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 less than $50 on dinner?
A)0.4121
B)0.9868
C)0.0132
D)0.5879
E)0.9614

Question

Quizplus · Accepted Answer

We need to calculate the probability that the sample mean of 100 customers' dinners is less than $50. By Central Limit Theorem, the distribution of sample means is approximately normal with a mean of $54 and a standard deviation of $18/sqrt(100) = $1.8.

We can standardize this distribution using the formula z = (x - mu) / (sigma / sqrt(n)), where x is the sample mean, mu is the population mean, sigma is the population standard deviation, and n is the sample size.

z = (50 - 54) / (1.8) = -2.22

Using a standard normal table or calculator, we can find that the probability of getting a z-score less than -2.22 is 0.0132. Therefore, the probability that the next 100 customers will spend an average of less than $50 on dinner is 0.0132 or 1.32%.

The closest choice is C.

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