Question

According to the National Health Survey, the heights of adult males in the United States are normally distributed with mean 69.0 inches and standard deviation 2.8 inches. (a) What is the probability that an adult male chosen at random is between 64 and 74 inches tall? (Round your answer to three decimal places.) (b) What percentage of the adult male population is more than 6 feet tall? (Round your answer to one decimal place.)

Answer 1

Answer:  (a)  0.927

(b) 14.2%

Explanation:

Given : According to the National Health Survey, the heights of adult males in the United States are normally distributed .

Mean : $\mu=69$

Standard deviation : $\sigma=2.8$

Let X be a random variable that represents the heights of adult males in the United States .

Z-score : $z=\dfrac{x-\mu}{\sigma}$

(a) For x= 64

$z=\dfrac{64-69}{2.8}\approx-1.79$

For x= 74

$z=\dfrac{74-69}{2.8}\approx1.79$

Using the standard normal distribution table , we have

The deviation 2.8 inches. (a) What is the probability that an adult male chosen at random is between 64 and 74 inches tall :-

$P(64<x<74)=P(-1.79<z<1.79)=P(z<1.79)-P(z<-1.79)\\\\=0.963273-0.036727=0.926546\approx0.927$

Hence, the probability that an adult male chosen at random is between 64 and 74 inches tall = 0.927

b) Since 1 feet = 12 inches

Then , 6 feet = $6\times12=72\text{ inches}$

The z-score corresponds to x=72 :

$z=\dfrac{72-69}{2.8}\approx1.07$

Then , the probability that the adult male population is more than 6 feet tall :-

$P(x>72)=P(z>1.07)=1-P(z<1.07)\\\\=1- 0.8576903=0.1423097\approx14.2\%$

Hence, the  percentage of the adult male population is more than 6 feet tall = 14.2%