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14 December, 22:54

A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 19 hours, with a standard deviation of 5.0 hours. It is desired to estimate the mean viewing time within 30 minutes. The 99% level of confidence is to be used. (Use z Distribution Table.)

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  1. 14 December, 23:17
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    We are given

    x = 19

    s = 5

    u = 30

    99% LOC

    Getting the z-score

    z = (19 - 30) / 5

    z = - 2.2

    From the z-score table,

    the probability is

    0.0139

    In percentage, the answer is 0.39%
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