Chapter 6.5, Problem 21P

Expand Your Knowledge: Totals Instead of Averages 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 μ = 2.7 minutes, with standard deviation σ = 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.

1. (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 μ = 2.7 minutes and standard deviation σ = 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.
2. (b) Use a little algebra and explain why w < 90 is mathematically equivalent to w/30 < 3. Since w is the total of the 30 x values, then $w/\text{3}0=\bar{x}$. Therefore, the statement $\bar{x}<\text{3}$ is equivalent to the statement w < 90. From this we conclude that the probabilities $P(\bar{x}<\text{3})$ and P(w < 90) are equal.
3. (c) What does the central limit theorem say about the probability distribution of $\bar{x}$? Is it approximately normal? What are the mean and standard deviation of the $\bar{x}$ distribution?
4. (d) Use the result of part (c) to compute $P(\bar{x}<\text{3})$. What does this result tell you about P(w < 90)?