# Given Information - The probability density function (PDF) of a random variable \( X \) is: \[ f(x) = \begin{cases} kx, & \leq x \leq 2 \\ , & \text{otherwise} \end{cases} \] # What To Find - The value of \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used - The total area under the PDF over its domain must be 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] - The probability that \( X \) falls within an interval \([a, b]\) is: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate \( f(x) \) over its domain and set the result to 1: \[ \int_^2 kx \, dx = 1 \] \[ k \int_^2 x \, dx = 1 \] \[ k \left[ \frac{x^2}{2} \right]_^2 = 1 \] \[ k \left( \frac{2^2}{2} - \frac{^2}{2} \right) = 1 \] \[ k \left( \frac{4}{2} \right) = 1 \] \[ k \cdot 2 = 1 \] \[ k = \frac{1}{2} \] To find \( P(1 \leq x \leq 2) \), integrate the PDF over \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substitute \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2}x \, dx \] \[ = \frac{1}{2} \int_1^2 x \, dx \] \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is: \[ f(x) = \begin{cases} k(2 - x), & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of \( k \). - The probability \( P(0 \leq x \leq 1) \). # Definition or Concept Used - The total area under the PDF over its domain must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] - The probability that \( X \) falls within an interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate \( f(x) \) over its defined domain and set the result to 1: \[ \int_0^2 k(2 - x) \, dx = 1 \] \[ k \int_0^2 (2 - x) \, dx = 1 \] Calculating the integral: \[ = k \left[ 2x - \frac{x^2}{2} \right]_0^2 \] \[ = k \left( 2(2) - \frac{2^2}{2} - (0) \right) \] \[ = k \left( 4 - 2 \right) \] \[ = k \cdot 2 = 1 \] Solving for \( k \): \[ k = \frac{1}{2} \] Next, to find \( P(0 \leq x \leq 1) \), integrate the PDF over \([0, 1]\): \[ P(0 \leq x \leq 1) = \int_0^1 k(2 - x) \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_0^1 \frac{1}{2}(2 - x) \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ 2x - \frac{x^2}{2} \right]_0^1 \] \[ = \frac{1}{2} \left( 2(1) - \frac{1^2}{2} - (0) \right) \] \[ = \frac{1}{2} \left( 2 - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(0 \leq x \leq 1) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} k(3 - x), & 0 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - Determine the value of \( k \). - Calculate the probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used - The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] - The probability that \( X \) lies within the interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution Begin by determining \( k \) by integrating \( f(x) \) over its valid range and setting the result equal to 1: \[ \int_0^3 k(3 - x) \, dx = 1 \] This leads to: \[ k \int_0^3 (3 - x) \, dx = 1 \] Calculating the integral: \[ = k \left[ 3x - \frac{x^2}{2} \right]_0^3 \] Evaluating the boundaries: \[ = k \left( 3(3) - \frac{3^2}{2} - (0) \right) \] \[ = k \left( 9 - \frac{9}{2} \right) \] \[ = k \left( \frac{18}{2} - \frac{9}{2} \right) = k \cdot \frac{9}{2} \] Setting this equal to 1 gives: \[ k \cdot \frac{9}{2} = 1 \] Solving for \( k \): \[ k = \frac{2}{9} \] Next, to find \( P(1 \leq x \leq 2) \), integrate the PDF over the interval \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 k(3 - x) \, dx \] Substituting \( k = \frac{2}{9} \): \[ = \int_1^2 \frac{2}{9}(3 - x) \, dx \] Calculating the integral: \[ = \frac{2}{9} \left[ 3x - \frac{x^2}{2} \right]_1^2 \] Evaluating at the boundaries: \[ = \frac{2}{9} \left( 3(2) - \frac{2^2}{2} - (3(1) - \frac{1^2}{2}) \right) \] \[ = \frac{2}{9} \left( 6 - 2 - (3 - \frac{1}{2}) \right) \] \[ = \frac{2}{9} \left( 4 - 2.5 \right) \] \[ = \frac{2}{9} \cdot 1.5 \] \[ = \frac{3}{9} = \frac{1}{3} \] # Summary - Value of \( k \): \( \frac{2}{9} \) - \( P(1 \leq x \leq 2) = \frac{1}{3} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} k(3 - x), & 0 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - Determine the value of \( k \). - Calculate the probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within the interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate \( f(x) \) over its valid range and set the result equal to 1: \[ \int_0^3 k(3 - x) \, dx = 1 \] This simplifies to: \[ k \int_0^3 (3 - x) \, dx = 1 \] Calculating the integral: \[ = k \left[ 3x - \frac{x^2}{2} \right]_0^3 \] Evaluating the boundaries: \[ = k \left( 3(3) - \frac{3^2}{2} \right) \] \[ = k \left( 9 - \frac{9}{2} \right) = k \left( \frac{18}{2} - \frac{9}{2} \right) = k \cdot \frac{9}{2} \] Setting this equal to 1 gives: \[ k \cdot \frac{9}{2} = 1 \] Solving for \( k \): \[ k = \frac{2}{9} \] Now, to find \( P(1 \leq x \leq 2) \), integrate the PDF over the interval \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 k(3 - x) \, dx \] Substituting \( k = \frac{2}{9} \): \[ = \int_1^2 \frac{2}{9}(3 - x) \, dx \] Calculating the integral: \[ = \frac{2}{9} \left[ 3x - \frac{x^2}{2} \right]_1^2 \] Evaluating at the boundaries: \[ = \frac{2}{9} \left( \left( 3(2) - \frac{2^2}{2} \right) - \left( 3(1) - \frac{1^2}{2} \right) \right) \] \[ = \frac{2}{9} \left( 6 - 2 - (3 - 0.5) \right) \] \[ = \frac{2}{9} \left( 4 - 2.5 \right) \] \[ = \frac{2}{9} \cdot 1.5 = \frac{3}{9} = \frac{1}{3} \] # Summary - Value of \( k \): \( \frac{2}{9} \) - \( P(1 \leq x \leq 2) = \frac{1}{3} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} kx, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within an interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To determine \( k \), integrate \( f(x) \) over its valid domain and set the result equal to 1: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral: \[ k \int_0^2 x \, dx = 1 \] This leads to: \[ k \left[ \frac{x^2}{2} \right]_0^2 = 1 \] Evaluating the boundaries: \[ k \left( \frac{2^2}{2} - 0 \right) = 1 \] \[ k \cdot 2 = 1 \] Solving for \( k \): \[ k = \frac{1}{2} \] Next, to find \( P(1 \leq x \leq 2) \), integrate the PDF over the interval \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the boundaries: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is expressed as: \[ f(x) = \begin{cases} kx, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of the constant \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The requirement that the total area under the PDF is equal to 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) falls within the interval \([a, b]\) can be computed as: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate the PDF over its specified range and set the area equal to 1: \[ \int_0^2 kx \, dx = 1 \] This expands to: \[ k \int_0^2 x \, dx = 1 \] Calculating the integral yields: \[ k \left[ \frac{x^2}{2} \right]_0^2 = 1 \] Substituting the limits gives: \[ k \left( \frac{2^2}{2} - 0 \right) = 1 \] \[ k \cdot 2 = 1 \] Solving for \( k \) results in: \[ k = \frac{1}{2} \] Next, to determine \( P(1 \leq x \leq 2) \), integrate the PDF over the interval from 1 to 2: \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting in the value of \( k \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating this integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the limits: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is described as: \[ f(x) = \begin{cases} kx, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of the constant \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within an interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find the value of \( k \), integrate the PDF over its valid range and equate the result to 1: \[ \int_0^2 kx \, dx = 1 \] This simplifies to: \[ k \int_0^2 x \, dx = 1 \] Calculating the integral: \[ = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the integral gives: \[ = k \left( \frac{2^2}{2} - 0 \right) = k \cdot 2 \] Setting this equal to 1 results in: \[ k \cdot 2 = 1 \] Thus, solving for \( k \) yields: \[ k = \frac{1}{2} \] Next, to calculate \( P(1 \leq x \leq 2) \), we integrate the PDF over the specified interval: \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the integral at the boundaries: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} \] Thus, we have: \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} kx, & \text{if } 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of the constant \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The total area under the PDF must be equal to 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within the interval \([a, b]\) is expressed as: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate the PDF across its valid range and set the resulting area equal to 1: \[ \int_0^2 kx \, dx = 1 \] This expands to: \[ k \int_0^2 x \, dx = 1 \] Calculating the integral: \[ = k \left[ \frac{x^2}{2} \right]_0^2 \] Substituting the limits gives: \[ = k \left( \frac{2^2}{2} - 0 \right) = k \cdot 2 \] Setting this equal to 1 results in: \[ k \cdot 2 = 1 \] Thus, solving for \( k \) provides: \[ k = \frac{1}{2} \] Next, to compute \( P(1 \leq x \leq 2) \), we integrate the PDF over the range from 1 to 2: \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating this integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the limits: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] This simplifies to: \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} \] Thus, we find: \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - **Machine A**: - Produces: \(30\%\) - Defect Rate: \(2\%\) - **Machine B**: - Produces: \(45\%\) - Defect Rate: \(3\%\) - **Machine C**: - Produces: \(25\%\) - Defect Rate: \(5\%\) # What To Find - The probability that a randomly selected defective product was produced by Machine C, denoted as \( P(C | D) \). # Definition or Concept Used - **Bayes' Theorem**: \[ P(C | D) = \frac{P(D | C) \cdot P(C)}{P(D)} \] - Where: - \( P(D | C) \) = Probability of a product being defective given it was produced by Machine C. - \( P(C) \) = Probability that a product is produced by Machine C. - \( P(D) \) = Total probability of a product being defective. # Solution Calculate the probabilities needed for Bayes' Theorem: 1. **Probability of Machine C**: \[ P(C) = 0.25 \] 2. **Probability of defect given Machine C**: \[ P(D | C) = 0.05 \] 3. **Total probability of a defect \( P(D) \)**: Using the law of total probability: \[ P(D) = P(D | A) \cdot P(A) + P(D | B) \cdot P(B) + P(D | C) \cdot P(C) \] - For Machine A: \[ P(D | A) = 0.02, \quad P(A) = 0.30 \] Therefore, \[ P(D | A) \cdot P(A) = 0.02 \cdot 0.30 = 0.006 \] - For Machine B: \[ P(D | B) = 0.03, \quad P(B) = 0.45 \] Therefore, \[ P(D | B) \cdot P(B) = 0.03 \cdot 0.45 = 0.0135 \] - For Machine C: \[ P(D | C) \cdot P(C) = 0.05 \cdot 0.25 = 0.0125 \] Now, summing these probabilities: \[ P(D) = 0.006 + 0.0135 + 0.0125 = 0.032 \] 4. **Now apply Bayes' Theorem**: \[ P(C | D) = \frac{P(D | C) \cdot P(C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] \[ = \frac{0.0125}{0.032} \approx 0.390625 \] # Summary - The probability that a randomly selected defective product was produced by Machine C is approximately \( 0.391 \) or \( 39.1\% \).

