Question

Men in the U.S have heights which are normally distributed with a mean of 68 inches and a standard deviation of 2.5 inches. What percentage of men have heights between 66 inches and 69.5 inches? What height separates the shortest 6% of men from the 94% tallest men?

Answer 1

Here, u = 68, = 2.5

z = (x-u)/

1) p(66 < x < 69.5)

= P( (66-68)/2.5 < z < (69.5-68)/2.5 )

= p(-0.8 < z < 0.6)

= 0.7258 - 0.2118

= 0.5140

2) shortest 6% implies p = 0.06

Corresponding z score = -1.55

z = (x-u)/

Thus, -1.55 = (x - 68)/2.5

X = 68 - 1.55*2.5

X = 68- 3.88

X = 64.12 inches

Rounding off to 64.1

Thus, this height separates short 6% from tall 94%