estimation for the Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) be independent and identically distributed (iid) samples from an exponential distribution with the parameter \(\lambda\). The probability density function (pdf) of this distribution is expressed as: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Find the maximum likelihood estimator (MLE) for \(\lambda\). 2. Assess whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for the exponential distribution is defined as the product of the individual probabilities corresponding to the observed data: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function yields the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] To find the MLE, we differentiate the log-likelihood function with respect to \(\lambda\) and set the result to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative of the log-likelihood with respect to \(\lambda\): \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero Setting the derivative equal to zero gives us the condition for maximization: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this equation leads to: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To evaluate whether \(\hat{\lambda}\) is biased or unbiased, we calculate the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since the random variable \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the sum is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] Consequently, we have: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] It is known that the expected value of the reciprocal of a gamma-distributed variable (since the sum of exponential random variables follows a gamma distribution) does not yield a simple result, but can be expressed as: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) is biased. --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as shown by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE tends to overestimate the true parameter \(\lambda\), leading to a systematic upward bias.

# Maximum Likelihood Estimation for Exponential Distribution ## Given Information Let \(X_1, X_2, \ldots, X_n\) denote a set of independent and identically distributed (iid) samples drawn from an exponential distribution with the parameter \(\lambda\). The probability density function (pdf) for this distribution is represented as: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x > 0 \] ## Objective 1. Derive the maximum likelihood estimator (MLE) for \(\lambda\). 2. Determine whether this estimator is biased or unbiased. ## Concept Used The likelihood function \(L(\lambda)\) for this sample is given by the product of the individual probabilities: \[ L(\lambda) = \prod_{i=1}^{n} f(X_i; \lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda X_i} \] Taking the natural logarithm of the likelihood function gives the log-likelihood function: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] We will find the MLE by differentiating the log-likelihood function with respect to \(\lambda\) and setting the derivative to zero. --- ## Step-by-Step Solution ### Log-Likelihood Function The log-likelihood function can be expressed as: \[ \log L(\lambda) = n \log(\lambda) - \lambda \sum_{i=1}^{n} X_i \] ### Differentiate the Log-Likelihood Taking the derivative of the log-likelihood with respect to \(\lambda\) results in: \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^{n} X_i \] ### Set the Derivative to Zero To find the maximum likelihood estimate, we set the derivative equal to zero: \[ \frac{n}{\lambda} - \sum_{i=1}^{n} X_i = 0 \] Rearranging this equation yields: \[ \frac{n}{\lambda} = \sum_{i=1}^{n} X_i \quad \Rightarrow \quad \lambda = \frac{n}{\sum_{i=1}^{n} X_i} \] ### Maximum Likelihood Estimator Thus, the maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] --- ## Is the Estimator Biased or Unbiased? To evaluate whether \(\hat{\lambda}\) is biased or unbiased, we need to find the expected value of \(\hat{\lambda}\): \[ E(\hat{\lambda}) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] Since \(X_i\) follows an exponential distribution, the expected value of each sample is: \[ E[X_i] = \frac{1}{\lambda} \] Thus, the expected value of the sum of the observations is: \[ E\left(\sum_{i=1}^{n} X_i\right) = n \cdot E[X] = n \cdot \frac{1}{\lambda} \] This leads to: \[ E\left(\hat{\lambda}\right) = E\left(\frac{n}{\sum_{i=1}^{n} X_i}\right) \] The expected value of the reciprocal of a random variable that follows a gamma distribution does not yield a direct relationship. However, it is known that the maximum likelihood estimator for the exponential distribution is biased: \[ E\left(\hat{\lambda}\right) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This indicates that the MLE \(\hat{\lambda}\) tends to overestimate \(\lambda\). --- ## Final Summary 1. The maximum likelihood estimator for \(\lambda\) is: \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} X_i} \] 2. The estimator \(\hat{\lambda}\) is **biased**, as indicated by: \[ E(\hat{\lambda}) = \frac{n\lambda}{n-1} \quad \text{(for \(n > 1\))} \] This suggests that the MLE does not accurately estimate the true parameter \(\lambda\) on average, resulting in a systematic upward bias.

# Probability Analysis of Defective Item Origin ## Given Information A factory operates with three machines producing items with the following proportions and defect rates: - **Machine A**: - Production percentage: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production percentage: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production percentage: \(25\%\) - Defect rate: \(5\%\) Let \(D\) be the event that an item is found defective, and let \(M_A\), \(M_B\), and \(M_C\) represent the events that an item comes from Machine A, Machine B, and Machine C, respectively. ## Objective Calculate the probability \(P(M_C | D)\), which is the probability that a defective item came from Machine C. ## Concept Used To find \(P(M_C | D)\), we will use **Bayes' Theorem**: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The conditional probabilities of an item being defective given it comes from each machine are: - \(P(D | M_A) = 0.02\) (2% for Machine A) - \(P(D | M_B) = 0.03\) (3% for Machine B) - \(P(D | M_C) = 0.05\) (5% for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of picking an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we can find \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating each term: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now we can substitute into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this gives: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This indicates that if an item is found to be defective, there is about a 39.06% chance that it originated from Machine C.

