Human heights are one of many biological random variables that can be modeled by the normal distribution. a 2002 study in the us found the mean height for 19 year old males is 176.7 cm with a standard deviation of 9.9 cm. a basketball play is usually over 188 cm - what proportion of the 19 year old male population is greater than this height?

To answer this item, find first the z-score of the given data.

    z-score = (x - m) / d

where x is data point, m is the mean and d is the standard deviation

Substituting the given values,

    z-score = (188 - 176.7) / 9.9  = 1.14

This translates to a percentage of 87.29%. 

Because we are asked of the percentage of those with height greater than this, that would be,
     100 - 87.29% = 12.71%

12.71%

Answer is a not really sure tho