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8 February, 03:07

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in , and a standard deviation given by sigma equals 2.6 in. (a) if 1 woman is randomly selected, find the probability that her height is less than 63 in. (b) if 44 women are randomly selected, find the probability that they have a mean height less than 63 in.

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  1. 8 February, 03:20
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    If 1 woman is randomly selected, the probability that her height is less than 63 in will be:

    The z-score is given by:

    z = (x-mu) / sig

    z = (63-62.5) / 2.6

    z=0.1923

    Thus:

    P (x<63) = P (z<0.1923) = 0.5753

    (b) if 44 women are randomly selected, find the probability that they have a mean height less than 63 in.

    In this scenario the desired probability is for the mean of a sample of 44 women, therefore we use the central limit theorem. The standard deviation of the sample means is:

    2.6/sqrt44=0.392

    since n>30, use the z-distribution

    z = (63-62.5) / 0.392

    z=1.28

    Then

    P (x-bar <63) = P (z<1.28) = 0.8997
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