# Given Information - We roll a fair six-sided die repeatedly. - We stop rolling when a 6 appears for the first time. - Let \( X \) be the number of rolls until the first 6 appears. - The die is fair, so the probability of rolling a 6 on any roll is \( p = \frac{1}{6} \). # Definition or Concept Used This scenario is modeled by a **Geometric Distribution**: - The probability that the first success occurs on the \( k \)-th trial: \( P(X = k) = (1-p)^{k-1}p \), where \( p \) is the probability of success in each trial. - For a geometric random variable \( X \) with success probability \( p \): - **Expected value:** \( E[X] = \frac{1}{p} \) - **Variance:** \( \operatorname{Var}(X) = \frac{1-p}{p^2} \) # Solution For a fair die, the probability of rolling a 6 in one trial is \( p = \frac{1}{6} \). The expected number of rolls until the first 6 appears: \[ E[X] = \frac{1}{p} = \frac{1}{1/6} = 6 \] The variance of the number of rolls: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substitute \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5}{6} \times 36 = 30 \] # Final Answer - The expected number of rolls until a 6 appears is **6**. - The variance of the number of rolls is **30**.

# Given Information - We are rolling a fair six-sided die repeatedly. - The rolling stops when a 6 is rolled for the first time. - Let \( X \) represent the number of rolls until the first 6 appears. - The probability of rolling a 6 on any given roll is \( p = \frac{1}{6} \). # Definition or Concept Used This situation follows a **Geometric Distribution**: - The probability that the first success (rolling a 6) occurs on the \( k \)-th trial is given by: \( P(X = k) = (1-p)^{k-1}p \) - For a geometric random variable \( X \) with success probability \( p \): - **Expected Value:** \( E[X] = \frac{1}{p} \) - **Variance:** \( \operatorname{Var}(X) = \frac{1-p}{p^2} \) # Solution The probability of rolling a 6 in one trial is \( p = \frac{1}{6} \). To find the expected number of rolls until the first 6 appears: \[ E[X] = \frac{1}{p} = \frac{1}{1/6} = 6 \] Next, we calculate the variance of the number of rolls: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5}{6} \times 36 = 30 \] # Final Answer - The expected number of rolls until a 6 appears is **6**. - The variance of the number of rolls is **30**.

# Given Information - A fair six-sided die is rolled repeatedly. - The process stops when a 6 is rolled for the first time. - Let \( X \) denote the total number of rolls until the first occurrence of a 6. - The probability of rolling a 6 in any individual roll is \( p = \frac{1}{6} \). # Definition or Concept Used This scenario is described using a **Geometric Distribution**, which models the number of trials until the first success: - The probability mass function is given by: \[ P(X = k) = (1-p)^{k-1}p \] - For a geometric random variable \( X \): - **Expected Value:** \[ E[X] = \frac{1}{p} \] - **Variance:** \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution The probability of rolling a 6 in a single trial is \( p = \frac{1}{6} \). To determine the expected number of rolls until the first 6 appears: \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, we calculate the variance of the number of rolls: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \times 36}{6} = 30 \] # Final Answer - The expected number of rolls until a 6 appears is **6**. - The variance of the number of rolls is **30**.

# Given Information A fair six-sided die is rolled repeatedly until a 6 appears for the first time. Let \( X \) represent the number of rolls needed. The probability of rolling a 6 on any roll is \( p = \frac{1}{6} \). # Definition or Concept Used This situation can be modeled by a **Geometric Distribution**, which describes the number of trials until the first success occurs. The relevant formulas are: The probability that the first success occurs on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] For a geometric random variable \( X \): The expected value is: \[ E[X] = \frac{1}{p} \] The variance is: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution First, we calculate the expected number of rolls until the first 6 appears. Given \( p = \frac{1}{6} \): \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, we compute the variance of the number of rolls. Using the formula for variance: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \times 36}{6} = 30 \] # Final Answer The expected number of rolls until a 6 appears is **6**, and the variance of the number of rolls is **30**.

# Given Information In this scenario, we roll a fair six-sided die repeatedly until we roll a 6. Let \( X \) denote the number of rolls required to get the first 6. The probability of rolling a 6 in a single attempt is \( p = \frac{1}{6} \) because there is one favorable outcome (rolling a 6) out of six possible outcomes (1, 2, 3, 4, 5, and 6). # Definition or Concept Used This process follows a **Geometric Distribution**. This distribution is appropriate for modeling the number of trials until the first success (in this case, rolling a 6). The key properties of a geometric random variable \( X \) include: The probability of achieving the first success on the \( k \)-th trial is expressed as: \[ P(X = k) = (1-p)^{k-1}p \] For a geometric random variable, the expected value and variance are defined as follows: - The expected value (mean) is given by: \[ E[X] = \frac{1}{p} \] - The variance is calculated as: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution To find the expected number of rolls until the first occurrence of a 6, we substitute \( p = \frac{1}{6} \) into the expected value formula: \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, we compute the variance of the number of rolls. Using the variance formula for a geometric distribution, we substitute \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1-p}{p^2} = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} \] Simplifying this further: \[ \operatorname{Var}(X) = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \cdot 36}{6} = 30 \] # Final Answer The expected number of rolls until a 6 appears is **6**, and the variance of the number of rolls is **30**. The probability \( p = \frac{1}{6} \) arises because there is one successful outcome (rolling a 6) among six equally likely outcomes on a fair die.

# Given Information In this problem, we are rolling a fair six-sided die continuously until a 6 is rolled for the first time. Let \( X \) represent the total number of rolls required to achieve this. The probability of rolling a 6 in one attempt is \( p = \frac{1}{6} \), as there is one favorable outcome (rolling a 6) out of six possible outcomes (1, 2, 3, 4, 5, and 6). # Definition or Concept Used This situation can be analyzed using a **Geometric Distribution**, which is suitable for modeling the number of trials until the first success occurs. The relevant formulas for a geometric random variable \( X \) are as follows: The probability of achieving the first success on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] For a geometric random variable, the expected value and variance are defined as: - **Expected Value:** \[ E[X] = \frac{1}{p} \] - **Variance:** \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution To determine the expected number of rolls until the first 6 appears, we substitute \( p = \frac{1}{6} \) into the expected value formula: \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, we calculate the variance of the number of rolls using the variance formula: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \cdot 36}{6} = 30 \] # Final Answer The expected number of rolls until a 6 appears is **6**, and the variance of the number of rolls is **30**. The probability \( p \) is \( \frac{1}{6} \) because there is one successful outcome (rolling a 6) among the six equally likely outcomes of a fair die.

# Given Information We are rolling a fair six-sided die repeatedly until the first occurrence of a 6. Let \( X \) represent the number of rolls needed to roll a 6. The probability of rolling a 6 on any single roll is \( p = \frac{1}{6} \) since there is one favorable outcome (rolling a 6) among six equally likely outcomes. # Definition or Concept Used This problem is modeled by a **Geometric Distribution**, which describes the number of trials required to achieve the first success. For a geometric random variable \( X \): The probability of the first success occurring on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] The key properties of a geometric distribution include: - **Expected Value:** \[ E[X] = \frac{1}{p} \] - **Variance:** \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution First, to find the expected number of rolls until we get the first 6, we use the expected value formula with \( p = \frac{1}{6} \): \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, we calculate the variance of the number of rolls using the variance formula: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \cdot 36}{6} = 30 \] # Final Answer The expected number of rolls until a 6 appears is **6**, and the variance of the number of rolls is **30**. The probability \( p \) is \( \frac{1}{6} \) because there is one successful outcome (rolling a 6) in the context of six equally likely outcomes when rolling a fair die.

