Hi, 

variable X with a normal distribution with mean 36 inches and standard deviation 2 inches       

finding the X value approximating the 84th percentile of this distribution

*Note: 

    (X - 36)/2 = .9945    |Using Excel function: NORMSINV(0.84) 

        X = 2*.9945 +36

        X = 37.989 or 38 will approxiamate the 84th percentile 

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