Question

The heights of adult men in America are normally distributed, with a mean of 69.4 inches and a standard deviation of 2.66 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.4 inches and a standard deviation of 2.59 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? z =

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)? z =

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

Answer 1

a)
Here, μ = 69.4, σ = 2.66 and x = 75. We need to compute P(X <= 75). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (75 - 69.4)/2.66 = 2.11

b)

Therefore,
P(X <= 75) = P(z <= (75 - 69.4)/2.66)
= P(z <= 2.11)
= 0.9826 = 98.3%

c)

Here, μ = 64.4, σ = 2.59 and x = 71. We need to compute P(X >= 71). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (71 - 64.4)/2.59 = 2.55

d)

Therefore,
P(X >= 71) = P(z <= (71 - 64.4)/2.59)
= P(z >= 2.55)
= 1 - 0.9946 = 0.0055 = 0.6%