# Given Information In this scenario, we repeatedly roll a fair six-sided die until we roll a 6 for the first time. Let \( X \) denote the total number of rolls taken to achieve this outcome. The probability of rolling a 6 in a single attempt is \( p = \frac{1}{6} \), since there is one successful outcome (rolling a 6) among six possible outcomes (the numbers 1 through 6). # Definition or Concept Used This situation is best described by a **Geometric Distribution**, which models the number of trials needed to achieve the first success. For a geometric random variable \( X \): The probability of achieving the first success on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] The properties of a geometric distribution include: - **Expected Value:** \[ E[X] = \frac{1}{p} \] - **Variance:** \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution To calculate the expected number of rolls until the first 6 appears, we substitute \( p = \frac{1}{6} \) into the expected value formula: \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, we compute the variance of the number of rolls using the variance formula: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \cdot 36}{6} = 30 \] # Final Answer The expected number of rolls until a 6 appears is **6**, and the variance of the number of rolls is **30**. The probability \( p \) is \( \frac{1}{6} \) because there is one successful outcome (rolling a 6) out of six equally likely outcomes when rolling a fair die.

# Given Information We are conducting an experiment where a fair six-sided die is rolled repeatedly until a 6 is rolled for the first time. Let \( X \) represent the number of rolls needed to achieve this outcome. The probability of rolling a 6 on a single roll is \( p = \frac{1}{6} \), reflecting that there is one successful outcome (rolling a 6) among six possible outcomes (1, 2, 3, 4, 5, 6). # Definition or Concept Used This scenario can be modeled using a **Geometric Distribution**, which applies to the number of trials required to obtain the first success. The relevant properties of a geometric random variable \( X \) include: The probability that the first success occurs on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] Important characteristics of a geometric distribution are: - **Expected Value:** \[ E[X] = \frac{1}{p} \] - **Variance:** \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] # Solution To find the expected number of rolls until the first 6 appears, we substitute \( p = \frac{1}{6} \) into the expected value formula: \[ E[X] = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6 \] Next, to calculate the variance of the number of rolls, we use the variance formula for a geometric distribution: \[ \operatorname{Var}(X) = \frac{1-p}{p^2} \] Substituting \( p = \frac{1}{6} \): \[ \operatorname{Var}(X) = \frac{1 - \frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = \frac{5 \cdot 36}{6} = 30 \] # Final Answer The expected number of rolls until a 6 appears is **6**, and the variance of the number of rolls is **30**. The probability \( p \) is \( \frac{1}{6} \) because there is one successful outcome (rolling a 6) out of the six equally likely outcomes when rolling a fair die.

# Given Information Two dice are rolled repeatedly until a sum of 8 appears. We want to find the probability that the first occurrence of a sum of 8 happens on the 5th throw. # Definition or Concept Used This scenario can be modeled using the **Geometric Distribution**. The probability that the first success (rolling a sum of 8) occurs on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] where \( p \) is the probability of rolling an 8 with two dice. To find \( p \), we first determine the number of ways to roll a sum of 8 with two dice. The possible combinations are: - (2, 6) - (3, 5) - (4, 4) - (5, 3) - (6, 2) Thus, there are 5 successful outcomes. The total number of outcomes when rolling two dice is \( 6 \times 6 = 36 \). Therefore, the probability \( p \) is: \[ p = \frac{5}{36} \] The probability of not rolling a sum of 8 (failure) is: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \] # Solution To find the probability that the first sum of 8 appears on the 5th throw, we apply the geometric distribution formula: \[ P(X = 5) = (1-p)^{4}p \] Substituting the values of \( p \) and \( 1-p \): \[ P(X = 5) = \left(\frac{31}{36}\right)^{4} \cdot \frac{5}{36} \] Now we calculate: 1. Calculate \( \left(\frac{31}{36}\right)^{4} \): \[ \left(\frac{31}{36}\right)^{4} = \frac{31^4}{36^4} \] Using \( 31^4 = 923521 \) and \( 36^4 = 1296^2 = 1679616 \): \[ \left(\frac{31}{36}\right)^{4} \approx \frac{923521}{1679616} \] 2. Multiply by \( \frac{5}{36} \): \[ P(X = 5) = \frac{923521}{1679616} \cdot \frac{5}{36} = \frac{4617605}{60466176} \] # Final Answer The probability that a sum of 8 appears for the first time on the 5th throw is approximately **0.0761** or in fractional form **\(\frac{4617605}{60466176}\)**.

# Given Information Two dice are rolled repeatedly until a sum of 8 appears. We want to find the probability that the first occurrence of a sum of 8 happens on the 5th throw. # Definition or Concept Used This situation can be modeled using the **Geometric Distribution**. The probability that the first success (rolling a sum of 8) occurs on the \( k \)-th trial is given by: \[ P(X = k) = (1-p)^{k-1}p \] where \( p \) is the probability of rolling an 8 with two dice. To find \( p \), we determine the number of ways to roll a sum of 8 with two dice. The combinations that yield a sum of 8 are: - (2, 6) - (3, 5) - (4, 4) - (5, 3) - (6, 2) Thus, there are 5 successful outcomes. The total number of outcomes when rolling two dice is \( 6 \times 6 = 36 \). Therefore, the probability \( p \) is calculated as: \[ p = \frac{5}{36} \] The probability of not rolling a sum of 8 (failure) is: \[ 1 - p = 1 - \frac{5}{36} = \frac{31}{36} \] # Solution To find the probability that the first sum of 8 appears on the 5th throw, we apply the geometric distribution formula: \[ P(X = 5) = (1-p)^{4}p \] Substituting the values of \( p \) and \( 1-p \): \[ P(X = 5) = \left(\frac{31}{36}\right)^{4} \cdot \frac{5}{36} \] Now we calculate: 1. Calculate \( \left(\frac{31}{36}\right)^{4} \): \[ \left(\frac{31}{36}\right)^{4} = \frac{31^4}{36^4} \] Using \( 31^4 = 923521 \) and \( 36^4 = 1296^2 = 1679616 \): \[ \left(\frac{31}{36}\right)^{4} \approx \frac{923521}{1679616} \] 2. Multiply by \( \frac{5}{36} \): \[ P(X = 5) = \frac{923521}{1679616} \cdot \frac{5}{36} = \frac{4617605}{60466176} \] Now calculating the decimal approximation: \[ P(X = 5) \approx \frac{4617605}{60466176} \approx 0.0764 \] # Final Answer The probability that a sum of 8 appears for the first time on the 5th throw is approximately **0.0764**.