# Probability Analysis of Defective Item Origin ## Given Information A factory operates three machines, each contributing to the total production and having varying defect rates: - **Machine A**: - Production percentage: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production percentage: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production percentage: \(25\%\) - Defect rate: \(5\%\) Let \(D\) denote the event that an item is defective, and \(M_A\), \(M_B\), and \(M_C\) represent the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective Calculate the probability \(P(M_C | D)\), which is the likelihood that a defective item originated from Machine C. ## Concept Used To determine \(P(M_C | D)\), we will utilize **Bayes' Theorem**: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The conditional probabilities of an item being defective given it comes from each machine are as follows: - \(P(D | M_A) = 0.02\) (2% defect rate for Machine A) - \(P(D | M_B) = 0.03\) (3% defect rate for Machine B) - \(P(D | M_C) = 0.05\) (5% defect rate for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we can calculate \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating each term: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now we substitute into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this gives: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This result indicates that if an item is found to be defective, there is about a 39.06% chance that it originated from Machine C.

# Probability of Defective Item Origin ## Given Information A factory operates three machines, each contributing differently to the overall production and exhibiting various defect rates: - **Machine A**: - Production proportion: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production proportion: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production proportion: \(25\%\) - Defect rate: \(5\%\) Let \(D\) represent the event that an item is defective, while \(M_A\), \(M_B\), and \(M_C\) denote the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective Determine the probability \(P(M_C | D)\), which is the probability that a defective item was produced by Machine C. ## Concept Used To find \(P(M_C | D)\), we will apply **Bayes' Theorem**: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The probabilities of an item being defective given its machine origin are: - \(P(D | M_A) = 0.02\) (2% for Machine A) - \(P(D | M_B) = 0.03\) (3% for Machine B) - \(P(D | M_C) = 0.05\) (5% for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we can determine \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating gives: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now we substitute into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this yields: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This result indicates that if an item is identified as defective, there is about a 39.06% chance that it originated from Machine C.

# Probability of Defective Item Origin ## Given Information In a factory, there are three machines, each contributing to the overall production with different rates and defect percentages: - **Machine A**: - Production share: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production share: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production share: \(25\%\) - Defect rate: \(5\%\) We define the event \(D\) as an item being defective, while \(M_A\), \(M_B\), and \(M_C\) denote the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective We want to find the probability \(P(M_C | D)\), which represents the likelihood that a defective item was produced by Machine C. ## Concept Used To calculate \(P(M_C | D)\), we will utilize **Bayes' Theorem**, which is formulated as follows: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The conditional probabilities of an item being defective based on its machine origin are: - \(P(D | M_A) = 0.02\) (for Machine A) - \(P(D | M_B) = 0.03\) (for Machine B) - \(P(D | M_C) = 0.05\) (for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we can find \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating this yields: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now, we can substitute into Bayes' theorem to find the desired probability: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating further: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This indicates that if an item is found defective, there is a 39.06% chance it originated from Machine C.

# Probability of Defective Item Origin ## Given Information In a manufacturing facility, there are three machines contributing to the overall item production, each with distinct production shares and defect rates: - **Machine A**: - Production percentage: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production percentage: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production percentage: \(25\%\) - Defect rate: \(5\%\) Let \(D\) signify the event that an item is defective, while \(M_A\), \(M_B\), and \(M_C\) represent the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective Calculate the probability \(P(M_C | D)\), which represents the chance that a defective item was produced by Machine C. ## Concept Used To find \(P(M_C | D)\), we will use **Bayes' Theorem**: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The probabilities of an item being defective given its machine origin are: - \(P(D | M_A) = 0.02\) (for Machine A) - \(P(D | M_B) = 0.03\) (for Machine B) - \(P(D | M_C) = 0.05\) (for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we express \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the respective values gives: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating this results in: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now we can substitute into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Performing the calculation yields: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This result indicates that if an item is found to be defective, there is about a 39.06% chance that it originated from Machine C.

