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8 November, 07:46

Assume the random variable X is normally distributed with a mean of 50 and a standard deviation of 7. Find the 96th percentile.

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  1. 8 November, 08:03
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    90% then z = invNorm (.90) = 1.2816

    z = 1.2816 = (X-50) / 77 (1.2816) + 50 = 8.97 + 50 = X = 59
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