# Given Information The scores on a standardized intelligence test are normally distributed with: - Mean (\( \mu \)) = 100 - Standard deviation (\( \sigma \)) = 15 # Definition or Concept Used To solve the problems, we will use the properties of the **Normal Distribution**. We can standardize a score using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will use Z-tables or a calculator to find probabilities associated with the Z-scores. # Solution ### Part (a): Probability that a randomly selected individual scores below 85 First, we calculate the Z-score for \( X = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Now, we look up the Z-score of -1 in the Z-table, which gives us: \[ P(Z < -1) \approx 0.1587 \] ### Part (b): Probability that a randomly selected individual scores above 130 Next, we calculate the Z-score for \( X = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Now, we look up the Z-score of 2 in the Z-table, which gives us: \[ P(Z < 2) \approx 0.9772 \] To find the probability of scoring above 130: \[ P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \] ### Part (c): Probability that a randomly selected individual scores between 85 and 130 We already have the Z-scores for 85 and 130. We can use the probabilities calculated above: 1. For \( X = 85 \) (Z = -1): \( P(Z < -1) \approx 0.1587 \) 2. For \( X = 130 \) (Z = 2): \( P(Z < 2) \approx 0.9772 \) To find the probability of scoring between 85 and 130: \[ P(85 < X < 130) = P(Z < 2) - P(Z < -1) \approx 0.9772 - 0.1587 = 0.8185 \] # Final Answer (a) The probability that a randomly selected individual scores below 85 is approximately **0.1587**. (b) The probability that a randomly selected individual scores above 130 is approximately **0.0228**. (c) The probability that a randomly selected individual scores between 85 and 130 is approximately **0.8185**.

# Given Information The diameters of manufactured bolts are normally distributed with: - Mean (\( \mu \)) = 10 mm - Standard deviation (\( \sigma \)) = 0.8 mm # Definition or Concept Used To solve the problems, we will use the properties of the **Normal Distribution** and the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also use the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to find probabilities. # Solution ### Part (a): Percentage of bolts with diameter less than 9 mm First, we calculate the Z-score for \( X = 9 \): \[ Z = \frac{9 - 10}{0.8} = \frac{-1}{0.8} = -1.25 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(-1.25, 0, 1, TRUE) \] This gives approximately: \[ P(Z < -1.25) \approx 0.1056 \] To convert this to a percentage: \[ \text{Percentage} = 0.1056 \times 100 \approx 10.56\% \] ### Part (b): Percentage of bolts with diameter greater than 11.5 mm Next, we calculate the Z-score for \( X = 11.5 \): \[ Z = \frac{11.5 - 10}{0.8} = \frac{1.5}{0.8} = 1.875 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(1.875, 0, 1, TRUE) \] This gives approximately: \[ P(Z < 1.875) \approx 0.9699 \] To find the probability of greater than 11.5 mm: \[ P(Z > 1.875) = 1 - P(Z < 1.875) \approx 1 - 0.9699 = 0.0301 \] Converting this to a percentage: \[ \text{Percentage} = 0.0301 \times 100 \approx 3.01\% \] ### Part (c): Proportion of bolts with diameters between 9 mm and 11.5 mm We already have the probabilities for the two Z-scores: - For \( X = 9 \) (Z = -1.25): \( P(Z < -1.25) \approx 0.1056 \) - For \( X = 11.5 \) (Z = 1.875): \( P(Z < 1.875) \approx 0.9699 \) To find the proportion of bolts with diameters between 9 mm and 11.5 mm: \[ P(9 < X < 11.5) = P(Z < 1.875) - P(Z < -1.25) \approx 0.9699 - 0.1056 = 0.8643 \] # Final Answer (a) The percentage of bolts with diameter less than 9 mm is approximately **10.56%**. (b) The percentage of bolts with diameter greater than 11.5 mm is approximately **3.01%**. (c) The proportion of bolts with diameters between 9 mm and 11.5 mm is approximately **86.43%**.

# Given Information The diameters of manufactured bolts follow a normal distribution characterized by: - Mean (\( \mu \)) = 10 mm - Standard deviation (\( \sigma \)) = 0.8 mm # Definition or Concept Used To solve the problem, we will employ the **Normal Distribution** properties and the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also utilize the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to calculate probabilities. # Solution ### Part (a): Percentage of bolts with diameter less than 9 mm First, we calculate the Z-score for \( X = 9 \): \[ Z = \frac{9 - 10}{0.8} = \frac{-1}{0.8} = -1.25 \] Using the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(-1.25, 0, 1, TRUE) \] This gives approximately: \[ P(Z < -1.25) \approx 0.1056 \] Converting this to a percentage: \[ \text{Percentage} = 0.1056 \times 100 \approx 10.56\% \] ### Part (b): Percentage of bolts with diameter greater than 11.5 mm Next, we calculate the Z-score for \( X = 11.5 \): \[ Z = \frac{11.5 - 10}{0.8} = \frac{1.5}{0.8} = 1.875 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(1.875, 0, 1, TRUE) \] This gives approximately: \[ P(Z < 1.875) \approx 0.9699 \] To find the probability of a diameter greater than 11.5 mm: \[ P(Z > 1.875) = 1 - P(Z < 1.875) \approx 1 - 0.9699 = 0.0301 \] Converting this to a percentage: \[ \text{Percentage} = 0.0301 \times 100 \approx 3.01\% \] ### Part (c): Proportion of bolts with diameters between 9 mm and 11.5 mm Using the probabilities calculated earlier: - For \( X = 9 \) (Z = -1.25): \( P(Z < -1.25) \approx 0.1056 \) - For \( X = 11.5 \) (Z = 1.875): \( P(Z < 1.875) \approx 0.9699 \) The proportion of bolts with diameters between 9 mm and 11.5 mm is found by: \[ P(9 < X < 11.5) = P(Z < 1.875) - P(Z < -1.25) \approx 0.9699 - 0.1056 = 0.8643 \] # Final Answer (a) The percentage of bolts with a diameter less than 9 mm is approximately **10.56%**. (b) The percentage of bolts with a diameter greater than 11.5 mm is approximately **3.01%**. (c) The proportion of bolts with diameters between 9 mm and 11.5 mm is approximately **86.43%**.

