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a. Which of the probability density functions of waiting time is applicable at Kroger? 1 a. f(x) = е 26 for a 20 %3D 26 1 b. f(x) for a >0 e e 26 - 26 1 c. f(x) = 1 e 26 for a <0 26 - 1 d. f(x)= 1 е 26 26 for a >0 %3D - Select your answer - b. What is the probability that a customer will have to wait between 15 and 30 seconds (to 4 decimals)? c. What is the probability that a customer will have to wait more than 2 minutes (to 4 decimals)?

a. Which of the probability density functions of waiting time is applicable at Kroger? 1 a. f(x) = е 26 for a 20 %3D 26 1 b. f(x) for a >0 e e 26 - 26 1 c. f(x) = 1 e 26 for a <0 26 - 1 d. f(x)= 1 е 26 26 for a >0 %3D - Select your answer - b. What is the probability that a customer will have to wait between 15 and 30 seconds (to 4 decimals)? c. What is the probability that a customer will have to wait more than 2 minutes (to 4 decimals)?

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**Understanding Waiting Times at Kroger** Do you dislike waiting in line? The supermarket chain Kroger has employed computer simulation and information technology to significantly reduce the average waiting time for customers across its 2,300 stores. Utilizing a novel system called QueVision, Kroger can better predict when customers will be checking out, reducing the average waiting time to just 26 seconds. **Assumptions:** - Waiting times at Kroger are assumed to be exponentially distributed. **Questions:** **a. Probability Density Functions** - Which of the following probability density functions of waiting time is applicable at Kroger? 1. \( f(x) = \frac{1}{\mu} e^{\frac{x}{\mu}} = \frac{1}{26} e^{-\frac{x}{26}} \) for \( x \geq 0 \) 2. \( f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} = \frac{1}{26} e^{-\frac{x}{26}} \) for \( x \geq 0 \) 3. \( f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} = \frac{1}{26} e^{-\frac{x}{26}} \) for \( x \leq 0 \) 4. \( f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} = \frac{1}{26} e^{\frac{x}{26}} \) for \( x \geq 0 \) **b. Probability of Waiting Between 15 and 30 Seconds** - What is the probability that a customer will have to wait between 15 and 30 seconds (to 4 decimals)? **c. Probability of Waiting More Than 2 Minutes** - What is the probability that a customer will have to wait more than 2 minutes (to 4 decimals)? **Note:** - The correct solution for each part requires knowledge of exponential distributions and how to compute probabilities based on them.
Transcribed Image Text:**Understanding Waiting Times at Kroger** Do you dislike waiting in line? The supermarket chain Kroger has employed computer simulation and information technology to significantly reduce the average waiting time for customers across its 2,300 stores. Utilizing a novel system called QueVision, Kroger can better predict when customers will be checking out, reducing the average waiting time to just 26 seconds. **Assumptions:** - Waiting times at Kroger are assumed to be exponentially distributed. **Questions:** **a. Probability Density Functions** - Which of the following probability density functions of waiting time is applicable at Kroger? 1. \( f(x) = \frac{1}{\mu} e^{\frac{x}{\mu}} = \frac{1}{26} e^{-\frac{x}{26}} \) for \( x \geq 0 \) 2. \( f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} = \frac{1}{26} e^{-\frac{x}{26}} \) for \( x \geq 0 \) 3. \( f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} = \frac{1}{26} e^{-\frac{x}{26}} \) for \( x \leq 0 \) 4. \( f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} = \frac{1}{26} e^{\frac{x}{26}} \) for \( x \geq 0 \) **b. Probability of Waiting Between 15 and 30 Seconds** - What is the probability that a customer will have to wait between 15 and 30 seconds (to 4 decimals)? **c. Probability of Waiting More Than 2 Minutes** - What is the probability that a customer will have to wait more than 2 minutes (to 4 decimals)? **Note:** - The correct solution for each part requires knowledge of exponential distributions and how to compute probabilities based on them.
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