Question

(1 point) The amount of time (in minutes) to serve a customer at a bank's drive-thru window is a random variable following a certain distribution with a mean of 1 minute and a standard deviation of 1 minute. 66 customers using the bank's drive-thru service are randomly chosen and the amount of time to serve each customer was recorded. What is the probability that the mean amount of time it takes to serve the 66 customers is between 0.9 minutes to 1 minutes?

Answer 1

Since the sample size is large (n > 30), so the sampling distribution of the sample mean will be normally distributed.

P(0.9 < < 1)

= P((0.9 - )/() < ( - )/() < (1 - )/())

= P((0.9 - 1)/(1/) < Z < (1 - 1)/(1/))

= P(-0.81 < Z < 0)

= P(Z < 0) - P(Z < -0.81)

= 0.5 - 0.2090

= 0.291