Let x be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the x distribution is about ? = 2.7 minutes, with standard deviation ? = 0.6 minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of x values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let xi (for i = 1, 2, 3, . . . , 30) represent the checkout time for each customer. For example, x1 is the checkout time for the first customer, x2 is the checkout time for the second customer, and so forth. Each xi has mean ? = 2.7 minutes and standard deviation ? = 0.6 minute. Let w = x1 + x2 + . . . + x30. Explain why the problem is asking us to compute the probability that w is less than 90. w is the sum of the waiting times for 30 customers, so we want to find P(w < 90).w is the total waiting time for all customers, so we want to find P(w < 90). w is the average of the waiting times for 30 customers, so we want to find P(w < 90).w is the average waiting time for 30 customers, so we want to find P(w < 90). (b) Use a little algebra and explain why w < 90 is mathematically equivalent to w/30 < 3. add 30subtract 30 divide by 3subtract 90divide by 30 Since w is the total of the 30 x values, then w/30 = x. Therefore, the statement x < 3 is equivalent to the statement w < 90. From this we conclude that the probabilities P(x < 3) and P(w < 90) are equal. (c) What does the central limit theorem say about the probability distribution of x? Is it approximately normal? What are the mean and standard deviation of the x distribution? The probability distribution of x is approximately normal with mean ?x = 2.7 and standard deviation ?x = 0.02.The probability distribution of x is approximately normal with mean ?x = 2.7 and standard deviation ?x = 0.11. The probability distribution of x is not normal.The probability distribution of x is approximately normal with mean ?x = 2.7 and standard deviation ?x = 0.6. (d) Use the result of part (c) to compute P(x < 3). (Round your answer to three decimal places.) What does this result tell you about P(w < 90)? (Round your answer to three decimal places.) The probability that the total checkout time for the next 30 customers is less than 90 minutes is

Let x be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the x distribution is about ? = 2.7 minutes, with standard deviation ? = 0.6 minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of x values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let xi (for i = 1, 2, 3, . . . , 30) represent the checkout time for each customer. For example, x1 is the checkout time for the first customer, x2 is the checkout time for the second customer, and so forth. Each xi has mean ? = 2.7 minutes and standard deviation ? = 0.6 minute. Let w = x1 + x2 + . . . + x30. Explain why the problem is asking us to compute the probability that w is less than 90. w is the sum of the waiting times for 30 customers, so we want to find P(w < 90).w is the total waiting time for all customers, so we want to find P(w < 90). w is the average of the waiting times for 30 customers, so we want to find P(w < 90).w is the average waiting time for 30 customers, so we want to find P(w < 90). (b) Use a little algebra and explain why w < 90 is mathematically equivalent to w/30 < 3. add 30subtract 30 divide by 3subtract 90divide by 30 Since w is the total of the 30 x values, then w/30 = x. Therefore, the statement x < 3 is equivalent to the statement w < 90. From this we conclude that the probabilities P(x < 3) and P(w < 90) are equal. (c) What does the central limit theorem say about the probability distribution of x? Is it approximately normal? What are the mean and standard deviation of the x distribution? The probability distribution of x is approximately normal with mean ?x = 2.7 and standard deviation ?x = 0.02.The probability distribution of x is approximately normal with mean ?x = 2.7 and standard deviation ?x = 0.11. The probability distribution of x is not normal.The probability distribution of x is approximately normal with mean ?x = 2.7 and standard deviation ?x = 0.6. (d) Use the result of part (c) to compute P(x < 3). (Round your answer to three decimal places.) What does this result tell you about P(w < 90)? (Round your answer to three decimal places.) The probability that the total checkout time for the next 30 customers is less than 90 minutes is

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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Question

(a) Let x_i (for i = 1, 2, 3, . . . , 30) represent the checkout time for each customer. For example, x₁ is the checkout time for the first customer, x₂ is the checkout time for the second customer, and so forth. Each x_i has mean ? = 2.7 minutes and standard deviation ? = 0.6 minute. Let w = x₁ + x₂ + . . . + x₃₀. Explain why the problem is asking us to compute the probability that w is less than 90.

w is the sum of the waiting times for 30 customers, so we want to find P(w < 90).w is the total waiting time for all customers, so we want to find P(w < 90). w is the average of the waiting times for 30 customers, so we want to find P(w < 90).w is the average waiting time for 30 customers, so we want to find P(w < 90).

(b) Use a little algebra and explain why w < 90 is mathematically equivalent to w/30 < 3.

add 30subtract 30 divide by 3subtract 90divide by 30

Since w is the total of the 30 x values, then w/30 = x. Therefore, the statement x < 3 is equivalent to the statement w < 90. From this we conclude that the probabilities P(x < 3) and P(w < 90) are equal.

(c) What does the central limit theorem say about the probability distribution of x? Is it approximately normal? What are the mean and standard deviation of the x distribution?

The probability distribution of x is approximately normal with mean ?_x = 2.7 and standard deviation ?_x = 0.02.The probability distribution of x is approximately normal with mean ?_x = 2.7 and standard deviation ?_x = 0.11. The probability distribution of x is not normal.The probability distribution of x is approximately normal with mean ?_x = 2.7 and standard deviation ?_x = 0.6.

(d) Use the result of part (c) to compute P(x < 3). (Round your answer to three decimal places.)

What does this result tell you about P(w < 90)? (Round your answer to three decimal places.)

The probability that the total checkout time for the next 30 customers is less than 90 minutes is

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