Question

Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketball player is 79 inches tall. Approximately what percent of the adult male population is taller than the average basketball player?
0.13%

0.87%

49.87%

99.87%

Answer 1

Answer: First Option is correct.

Step-by-step explanation:

Since we have given that

Mean $(\mu)$= 70 inches

Standard deviation $(\sigma)$= 3 inches

Average basket ball player $(\bar{X})$ = 79 inches

Since it is normally distributed assume with 5% significance .

So, it becomes,

$Z>\frac{\bar{X}-\mu}{\sigma}\\\\Z>\frac{79-70}{3}\\\\Z>\frac{9}{3}\\\\Z>3\\\\\therefore\ P(Z>3)=1-0.9987=0.0013$

At 0.05 level of significance , 0.0013 = 0.13% of the adult male population is taller than the average basketball player .

Hence, First Option is correct.

Answer 2

Kindly check this link for more details:

Therefore the answer is 0.13% is the approximate percentae of the adult male population who are taller than the average basketball player