Question

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic random sampling of 100 filings. These 100 filings showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

A) What is the standard error of the mean?

B) What is the probability that the mean completion time will be more than 120 minutes?

I believe the answer to A is 14 minutes, I need to know how to get that in long hand please...

I believe the answer to B is 0.0162, but I need to know how to figure this out in long hand as well please.

Answer 1

n=100

μ = 90

σ = 140

A) The standard error is given by σ/√n = 140/√100 = 140/10 = 14

B) note that xbar = sample mean. We are working with a sampling distribution now so we must use the standard error to normalize xbar. First we will subtract the mean from xbar to get a mean of zero and then we will divide that by the standard error to get a standard deviation of 1. THis will leave us with a mean of 0 and a standard deviation of 1, which is the standard normal distribution.

P(xbar>120) = P((xbar-μ)/(σ/√n) > (120-90)/(140/√100)) = P(z > 2.14) = .0162