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 m 5 2.7 minutes and standard deviation s 5 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.

(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/30 = x̅. Therefore, the statement x̅ < 3 is equivalent to the statement w < 90. From this we conclude that the probabilities P(x̅ < 3) and P (w < 902) 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 x̅ distribution?

(d) Use the result of part (c) to compute P(x̅ < 3). What does this result tell you about P(w < 90)?