# Probability of Defective Item Origin ## Given Information In a manufacturing environment, three machines contribute to the production of items, each with distinct production rates and defect percentages: - **Machine A**: - Production share: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production share: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production share: \(25\%\) - Defect rate: \(5\%\) Let \(D\) denote the event that an item is defective, while \(M_A\), \(M_B\), and \(M_C\) represent the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective To find the probability \(P(M_C | D)\), which indicates the likelihood that a defective item was produced by Machine C. ## Concept Used To calculate \(P(M_C | D)\), we will apply **Bayes' Theorem**, which is given by: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] The conditional probabilities of an item being defective based on the machine are as follows: - \(P(D | M_A) = 0.02\) (2% for Machine A) - \(P(D | M_B) = 0.03\) (3% for Machine B) - \(P(D | M_C) = 0.05\) (5% for Machine C) Next, the prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) Using the law of total probability, we can determine \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the known values yields: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating this gives: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] Now we can substitute our values into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this results in: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This result indicates that if an item is determined to be defective, there is about a 39.06% chance that it originated from Machine C.

# Probability of Defective Item Origin ## Given Information A factory operates three machines, each with different production shares and defect rates: - **Machine A**: - Production percentage: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production percentage: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production percentage: \(25\%\) - Defect rate: \(5\%\) Let \(D\) represent the event that an item is defective, while \(M_A\), \(M_B\), and \(M_C\) denote the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective To find the probability \(P(M_C | D)\), which indicates the likelihood that a defective item was produced by Machine C. ## Concept Used We will apply **Bayes' Theorem** to determine \(P(M_C | D)\): \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The conditional probabilities of an item being defective based on its machine origin are: - \(P(D | M_A) = 0.02\) for Machine A - \(P(D | M_B) = 0.03\) for Machine B - \(P(D | M_C) = 0.05\) for Machine C ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we find \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating each term gives: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now substituting into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this results in: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This calculation indicates that if an item is found to be defective, there is about a 39.06% chance that it originated from Machine C.

# Probability of Defective Item Origin ## Given Information In a factory, three machines are responsible for producing items, each with distinct production shares and defect rates: - **Machine A**: - Production share: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production share: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production share: \(25\%\) - Defect rate: \(5\%\) We define \(D\) as the event that an item is defective, while \(M_A\), \(M_B\), and \(M_C\) denote the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective To find the probability \(P(M_C | D)\), which represents the likelihood that a defective item was produced by Machine C. ## Concept Used To compute \(P(M_C | D)\), we will utilize **Bayes' Theorem**, given by: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) We need to determine the probabilities of an item being defective based on its machine origin: - \(P(D | M_A) = 0.02\) (for Machine A) - \(P(D | M_B) = 0.03\) (for Machine B) - \(P(D | M_C) = 0.05\) (for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we find \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the known values: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating gives: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now we substitute into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this results in: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This outcome indicates that if an item is identified as defective, there is roughly a 39.06% chance that it came from Machine C.

# Probability of Defective Item Origin ## Given Information In a manufacturing facility, three machines produce items, each with different production contributions and defect rates: - **Machine A**: - Production proportion: \(30\%\) - Defect rate: \(2\%\) - **Machine B**: - Production proportion: \(45\%\) - Defect rate: \(3\%\) - **Machine C**: - Production proportion: \(25\%\) - Defect rate: \(5\%\) Let \(D\) be the event that an item is defective, while \(M_A\), \(M_B\), and \(M_C\) denote the events that an item is produced by Machine A, Machine B, and Machine C, respectively. ## Objective We need to determine the probability \(P(M_C | D)\), which indicates the likelihood that a defective item was produced by Machine C. ## Concept Used To find \(P(M_C | D)\), we will apply **Bayes' Theorem**: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} \] ### Step 1: Calculate \(P(D | M_i)\) The probabilities of an item being defective given its machine origin are: - \(P(D | M_A) = 0.02\) (for Machine A) - \(P(D | M_B) = 0.03\) (for Machine B) - \(P(D | M_C) = 0.05\) (for Machine C) ### Step 2: Calculate \(P(M_i)\) The prior probabilities of selecting an item from each machine are: - \(P(M_A) = 0.30\) - \(P(M_B) = 0.45\) - \(P(M_C) = 0.25\) ### Step 3: Calculate \(P(D)\) Using the law of total probability, we can express \(P(D)\): \[ P(D) = P(D | M_A) \cdot P(M_A) + P(D | M_B) \cdot P(M_B) + P(D | M_C) \cdot P(M_C) \] Substituting the values into the equation: \[ P(D) = (0.02 \cdot 0.30) + (0.03 \cdot 0.45) + (0.05 \cdot 0.25) \] Calculating the result: \[ = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Step 4: Apply Bayes' Theorem Now we can substitute into Bayes' theorem: \[ P(M_C | D) = \frac{P(D | M_C) \cdot P(M_C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] Calculating this yields: \[ P(M_C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] ## Final Answer The probability that a defective item was produced by Machine C is approximately: \[ \boxed{0.3906} \] This result indicates that if an item is found to be defective, there is about a 39.06% chance that it originated from Machine C.