Question

The population of NBA players is Normally distributed

with a mean of 6'7" and a standard deviation of 3. 9

inches. (Wikipedia) Greg is considered unusually tall for

his high school at 6' 3".

1. About what percent of NBA players are taller than Greg?

2. About what percent are shorter?

3. What minimum height would Greg have to be (IN INCHES) in order to be in the top 2. 5% of

NBA player heights?

Answer 1

Using the normal distribution, it is found that:

1. 85% of NBA players are taller than Greg.

2. 15% are shorter.

3. A height of 6'11'' would be needed.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• In inches, considering that a feet has 12 inches, the mean is of $\mu = 6 \times 12 + 7 = 79$.

• The standard deviation is of $\sigma = 3.9$.

• Greg is 6'3'', that is, X = 6 x 12 + 3 = 75.

Item 1:

The proportion is one subtracted by the p-value of Z, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75 - 79}{3.9}$

$Z = -1.03$

$Z = -1.03$ has a p-value of 0.15

1 - 0.15 = 0.85.

85% of NBA players are taller than Greg.

Item 2:

100 - 85 = 15% are shorter.

Item 3:

This is X when Z has a p-value of 1 - 0.025 = 0.975, so X when Z = 1.96.

$Z = \frac{X - \mu}{\sigma}$

$1.96 = \frac{X - 75}{3.9}$

$X - 75 = 1.96(3.9)$

$X = 83$

83 = 6 x 12 + 11.

A height of 6'11'' would be needed.

More can be learned about the normal distribution at https://brainly.com/question/24663213