# Given Information The diameters of manufactured bolts are normally distributed with: - Mean (\( \mu \)) = 10 mm - Standard deviation (\( \sigma \)) = 0.8 mm # Definition or Concept Used We will use the **Normal Distribution** and the Z-score formula to find the required probabilities: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. The Excel function `=NORM.DIST(z, 0, 1, TRUE)` will be used to calculate cumulative probabilities. # Solution ### Part (a): Percentage of bolts with a diameter less than 9 mm First, we calculate the Z-score for \( X = 9 \): \[ Z = \frac{9 - 10}{0.8} = \frac{-1}{0.8} = -1.25 \] Using the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(-1.25, 0, 1, TRUE) \] This evaluates to approximately: \[ P(Z < -1.25) \approx 0.1056 \] To express this as a percentage: \[ \text{Percentage} = 0.1056 \times 100 \approx 10.56\% \] ### Part (b): Percentage of bolts with a diameter greater than 11.5 mm Next, we calculate the Z-score for \( X = 11.5 \): \[ Z = \frac{11.5 - 10}{0.8} = \frac{1.5}{0.8} = 1.875 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(1.875, 0, 1, TRUE) \] This evaluates to approximately: \[ P(Z < 1.875) \approx 0.9699 \] To find the probability of a diameter greater than 11.5 mm: \[ P(Z > 1.875) = 1 - P(Z < 1.875) \approx 1 - 0.9699 = 0.0301 \] Converting this to a percentage: \[ \text{Percentage} = 0.0301 \times 100 \approx 3.01\% \] ### Part (c): Proportion of bolts with diameters between 9 mm and 11.5 mm We will use the probabilities calculated for the two Z-scores: - For \( X = 9 \) (Z = -1.25): \( P(Z < -1.25) \approx 0.1056 \) - For \( X = 11.5 \) (Z = 1.875): \( P(Z < 1.875) \approx 0.9699 \) The proportion of bolts with diameters between 9 mm and 11.5 mm is: \[ P(9 < X < 11.5) = P(Z < 1.875) - P(Z < -1.25) \approx 0.9699 - 0.1056 = 0.8643 \] # Final Answer (a) The percentage of bolts with a diameter less than 9 mm is approximately **10.56%**. (b) The percentage of bolts with a diameter greater than 11.5 mm is approximately **3.01%**. (c) The proportion of bolts with diameters between 9 mm and 11.5 mm is approximately **86.43%**.

# Given Information The scores on a standardized intelligence test are normally distributed with: - Mean (\( \mu \)) = 100 - Standard deviation (\( \sigma \)) = 15 # Definition or Concept Used To find the probabilities requested, we will use the **Normal Distribution** and the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also employ the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to find the cumulative probabilities. # Solution ### Part (a): Probability that a randomly selected individual scores below 85 First, calculate the Z-score for \( X = 85 \): \[ Z = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \] Using the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(-1, 0, 1, TRUE) \] This yields: \[ P(Z < -1) \approx 0.1587 \] ### Part (b): Probability that a randomly selected individual scores above 130 Next, calculate the Z-score for \( X = 130 \): \[ Z = \frac{130 - 100}{15} = \frac{30}{15} = 2 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(2, 0, 1, TRUE) \] This yields: \[ P(Z < 2) \approx 0.9772 \] To find the probability of scoring above 130: \[ P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \] ### Part (c): Probability that a randomly selected individual scores between 85 and 130 Using the previously calculated probabilities: - For \( X = 85 \) (Z = -1): \( P(Z < -1) \approx 0.1587 \) - For \( X = 130 \) (Z = 2): \( P(Z < 2) \approx 0.9772 \) To find the probability of scoring between 85 and 130: \[ P(85 < X < 130) = P(Z < 2) - P(Z < -1) \approx 0.9772 - 0.1587 = 0.8185 \] # Final Answer (a) The probability that a randomly selected individual scores below 85 is approximately **0.1587**. (b) The probability that a randomly selected individual scores above 130 is approximately **0.0228**. (c) The probability that a randomly selected individual scores between 85 and 130 is approximately **0.8185**.

# Given Information The processing time (in minutes) for a machine to complete a task is normally distributed with: - Mean (\( \mu \)) = 25 minutes - Standard deviation (\( \sigma \)) = 4 minutes # Definition or Concept Used To calculate the probabilities, we will use the properties of the **Normal Distribution** and the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also use the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to find cumulative probabilities. # Solution ### Part (a): Probability that a task takes less than 20 minutes First, we calculate the Z-score for \( X = 20 \): \[ Z = \frac{20 - 25}{4} = \frac{-5}{4} = -1.25 \] Using the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(-1.25, 0, 1, TRUE) \] This yields: \[ P(Z < -1.25) \approx 0.1056 \] ### Part (b): Probability that a task takes more than 32 minutes Next, we calculate the Z-score for \( X = 32 \): \[ Z = \frac{32 - 25}{4} = \frac{7}{4} = 1.75 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(1.75, 0, 1, TRUE) \] This yields: \[ P(Z < 1.75) \approx 0.9599 \] To find the probability of a task taking more than 32 minutes: \[ P(Z > 1.75) = 1 - P(Z < 1.75) = 1 - 0.9599 = 0.0401 \] ### Part (c): Probability that a task takes between 20 and 32 minutes We use the probabilities calculated for the two Z-scores: - For \( X = 20 \) (Z = -1.25): \( P(Z < -1.25) \approx 0.1056 \) - For \( X = 32 \) (Z = 1.75): \( P(Z < 1.75) \approx 0.9599 \) To find the probability of a task taking between 20 and 32 minutes: \[ P(20 < X < 32) = P(Z < 1.75) - P(Z < -1.25) \approx 0.9599 - 0.1056 = 0.8543 \] # Final Answer (a) The probability that a task takes less than 20 minutes is approximately **0.1056** (or **10.56%**). (b) The probability that a task takes more than 32 minutes is approximately **0.0401** (or **4.01%**). (c) The probability that a task takes between 20 and 32 minutes is approximately **0.8543** (or **85.43%**).

# Given Information The processing time (in minutes) for a machine to complete a task follows a normal distribution with: - Mean (\( \mu \)) = 25 minutes - Standard deviation (\( \sigma \)) = 4 minutes # Definition or Concept Used To find the probabilities, we will use the properties of the **Normal Distribution** along with the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the value we are interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also use the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to determine cumulative probabilities. # Solution ### Part (a): Probability that a task takes less than 20 minutes First, we calculate the Z-score for \( X = 20 \): \[ Z = \frac{20 - 25}{4} = \frac{-5}{4} = -1.25 \] Using the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(-1.25, 0, 1, TRUE) \] This gives: \[ P(Z < -1.25) \approx 0.1056 \] ### Part (b): Probability that a task takes more than 32 minutes Next, we calculate the Z-score for \( X = 32 \): \[ Z = \frac{32 - 25}{4} = \frac{7}{4} = 1.75 \] Using the Excel function: \[ \text{Probability} = =NORM.DIST(1.75, 0, 1, TRUE) \] This results in: \[ P(Z < 1.75) \approx 0.9599 \] To find the probability of a task taking more than 32 minutes: \[ P(Z > 1.75) = 1 - P(Z < 1.75) = 1 - 0.9599 = 0.0401 \] ### Part (c): Probability that a task takes between 20 and 32 minutes We will use the probabilities for the two Z-scores calculated earlier: - For \( X = 20 \) (Z = -1.25): \( P(Z < -1.25) \approx 0.1056 \) - For \( X = 32 \) (Z = 1.75): \( P(Z < 1.75) \approx 0.9599 \) To find the probability that a task takes between 20 and 32 minutes: \[ P(20 < X < 32) = P(Z < 1.75) - P(Z < -1.25) \approx 0.9599 - 0.1056 = 0.8543 \] # Final Answer (a) The probability that a task takes less than 20 minutes is approximately **0.1056** (or **10.56%**). (b) The probability that a task takes more than 32 minutes is approximately **0.0401** (or **4.01%**). (c) The probability that a task takes between 20 and 32 minutes is approximately **0.8543** (or **85.43%**).

