The manager of a grocery store has taken a random sample of 144 customers. The average length of time it took these 144 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a .95 probability, the sample mean will provide a margin of error of
A) 1.63.
B) .137.
C) .163.
D) 1.37.

Question

Quizplus · Accepted Answer

First, we need to use the formula for the margin of error:

Margin of Error = z*(standard deviation/square root of sample size)

Since we want to be 95% confident, we need to use the z-score that corresponds to a 95% confidence level. This is 1.96.

Therefore, the margin of error is:

Margin of Error = 1.96*(1/square root of 144) = 0.163

So, the answer is (C) 0.163.

The Manager of a Grocery Store Has Taken a Random