Question

The manager of a grocery store has taken a random sample of 196 customers. The average length of time it took these 196 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. The margin of error is 0.140. Find the 95% confidence interval for the true average checkout time (in minutes).

A. 1 to 5

B. 2.86 to 3.14

C. 2.883 to 3.117

D. 2.990 to 3.010

Answer #1

n = 196 sample size

minute sample mean

s =1 minute sample standard deviations

level of significance

Margin of error = 0.140

The 95% confidence interval for the true average is

. Here n >30

from Z table

( 2.86 , 3.14)

The 95 % confidence interval for the true average check out time in minutes is (2.86 to 3.14)

Option B is correct

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