# Given Information The processing time (in minutes) for a machine to complete a task is normally distributed with: - Mean (\( \mu \)) = 25 minutes - Standard deviation (\( \sigma \)) = 4 minutes # Definition or Concept Used To determine the required probabilities, we will utilize the **Normal Distribution** and calculate the Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the value we are assessing, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also use the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to find cumulative probabilities. # Solution ### Part (a): Probability that a task takes less than 20 minutes First, we compute the Z-score for \( X = 20 \): \[ Z = \frac{20 - 25}{4} = \frac{-5}{4} = -1.25 \] Next, we use the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(-1.25, 0, 1, TRUE) \] This results in: \[ P(Z < -1.25) \approx 0.1056 \] ### Part (b): Probability that a task takes more than 32 minutes Now, we calculate the Z-score for \( X = 32 \): \[ Z = \frac{32 - 25}{4} = \frac{7}{4} = 1.75 \] Using the Excel function to find the cumulative probability: \[ \text{Probability} = =NORM.DIST(1.75, 0, 1, TRUE) \] This yields: \[ P(Z < 1.75) \approx 0.9599 \] To find the probability of a task taking more than 32 minutes: \[ P(Z > 1.75) = 1 - P(Z < 1.75) = 1 - 0.9599 = 0.0401 \] ### Part (c): Probability that a task takes between 20 and 32 minutes Using the probabilities from the previous calculations: - For \( X = 20 \) (Z = -1.25): \( P(Z < -1.25) \approx 0.1056 \) - For \( X = 32 \) (Z = 1.75): \( P(Z < 1.75) \approx 0.9599 \) The probability that a task takes between 20 and 32 minutes is calculated as: \[ P(20 < X < 32) = P(Z < 1.75) - P(Z < -1.25) \approx 0.9599 - 0.1056 = 0.8543 \] # Final Answer (a) The probability that a task takes less than 20 minutes is approximately **0.1056** (or **10.56%**). (b) The probability that a task takes more than 32 minutes is approximately **0.0401** (or **4.01%**). (c) The probability that a task takes between 20 and 32 minutes is approximately **0.8543** (or **85.43%**).

# Given Information We have two discrete random variables: - \( X \): Number of units produced by a small machine during a fixed time interval, which can take values \( 0, 1, 2, 3, 4 \). - \( Y \): Indicates whether the machine overheats (0 for no overheating, 1 for overheating). The probabilities are provided as follows: - \( P(X=0) = 0.05 \) - \( P(X=1) = 0.15 \) - \( P(X=2) = 0.25 \) - \( P(X=3) = 0.30 \) - \( P(X=4) = 0.25 \) The conditional probability of overheating given \( X \) is defined as: \[ P(Y=1 | X=x) = \frac{x+1}{10} \] # Solution ### (a) Construct the full joint probability mass table for \( (X, Y) \) To find the joint probabilities, we apply the formula: \[ P(X=x, Y=y) = P(Y=y | X=x) P(X=x) \] For \( Y = 1 \): - \( P(X=0, Y=1) = P(Y=1 | X=0) P(X=0) = \frac{0+1}{10} \cdot 0.05 = 0.005 \) - \( P(X=1, Y=1) = P(Y=1 | X=1) P(X=1) = \frac{1+1}{10} \cdot 0.15 = 0.03 \) - \( P(X=2, Y=1) = P(Y=1 | X=2) P(X=2) = \frac{2+1}{10} \cdot 0.25 = 0.075 \) - \( P(X=3, Y=1) = P(Y=1 | X=3) P(X=3) = \frac{3+1}{10} \cdot 0.30 = 0.12 \) - \( P(X=4, Y=1) = P(Y=1 | X=4) P(X=4) = \frac{4+1}{10} \cdot 0.25 = 0.125 \) For \( Y = 0 \): - \( P(X=0, Y=0) = P(Y=0 | X=0) P(X=0) = \left(1 - \frac{0+1}{10}\right) \cdot 0.05 = 0.045 \) - \( P(X=1, Y=0) = P(Y=0 | X=1) P(X=1) = \left(1 - \frac{1+1}{10}\right) \cdot 0.15 = 0.12 \) - \( P(X=2, Y=0) = P(Y=0 | X=2) P(X=2) = \left(1 - \frac{2+1}{10}\right) \cdot 0.25 = 0.175 \) - \( P(X=3, Y=0) = P(Y=0 | X=3) P(X=3) = \left(1 - \frac{3+1}{10}\right) \cdot 0.30 = 0.18 \) - \( P(X=4, Y=0) = P(Y=0 | X=4) P(X=4) = \left(1 - \frac{4+1}{10}\right) \cdot 0.25 = 0.125 \) #### Joint Probability Mass Table | \( X \) | \( Y=0 \) | \( Y=1 \) | Total \( P(X) \) | |---------|-----------|-----------|-------------------| | 0 | 0.045 | 0.005 | 0.050 | | 1 | 0.120 | 0.030 | 0.150 | | 2 | 0.175 | 0.075 | 0.250 | | 3 | 0.180 | 0.120 | 0.300 | | 4 | 0.125 | 0.125 | 0.250 | | Total \( P(Y) \) | 0.645 | 0.355 | 1.000 | ### (b) Determine the marginal probability mass function of \( Y \) From the joint probability table, we can compute the marginal distribution of \( Y \): - \( P(Y=0) = 0.045 + 0.120 + 0.175 + 0.180 + 0.125 = 0.645 \) - \( P(Y=1) = 0.005 + 0.030 + 0.075 + 0.120 + 0.125 = 0.355 \) ### (c) Determine the conditional probability mass function \( P(X=x | Y=1) \) Using the joint probabilities: \[ P(X=x | Y=1) = \frac{P(X=x, Y=1)}{P(Y=1)} \] Calculating for each \( x \): - \( P(X=0 | Y=1) = \frac{0.005}{0.355} \approx 0.0141 \) - \( P(X=1 | Y=1) = \frac{0.030}{0.355} \approx 0.0845 \) - \( P(X=2 | Y=1) = \frac{0.075}{0.355} \approx 0.2113 \) - \( P(X=3 | Y=1) = \frac{0.120}{0.355} \approx 0.3387 \) - \( P(X=4 | Y=1) = \frac{0.125}{0.355} \approx 0.3521 \) ### (d) Determine whether \( X \) and \( Y \) are independent To determine independence, check if: \[ P(X=x, Y=y) = P(X=x) P(Y=y) \] for all combinations of \( X \) and \( Y \). For example, check \( P(X=0, Y=1) \): - \( P(X=0) P(Y=1) = 0.05 \times 0.355 = 0.01775 \) - \( P(X=0, Y=1) = 0.005 \) Since \( 0.005 \neq 0.01775 \), \( X \) and \( Y \) are not independent. ### (e) Compute the expected values \( E[X] \) and \( E[Y] \) \[ E[X] = \sum_{x=0}^{4} x P(X=x) = 0 \cdot 0.05 + 1 \cdot 0.15 + 2 \cdot 0.25 + 3 \cdot 0.30 + 4 \cdot 0.25 = 0 + 0.15 + 0.50 + 0.90 + 1.00 = 2.55 \] \[ E[Y] = \sum_{y=0}^{1} y P(Y=y) = 0 \cdot 0.645 + 1 \cdot 0.355 = 0 + 0.355 = 0.355 \] ### (f) Compute \( E[X^2] \) and \( Var(X) \) \[ E[X^2] = \sum_{x=0}^{4} x^2 P(X=x) = 0^2 \cdot 0.05 + 1^2 \cdot 0.15 + 2^2 \cdot 0.25 + 3^2 \cdot 0.30 + 4^2 \cdot 0.25 \] \[ = 0 + 0.15 + 1.00 + 2.70 + 4.00 = 7.85 \] Now, calculate \( Var(X) \): \[ Var(X) = E[X^2] - (E[X])^2 = 7.85 - (2.55)^2 = 7.85 - 6.5025 = 1.3475 \] ### (g) Compute \( Var(Y) \) Since \( Y \) is a Bernoulli random variable: \[ Var(Y) = E[Y](1 - E[Y]) = 0.355(1 - 0.355) = 0.355 \cdot 0.645 = 0.228975 \] ### (h) Compute \( E[XY] \) \[ E[XY] = \sum_{x=0}^{4} \sum_{y=0}^{1} xy P(X=x, Y=y) \] Calculating contributions: - For \( Y=0 \): Contributions are 0. - For \( Y=1 \): \[ 0 \cdot 0.005 + 1 \cdot 0.030 + 2 \cdot 0.075 + 3 \cdot 0.120 + 4 \cdot 0.125 = 0 + 0.030 + 0.150 + 0.360 + 0.500 = 1.040 \] ### (i) Compute the covariance \( Cov(X,Y) \) \[ Cov(X, Y) = E[XY] - E[X]E[Y] = 1.040 - (2.55)(0.355) = 1.040 - 0.90525 = 0.13475 \] ### (j) Compute the correlation coefficient between \( X \) and \( Y \) \[ \rho(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X) Var(Y)}} = \frac{0.13475}{\sqrt{1.3475 \cdot 0.228975}} \approx \frac{0.13475}{0.5483} \approx 0.245 \] ### Conclusion The relationship between \( X \) and \( Y \) is positive, as indicated by the positive correlation coefficient. This suggests that as the number of units produced increases, the likelihood of overheating also increases.

