The meanamount purchased by a typical customer at Churchill’s Grocery Store is $23.50 with a standard deviationof $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions.

a. What is the likelihood the sample mean is at least $25.00?

z = (x-bar – m)/SE = ( – )/ =

P(x-bar > ) = P(z > -) =

b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00?

z = (x-bar – m)/SE = ( – )/ =

P(x-bar > ) = P(z > -) =

c. Within what limits will 90 percent of the sample means occur?

Standard Error of the mean, SE =

Level of significance a =

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000.

a. Based on this sample information, develop a 90 percent confidence intervalfor the population mean yearly premium.

b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?

This solution shows step-by-step calculations in an Excel file to determine probabilities of the sample mean, confidence intervals and sample size.