The manager of a grocery store has taken a random sample of 100 customers.The average length of time it took these 100 customers to check out was 3 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.The 95% confidence interval for the true average checkout time (in minutes) is
A) 2.5 to 3.5.
B) 1.36 to 4.64.
C) 1 to 5.
D) 2.804 to 3.196.

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for the population mean:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean (3), σ is the population standard deviation (1), n is the sample size (100), and z is the critical value from the standard normal distribution for a 95% confidence level (1.96).

Plugging in the numbers, we get:

CI = 3 ± 1.96*(1/√100)
CI = 3 ± 0.196
CI = (2.804, 3.196)

Therefore, the 95% confidence interval for the true average checkout time is (2.804 to 3.196) minutes. Choice D is correct.

The Manager of a Grocery Store Has Taken a Random