# Summary of Final Answers ### (a) Joint Probability Mass Table for \( (X, Y) \) | \( X \) | \( Y=0 \) | \( Y=1 \) | Total \( P(X) \) | |---------|-----------|-----------|-------------------| | 0 | 0.045 | 0.005 | 0.050 | | 1 | 0.120 | 0.030 | 0.150 | | 2 | 0.175 | 0.075 | 0.250 | | 3 | 0.180 | 0.120 | 0.300 | | 4 | 0.125 | 0.125 | 0.250 | | Total \( P(Y) \) | 0.645 | 0.355 | 1.000 | ### (b) Marginal Probability Mass Function of \( Y \) - \( P(Y=0) = 0.645 \) - \( P(Y=1) = 0.355 \) ### (c) Conditional Probability Mass Function \( P(X=x | Y=1) \) - \( P(X=0 | Y=1) \approx 0.0141 \) - \( P(X=1 | Y=1) \approx 0.0845 \) - \( P(X=2 | Y=1) \approx 0.2113 \) - \( P(X=3 | Y=1) \approx 0.3387 \) - \( P(X=4 | Y=1) \approx 0.3521 \) ### (d) Independence of \( X \) and \( Y \) - \( X \) and \( Y \) are **not independent** as \( P(X=x, Y=y) \neq P(X=x) P(Y=y) \) for all combinations. ### (e) Expected Values - \( E[X] = 2.55 \) - \( E[Y] = 0.355 \) ### (f) Compute \( E[X^2] \) and \( Var(X) \) - \( E[X^2] = 7.85 \) - \( Var(X) = 1.3475 \) ### (g) Variance of \( Y \) - \( Var(Y) = 0.228975 \) ### (h) Compute \( E[XY] \) - \( E[XY] = 1.040 \) ### (i) Compute Covariance \( Cov(X,Y) \) - \( Cov(X,Y) = 0.13475 \) ### (j) Compute Correlation Coefficient - \( \rho(X, Y) \approx 0.245 \) ### Interpretation The relationship between \( X \) and \( Y \) is positive, indicating that as the number of units produced increases, the likelihood of overheating also increases.

# Given Information The random variable \( X \) is normally distributed with: - Mean (\( \mu \)) = 45 - Standard deviation (\( \sigma \)) = 8 # Definition or Concept Used To find the probabilities, we will use the properties of the **Normal Distribution** and the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also use the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to find cumulative probabilities. # Solution ### Part (a): Find \( P(X < 30) \) First, we calculate the Z-score for \( X = 30 \): \[ Z = \frac{30 - 45}{8} = \frac{-15}{8} = -1.875 \] Using the Excel function to find the cumulative probability: \[ P(X < 30) = =NORM.DIST(-1.875, 0, 1, TRUE) \] This yields: \[ P(X < 30) \approx 0.0301 \] ### Part (b): Find \( P(X > 60) \) Next, we calculate the Z-score for \( X = 60 \): \[ Z = \frac{60 - 45}{8} = \frac{15}{8} = 1.875 \] Using the Excel function to find the cumulative probability: \[ P(X < 60) = =NORM.DIST(1.875, 0, 1, TRUE) \] This yields: \[ P(X < 60) \approx 0.9699 \] To find the probability of \( X > 60 \): \[ P(X > 60) = 1 - P(X < 60) = 1 - 0.9699 = 0.0301 \] ### Part (c): Find \( P(30 < X < 60) \) We can use the previously calculated probabilities: \[ P(30 < X < 60) = P(X < 60) - P(X < 30) \] \[ P(30 < X < 60) = 0.9699 - 0.0301 = 0.9398 \] # Final Answer (a) The probability that \( P(X < 30) \) is approximately **0.0301**. (b) The probability that \( P(X > 60) \) is approximately **0.0301**. (c) The probability that \( P(30 < X < 60) \) is approximately **0.9398**.