# Given Information - **Machine A**: - Production Share: \(30\%\) - Defect Rate: \(2\%\) - **Machine B**: - Production Share: \(45\%\) - Defect Rate: \(3\%\) - **Machine C**: - Production Share: \(25\%\) - Defect Rate: \(5\%\) # What To Find - Determine the probability that a defective product was manufactured by Machine C, represented as \( P(C | D) \). # Definition or Concept Used - **Bayes' Theorem**: \[ P(C | D) = \frac{P(D | C) \cdot P(C)}{P(D)} \] - Where: - \( P(D | C) \) = Probability of a product being defective given it was produced by Machine C. - \( P(C) \) = Probability that a product is produced by Machine C. - \( P(D) \) = Total probability of a product being defective. # Solution To calculate \( P(C | D) \), we first need the individual probabilities: 1. **Probability of Machine C**: \[ P(C) = 0.25 \] 2. **Probability of defect given Machine C**: \[ P(D | C) = 0.05 \] 3. **Calculate Total Probability of Defect \( P(D) \)**: Using the law of total probability: \[ P(D) = P(D | A) \cdot P(A) + P(D | B) \cdot P(B) + P(D | C) \cdot P(C) \] - For Machine A: \[ P(D | A) = 0.02, \quad P(A) = 0.30 \] Therefore, \[ P(D | A) \cdot P(A) = 0.02 \cdot 0.30 = 0.006 \] - For Machine B: \[ P(D | B) = 0.03, \quad P(B) = 0.45 \] Therefore, \[ P(D | B) \cdot P(B) = 0.03 \cdot 0.45 = 0.0135 \] - For Machine C: \[ P(D | C) \cdot P(C) = 0.05 \cdot 0.25 = 0.0125 \] Now summing these contributions: \[ P(D) = 0.006 + 0.0135 + 0.0125 = 0.032 \] 4. **Utilize Bayes' Theorem**: \[ P(C | D) = \frac{P(D | C) \cdot P(C)}{P(D)} = \frac{0.05 \cdot 0.25}{0.032} \] \[ = \frac{0.0125}{0.032} \approx 0.390625 \] # Summary - The probability that a randomly selected defective product originates from Machine C is approximately \( 0.391 \) or \( 39.1\% \).

# Given Information - The duration of customer support calls, \( T \), is modeled as an exponential distribution with: - Mean duration: \( \mu = 6 \) minutes - Rate parameter: \[ \lambda = \frac{1}{\mu} = \frac{1}{6} \text{ minutes}^{-1} \] # What To Find 1. The probability that a randomly selected call lasts more than 10 minutes: \( P(T > 10) \). 2. The probability that a randomly selected call is resolved in less than 2 minutes: \( P(T < 2) \). 3. The probability that a call that has lasted 4 minutes exceeds 12 minutes: \( P(T > 12 | T > 4) \). 4. The probability that the total duration of three independent calls exceeds 20 minutes: \( P(S > 20) \). # Definition or Concept Used 1. For an exponential distribution: - The cumulative distribution function (CDF) is given by: \[ P(T \leq t) = 1 - e^{-\lambda t} \] - The complementary probability is: \[ P(T > t) = e^{-\lambda t} \] 2. For the sum of independent exponential random variables: - If \( S \) is the sum of \( n \) independent exponential random variables with rate \( \lambda \), then \( S \) follows a gamma distribution with shape parameter \( n \) and rate parameter \( \lambda \): \[ S \sim \Gamma(n, \lambda) \] - The probability density function for \( S \) is: \[ P(S > s) = 1 - \text{CDF}(s) = 1 - \left(1 - e^{-\lambda s} \sum_{k=0}^{n-1} \frac{(\lambda s)^k}{k!