Question

The heights of adult men in America are normally distributed, with a mean of 69.8 inches and a standard deviation of 2.69 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.53 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

Answer:

Step-by-step explanation:

Let X be the heights of adult men in America and Y that of women in America

X is N(69.8, 2.69) and Y is N(64.8,2.53)

a) Man is 6 feet 3 inches = 75 inches

Z score = $\frac{x-69.8}{2.69} \\=\frac{-5}{2.69} \\=-1.86$

b) Prob = 0.5-0.4686 = 0.0314

3.14% are shorter than him

c) Y score = 5 ft 11 inches = 71 inches

Z=$\frac{x-64.8}{2.53} \\=2.45$

d) Prob (Z>2.45) = 0.5-0.4929=0.0071

Nearly 0.71% people are taller than 5 ft 11 inches.