# Given Information We have two variables \( X \) and \( Y \) with the following data points: - \( X = [36, 12, 45, 30, 3, 45, 57, 21, 54, 30, 15] \) - \( Y = [12, 4, 15, 10, 1, 15, 19, 7, 18, 10, 5] \) The sample size is \( n = 11 \). # Definitions 1. **Sample Mean:** The sample mean for \( X \) and \( Y \) is calculated as: \[ \bar{X} = \frac{\sum X_i}{n}, \quad \bar{Y} = \frac{\sum Y_i}{n} \] 2. **Sample Covariance:** The sample covariance is defined as: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] # Solution ### Step 1: Calculate the Sample Means Calculating \( \bar{X} \): \[ \bar{X} = \frac{36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15}{11} = \frac{ 36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15}{11} = \frac{ 2 0 4 }{ 11 } = 27.63636 \approx 27.64 \] Calculating \( \bar{Y} \): \[ \bar{Y} = \frac{12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5}{11} = \frac{ 12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5}{ 11 } = \frac{ 1 1 6 }{ 11 } = 10.54545 \approx 10.55 \] ### Step 2: Calculate the Sample Covariance We will calculate \( S_{XY} \) using the formula given above. First, compute the products \( (X_i - \bar{X})(Y_i - \bar{Y}) \) for each pair. Calculating each term: - \( (36 - 27.64)(12 - 10.55) = (8.36)(1.45) \approx 12.122 \) - \( (12 - 27.64)(4 - 10.55) = (-15.64)(-6.55) \approx 102.062 \) - \( (45 - 27.64)(15 - 10.55) = (17.36)(4.45) \approx 77.182 \) - \( (30 - 27.64)(10 - 10.55) = (2.36)(-0.55) \approx -1.298 \) - \( (3 - 27.64)(1 - 10.55) = (-24.64)(-9.55) \approx 235.992 \) - \( (45 - 27.64)(15 - 10.55) = (17.36)(4.45) \approx 77.182 \) - \( (57 - 27.64)(19 - 10.55) = (29.36)(8.45) \approx 248.882 \) - \( (21 - 27.64)(7 - 10.55) = (-6.64)(-3.55) \approx 23.532 \) - \( (54 - 27.64)(18 - 10.55) = (26.36)(7.45) \approx 196.882 \) - \( (30 - 27.64)(10 - 10.55) = (2.36)(-0.55) \approx -1.298 \) - \( (15 - 27.64)(5 - 10.55) = (-12.64)(-5.55) \approx 70.042 \) Now we sum these products: \[ \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) = 12.122 + 102.062 + 77.182 - 1.298 + 235.992 + 77.182 + 248.882 + 23.532 + 196.882 - 1.298 + 70.042 \approx 823.528 \] Finally, calculate the sample covariance: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) = \frac{823.528}{10} \approx 82.3528 \approx 82.353 \] # Final Answer The sample covariance \( S_{XY} \) is approximately **82.353** (rounded to three decimal places).

# Given Information We have two variables \( X \) and \( Y \) with the following data points: - \( X = [36, 12, 45, 30, 3, 45, 57, 21, 54, 30, 15] \) - \( Y = [12, 4, 15, 10, 1, 15, 19, 7, 18, 10, 5] \) The sample size is \( n = 11 \). # Definitions 1. **Sample Mean:** The sample mean for \( X \) and \( Y \) is calculated as: \[ \bar{X} = \frac{\sum X_i}{n}, \quad \bar{Y} = \frac{\sum Y_i}{n} \] 2. **Sample Covariance:** The sample covariance is defined as: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] # Solution ### Step 1: Calculate the Sample Means Calculating \( \bar{X} \): \[ \bar{X} = \frac{36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15}{11} = \frac{ 273 }{ 11 } \approx 24.8182 \] Calculating \( \bar{Y} \): \[ \bar{Y} = \frac{12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5}{11} = \frac{ 105 }{ 11 } \approx 9.5455 \] ### Step 2: Calculate the Sample Covariance Now we will compute \( S_{XY} \) using the formula: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] Calculating each term: - For \( i = 1 \): \( (36 - 24.8182)(12 - 9.5455) = (11.1818)(2.4545) \approx 27.4137 \) - For \( i = 2 \): \( (12 - 24.8182)(4 - 9.5455) = (-12.8182)(-5.5455) \approx 71.1782 \) - For \( i = 3 \): \( (45 - 24.8182)(15 - 9.5455) = (20.1818)(5.4545) \approx 109.9629 \) - For \( i = 4 \): \( (30 - 24.8182)(10 - 9.5455) = (5.1818)(0.4545) \approx 2.3586 \) - For \( i = 5 \): \( (3 - 24.8182)(1 - 9.5455) = (-21.8182)(-8.5455) \approx 186.6291 \) - For \( i = 6 \): \( (45 - 24.8182)(15 - 9.5455) = (20.1818)(5.4545) \approx 109.9629 \) - For \( i = 7 \): \( (57 - 24.8182)(19 - 9.5455) = (32.1818)(9.4545) \approx 304.592 \) - For \( i = 8 \): \( (21 - 24.8182)(7 - 9.5455) = (-3.8182)(-2.5455) \approx 9.7386 \) - For \( i = 9 \): \( (54 - 24.8182)(18 - 9.5455) = (29.1818)(8.4545) \approx 246.7055 \) - For \( i = 10 \): \( (30 - 24.8182)(10 - 9.5455) = (5.1818)(0.4545) \approx 2.3586 \) - For \( i = 11 \): \( (15 - 24.8182)(5 - 9.5455) = (-9.8182)(-4.5455) \approx 44.6932 \) Now we sum these products: \[ \sum_{i=1}^{11} (X_i - \bar{X})(Y_i - \bar{Y}) = 27.4137 + 71.1782 + 109.9629 + 2.3586 + 186.6291 + 109.9629 + 304.592 + 9.7386 + 246.7055 + 2.3586 + 44.6932 \approx 455.8517 \] Finally, calculate the sample covariance: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) = \frac{455.8517}{10} \approx 45.5852 \] # Final Answer The sample covariance \( S_{XY} \) is approximately **45.585** (rounded to three decimal places).

# Given Information We have two variables \( X \) and \( Y \) with the following data points: - \( X = [36, 12, 45, 30, 3, 45, 57, 21, 54, 30, 15] \) - \( Y = [12, 4, 15, 10, 1, 15, 19, 7, 18, 10, 5] \) The sample size is \( n = 11 \). # Definitions 1. **Sample Mean:** The sample mean for \( X \) and \( Y \) is calculated as: \[ \bar{X} = \frac{\sum X_i}{n}, \quad \bar{Y} = \frac{\sum Y_i}{n} \] 2. **Sample Covariance:** The sample covariance is defined as: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] # Solution ### Step 1: Calculate the Sample Means #### For \( X \): \[ \bar{X} = \frac{36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15}{11} \] Calculating the sum: \[ \sum X = 36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15 = 36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15 = 36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15 = 369 \] Calculating the mean: \[ \bar{X} = \frac{369}{11} \approx 33.5454 \] #### For \( Y \): \[ \bar{Y} = \frac{12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5}{11} \] Calculating the sum: \[ \sum Y = 12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5 = 12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5 = 12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5 = 105 \] Calculating the mean: \[ \bar{Y} = \frac{105}{11} \approx 9.5455 \] ### Step 2: Calculate the Sample Covariance Now we will compute \( S_{XY} \) using the formula: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] Calculating each term: 1. For \( i = 1 \): \((36 - 33.5454)(12 - 9.5455) = (2.4546)(2.4545) \approx 6.034 \) 2. For \( i = 2 \): \((12 - 33.5454)(4 - 9.5455) = (-21.5454)(-5.5455) \approx 119.439 \) 3. For \( i = 3 \): \((45 - 33.5454)(15 - 9.5455) = (11.4546)(5.4545) \approx 62.388 \) 4. For \( i = 4 \): \((30 - 33.5454)(10 - 9.5455) = (-3.5454)(0.4545) \approx -1.608 \) 5. For \( i = 5 \): \((3 - 33.5454)(1 - 9.5455) = (-30.5454)(-8.5455) \approx 261.501 \) 6. For \( i = 6 \): \((45 - 33.5454)(15 - 9.5455) = (11.4546)(5.4545) \approx 62.388 \) 7. For \( i = 7 \): \((57 - 33.5454)(19 - 9.5455) = (23.4546)(9.4545) \approx 221.985 \) 8. For \( i = 8 \): \((21 - 33.5454)(7 - 9.5455) = (-12.5454)(-2.5455) \approx 31.993 \) 9. For \( i = 9 \): \((54 - 33.5454)(18 - 9.5455) = (20.4546)(8.4545) \approx 173.719 \) 10. For \( i = 10 \): \((30 - 33.5454)(10 - 9.5455) = (-3.5454)(0.4545) \approx -1.608 \) 11. For \( i = 11 \): \((15 - 33.5454)(5 - 9.5455) = (-18.5454)(-4.5455) \approx 84.319 \) Now we sum these products: \[ \sum_{i=1}^{11} (X_i - \bar{X})(Y_i - \bar{Y}) = 6.034 + 119.439 + 62.388 - 1.608 + 261.501 + 62.388 + 221.985 + 31.993 + 173.719 - 1.608 + 84.319 \approx 823.637 \] Finally, calculate the sample covariance: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) = \frac{823.637}{10} \approx 82.3637 \] # Final Answer The sample covariance \( S_{XY} \) is approximately **82.364** (rounded to three decimal places).

