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# Normal Distribution and Sample Information

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The mean amount purchased by a typical customer at Churchill's Grocery Store is \$23.50 with a standard deviation of \$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 interval for the population mean yearly premium.

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

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#### Solution Summary

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

\$2.19