Question 391758
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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: {{{z = blue(x - mu)/blue(sigma)}}}
    (X - 36)/2 = .9945    |Using Excel function: NORMSINV(0.84) 
        X = 2*.9945 +36
        X = 37.989 or 38 will approxiamate the 84th percentile