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8 January, 18:30

The average number of push-ups a united states marine does daily is 300, with a standard deviationof 50. a random sample of 36 marines is selected. what is the probability that the sample mean is atmost 320 push-ups?

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  1. 8 January, 21:19
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    To solve we need to calculate the z-score, but since we are given the sample size then we shall begin as follows:

    μ=300

    σ=50

    n=36

    x=320

    thus

    σ/√n

    =50/√36

    =8.33333

    thus the z-score will be:

    z = (320-300) / 8.33333

    z=2.4

    Thus

    P (x>320) = P (z<2.4) = 0.9918
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