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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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    https://brainmass.com/statistics/normal-distribution/normal-distribution-237551

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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.

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