}\right) \] # Solution 1. **Probability that a call lasts more than 10 minutes**: \[ P(T > 10) = e^{-\lambda \cdot 10} = e^{-\frac{10}{6}} \approx e^{-1.6667} \approx 0.1889 \] 2. **Probability that a call is resolved in less than 2 minutes**: \[ P(T < 2) = 1 - P(T > 2) = 1 - e^{-\lambda \cdot 2} = 1 - e^{-\frac{2}{6}} \approx 1 - e^{-0.3333} \approx 0.2546 \] 3. **Probability that a call lasting 4 minutes exceeds 12 minutes**: Using the memoryless property of the exponential distribution: \[ P(T > 12 | T > 4) = P(T > 12 - 4) = P(T > 8) = e^{-\lambda \cdot 8} = e^{-\frac{8}{6}} \approx e^{-1.3333} \approx 0.2636 \] 4. **Probability that the total duration of three independent calls exceeds 20 minutes**: For \( S \sim \Gamma(3, \frac{1}{6}) \): \[ P(S > 20) = 1 - P(S \leq 20) \] First, calculate \( P(S \leq 20) \): \[ P(S \leq 20) = 1 - e^{-\frac{20}{6}} \left( 1 + \frac{20}{6} + \frac{(20/6)^2}{2} \right) \] Calculate: \[ = 1 - e^{-3.3333} \left( 1 + 3.3333 + \frac{(3.3333)^2}{2} \right) \] \[ \approx 1 - e^{-3.3333} \left( 1 + 3.3333 + 5.5556 \right) \] \[ \approx 1 - e^{-3.3333} \cdot 10.8889 \approx 1 - 0.0355 \cdot 10.8889 \approx 1 - 0.3867 \approx 0.6133 \] Therefore: \[ P(S > 20) \approx 1 - 0.6133 \approx 0.3867 \] # Summary - Probability that a call lasts more than 10 minutes: \( P(T > 10) \approx 0.189 \) - Probability that a call is resolved in less than 2 minutes: \( P(T < 2) \approx 0.255 \) - Probability that a call lasting 4 minutes exceeds 12 minutes: \( P(T > 12 | T > 4) \approx 0.264 \) - Probability that the total duration of three calls exceeds 20 minutes: \( P(S > 20) \approx 0.387 \)

# Given Information - Delivery time \( X \) for parcels is modeled as a normal distribution with: - Mean (\( \mu \)): \( 18 \) hours - Standard deviation (\( \sigma \)): \( 3 \) hours # What To Find 1. The probability that a randomly selected parcel experiences a delayed delivery (\( P(X > 23) \)). 2. The probability that a randomly selected parcel arrives in less than \( 14 \) hours (\( P(X < 14) \)). 3. The probability that the average delivery time of \( 25 \) selected parcels exceeds \( 19 \) hours (\( P(\bar{X} > 19) \)). # Definition or Concept Used 1. **Standardization of a Normal Variable**: \[ Z = \frac{X - \mu}{\sigma} \] Where \( Z \) follows a standard normal distribution \( N(0, 1) \). 2. **Sampling Distribution of the Mean**: - The mean of the sampling distribution of \( \bar{X} \) is equal to the population mean, \( \mu \). - The standard deviation of the sampling distribution (standard error) is given by: \[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \] # Solution 1. **Probability of a delayed delivery** (\( P(X > 23) \)): - Standardize \( X \): \[ Z = \frac{23 - 18}{3} = \frac{5}{3} \approx 1.6667 \] - Use the standard normal distribution table or calculator to find \( P(Z > 1.6667) \): \[ P(X > 23) = 1 - P(Z \leq 1.6667) \approx 1 - 0.9525 = 0.0475 \] 2. **Probability of arriving in less than 14 hours** (\( P(X < 14) \)): - Standardize \( X \): \[ Z = \frac{14 - 18}{3} = \frac{-4}{3} \approx -1.3333 \] - Find \( P(Z < -1.3333) \): \[ P(X < 14) \approx 0.0918 \] 3. **Probability that the average delivery time of \( 25 \) parcels exceeds \( 19 \) hours** (\( P(\bar{X} > 19) \)): - Calculate the standard error: \[ \sigma_{\bar{X}} = \frac{3}{\sqrt{25}} = \frac{3}{5} = 0.6 \] - Standardize \( \bar{X} \): \[ Z = \frac{19 - 18}{0.6} = \frac{1}{0.6} \approx 1.6667 \] - Find \( P(Z > 1.6667) \): \[ P(\bar{X} > 19) = 1 - P(Z \leq 1.6667) \approx 1 - 0.9525 = 0.0475 \] # Summary - Probability of delayed delivery (\( P(X > 23) \)): \( \approx 0.0475 \) - Probability of arriving in less than 14 hours (\( P(X < 14) \)): \( \approx 0.0918 \) - Probability that the average delivery time exceeds 19 hours (\( P(\bar{X} > 19) \)): \( \approx 0.0475 \)