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? 0.135% 0.875% 49.875% 99.875%

Answer 1

Z = (x - mean) / SD = (79 - 70) / 3 = 3 
P (Z > 3)? = 1 - F (z) = 1 - F (3) = 0.00135

Answer 2

Answer:

A. 0.135%

Step-by-step explanation:

We have been given that 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.  

We need to find the area of normal curve above the raw score 79.

First of all let us find the z-score corresponding to our given raw score.

$z=\frac{x-\mu}{\sigma}$, where,

$z=\text{z-score}$,

$x=\text{Raw-score}$,

$\mu=\text{Mean}$,

$\sigma=\text{Standard deviation}$.

Upon substituting our given values in z-score formula we will get,

$z=\frac{79-70}{3}$

$z=\frac{9}{3}$

$z=3$

Now we will find the P(z>3) using formula:

$P(z>a)=1-P(z$

$P(z>3)=1-P(z$

Using normal distribution table we will get,

$P(z>3)=1-0.99865$

$P(z>3)=0.00135$

Let us convert our answer into percentage by multiplying 0.00135 by 100.

$0.00135\times 100=0.135%$

Therefore, approximately 0.135% of the adult male population is taller than the average basketball player and option A is the correct choice.