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?

Answer 1

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

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Answer 2

Percent of adult male population taller than ‘average basketball player’ is 0.13%

To find:

Percent of adult male population taller than ‘average basketball player’

Given:

Adult ‘male height’s are normally distributed with mean of ‘70 inches’ and standard deviation =3 inches.  

The height of average basketball player = 79 inches  

Mean of adult male heights normally distributed = 70 inches (i.e. the average of the given set of data)

and Standard deviation = 3 inches (i.e. the variability of the data set)

and average height of basketball player = 79 inches

Solution:

$Z>\frac{\text {Average}-\text {Mean}}{\text { Standard deviation }}$

$\mathrm{Z}>79-\frac{70}{3}$

$\mathrm{Z}>\frac{9}{3}$

Z > 3

Hence, P(Z>3) = 1 - 0.9987 = 0.0013  

Therefore, 0.13% of adult male population are taller then the average basketball player.