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27 January, 23:51

A normal distribution has a mean of 0.40 and standard deviation of 0.028. What percentage of observations will lie between 0.372 and 0.428?

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  1. 28 January, 00:47
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    Solution:

    To evaluate for P (0.372
    z-score is given by:

    z = (x-μ) / σ

    thus when x=0.372:

    z = (0.372-0.4) / 0.028

    z=-1

    thus

    P (x<0.372) = P (z<-1) = 0.1587

    when x=0.428

    z = (0.428-0.4) / (0.028) = 1

    P (x<0.428) = P (z<1) = 0.8413

    thus

    P (0.372
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