statistics

Assume that women's heights are normally distributed with a mean given by u=64.6 in
and a standard deviation given by 0=2.8in.

a) If 1 woman is randomly selected, find the probability that her height is less than 65in.

b) If 49 women are randomly selected, find the probability that they have a mean height less than 65 in.

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  1. great webpage for your problem, just plug in the values that you have

    http://davidmlane.com/hyperstat/z_table.html

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  2. a.
    z = (65-64.6)/(2.8/sqrt(1))

    z = 0.14
    b.
    z = (65-64.6)/(2.8/sqrt(49))

    z = 1

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  3. Assume that​ women's heights are normally distributed with a mean given by mu equals 64.9 inμ=64.9 in​, and a standard deviation given by sigma equals 2.7 inσ=2.7 in. Complete parts a and b.
    a. If 1 woman is randomly​ selected, find the probability that her height is between 64.264.2 in and 65.265.2 in.
    The probability is approximately
    0.14650.1465. ​(Round to four decimal places as​ needed.)
    b. If 6 women are randomly​ selected, find the probability that they have a mean height between 64.264.2 in and 65.265.2 in.
    The probability is approximately

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  4. Assume that​ women's heights are normally distributed with a mean given by mu equals 62.3 inμ=62.3 in​, and a standard deviation given by sigma equals 2.7 inσ=2.7 in.
    ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 6363 in.
    ​(b) If 4343 women are randomly​ selected, find the probability that they have a mean height less than 6363 in.

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  5. Assume that​ women's heights are normally distributed with a mean given by mu equals inμ=62.6 and a standard deviation given by sigma equals inσ=2.4 in.
    ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in.
    ​(b) If 36 women are randomly​ selected, find the probability that they have a mean height less than 63 in.
    ​(​a) The probability is approximately
    nothing.
    ​(Round to four decimal places as​ needed.)

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