Question

The heights of adult American males are normally distributed with a mean of 70 inches, and a standard deviation of 3 inches. The average NBA player is 6’7”, or 79 inches tall(3 standard deviations above the mean). If you select 1000 adult American males at random, how many of them would you expect to be taller than an average NBA-er?

Answer 1

X ~ N ( µ = 70 , σ = 3 )
P ( X > 79 ) = 1 - P ( X < 79 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 79 - 70 ) / 3
Z = 3
P ( ( X - µ ) / σ ) > ( 79 - 70 ) / 3 )
P ( Z > 3 )
P ( X > 79 ) = 1 - P ( Z < 3 )
P ( X > 79 ) = 1 - 0.9987
P ( X > 79 ) = 0.0013

Of the 1000 males we expect, 0.0013 * 1000 = 1.3

$\approx$ 1 male to be taller than an average NBA-er