# Given Information We have two variables \( X \) and \( Y \) with the following data points: - \( X = [36, 12, 45, 30, 3, 45, 57, 21, 54, 30, 15] \) - \( Y = [12, 4, 15, 10, 1, 15, 19, 7, 18, 10, 5] \) The sample size is \( n = 11 \). # Definitions 1. **Sample Mean:** The sample mean for \( X \) and \( Y \) is calculated as: \[ \bar{X} = \frac{\sum X}{n}, \quad \bar{Y} = \frac{\sum Y}{n} \] 2. **Sample Covariance:** The sample covariance is defined as: \[ S_{XY} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y}) \] # Solution ### Step 1: Calculate the Sample Means #### For \( X \): \[ \sum X = 36 + 12 + 45 + 30 + 3 + 45 + 57 + 21 + 54 + 30 + 15 = 348 \] Calculating the mean: \[ \bar{X} = \frac{348}{11} \approx 31.636 \] #### For \( Y \): \[ \sum Y = 12 + 4 + 15 + 10 + 1 + 15 + 19 + 7 + 18 + 10 + 5 = 116 \] Calculating the mean: \[ \bar{Y} = \frac{116}{11} \approx 10.545 \] ### Step 2: Create Deviation Table We will calculate \( (X_i - \bar{X}) \), \( (Y_i - \bar{Y}) \), and their product for each data point. | \( X_i \) | \( Y_i \) | \( X_i - \bar{X} \) | \( Y_i - \bar{Y} \) | Product \( (X_i - \bar{X})(Y_i - \bar{Y}) \) | |-----------|-----------|----------------------|----------------------|-----------------------------------------------| | 36 | 12 | \( 36 - 31.636 = 4.364 \) | \( 12 - 10.545 = 1.455 \) | \( 4.364 \times 1.455 \approx 6.347 \) | | 12 | 4 | \( 12 - 31.636 = -19.636 \) | \( 4 - 10.545 = -6.545 \) | \( -19.636 \times -6.545 \approx 128.518 \) | | 45 | 15 | \( 45 - 31.636 = 13.364 \) | \( 15 - 10.545 = 4.455 \) | \( 13.364 \times 4.455 \approx 59.540 \) | | 30 | 10 | \( 30 - 31.636 = -1.636 \) | \( 10 - 10.545 = -0.545 \) | \( -1.636 \times -0.545 \approx 0.892 \) | | 3 | 1 | \( 3 - 31.636 = -28.636 \) | \( 1 - 10.545 = -9.545 \) | \( -28.636 \times -9.545 \approx 273.350 \) | | 45 | 15 | \( 45 - 31.636 = 13.364 \) | \( 15 - 10.545 = 4.455 \) | \( 13.364 \times 4.455 \approx 59.540 \) | | 57 | 19 | \( 57 - 31.636 = 25.364 \) | \( 19 - 10.545 = 8.455 \) | \( 25.364 \times 8.455 \approx 214.409 \) | | 21 | 7 | \( 21 - 31.636 = -10.636 \) | \( 7 - 10.545 = -3.545 \) | \( -10.636 \times -3.545 \approx 37.702 \) | | 54 | 18 | \( 54 - 31.636 = 22.364 \) | \( 18 - 10.545 = 7.455 \) | \( 22.364 \times 7.455 \approx 166.665 \) | | 30 | 10 | \( 30 - 31.636 = -1.636 \) | \( 10 - 10.545 = -0.545 \) | \( -1.636 \times -0.545 \approx 0.892 \) | | 15 | 5 | \( 15 - 31.636 = -16.636 \) | \( 5 - 10.545 = -5.545 \) | \( -16.636 \times -5.545 \approx 92.164 \) | ### Step 3: Sum of Products Calculating the sum of the products: \[ \sum (X_i - \bar{X})(Y_i - \bar{Y}) = 6.347 + 128.518 + 59.540 + 0.892 + 273.350 + 59.540 + 214.409 + 37.702 + 166.665 + 0.892 + 92.164 \approx 1040.182 \] ### Step 4: Compute the Sample Covariance Now, calculate the sample covariance: \[ S_{XY} = \frac{1}{n-1} \sum (X_i - \bar{X})(Y_i - \bar{Y}) = \frac{1040.182}{10} = 104.0182 \] # Final Answer The sample covariance \( S_{XY} \) is approximately **104.018** (rounded to three decimal places).

# Given Information The random variable \( X \) is normally distributed with: - Mean (\( \mu \)) = 45 - Standard deviation (\( \sigma \)) = 8 # Definition or Concept Used To find the probabilities, we will use the properties of the **Normal Distribution** and the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will also use the Excel function `=NORM.DIST(z, 0, 1, TRUE)` to find cumulative probabilities. # Solution ### Part (a): Find \( P(X < 30) \) First, we calculate the Z-score for \( X = 30 \): \[ Z = \frac{30 - 45}{8} = \frac{-15}{8} = -1.875 \] Using the Excel function to find the cumulative probability: \[ P(X < 30) = =NORM.DIST(-1.875, 0, 1, TRUE) \] This yields: \[ P(X < 30) \approx 0.0301 \] ### Part (b): Find \( P(X > 60) \) Next, we calculate the Z-score for \( X = 60 \): \[ Z = \frac{60 - 45}{8} = \frac{15}{8} = 1.875 \] Using the Excel function to find the cumulative probability: \[ P(X < 60) = =NORM.DIST(1.875, 0, 1, TRUE) \] This yields: \[ P(X < 60) \approx 0.9699 \] To find the probability of \( X > 60 \): \[ P(X > 60) = 1 - P(X < 60) = 1 - 0.9699 = 0.0301 \] ### Part (c): Find \( P(30 < X < 60) \) We can use the previously calculated probabilities: \[ P(30 < X < 60) = P(X < 60) - P(X < 30) \] \[ P(30 < X < 60) = 0.9699 - 0.0301 = 0.9398 \] # Final Answer (a) The probability that \( P(X < 30) \) is approximately **0.0301**. (b) The probability that \( P(X > 60) \) is approximately **0.0301**. (c) The probability that \( P(30 < X < 60) \) is approximately **0.9398**.