A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of at least $58 on dinner?
A)0.0562
B)0.5879
C)0.4121
D)0.9868
E)0.0132

Question

Quizplus · Accepted Answer

The sampling distribution of the sample mean for 100 customers will be approximately normal due to the Central Limit Theorem. The standard deviation of the sample mean (standard error) is $18/sqrt(100) = $1.8. The z-score for a sample mean of $58 is (58-54)/1.8 = 2.22. Using the standard normal distribution, the probability of a z-score being greater than 2.22 is approximately 0.0132.

A Restaurant's Receipts Show That the Cost of Customers' Dinners