Question

Chapter 5: The ages at retirement for NFL players is normally distributed with a mean of 33 years and a standard deviation of 2 years.1. What percentage of NFL players retire by the age of 36? Round percentage to one decimal place.2. By what age will 40% of NFL players retire? Round to the nearest tenth of a year.3. If a sample of 36 NFL players are randomly selected, what is the probability that their mean retirement age will be less than 32 years? Round to three decimal places.

Answer 1

Using the normal distribution and the central limit theorem, it is found that:

1. 0.692 = 69.2% of NFL players retire by the age of 36.

2. 40% of NFL players will retire by an age of 32.5 years.

3. 0.001 = 0.1% probability that their mean retirement age will be less than 32 years.

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.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation given by $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of 33 years, thus $\mu = 33$.
• Standard deviation of 2 years, thus $\sigma = 2$.

Item 1:

This probability is the p-value of Z when X = 36, thus:

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

$Z = \frac{36 - 33}{2}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.692.

0.692 = 69.2% of NFL players retire by the age of 36.

Item 2:

This age is the 40th percentile, which is X when Z has a p-value of 0.4, so X when Z = -0.253.

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

$-0.253 = \frac{X - 33}{2}$

$X - 33 = -0.253(2)$

$X = 32.5$

40% of NFL players will retire by an age of 32.5 years.

Item 3:

Sample of 36, thus $n = 36, s = \frac{2}{\sqrt{36}} = 0.3333$

This probability is the p-value of Z when X = 32, thus:

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

By the Central Limit Theorem

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

$Z = \frac{32 - 33}{0.3333}$

$Z = -3$

$Z = -3$ has a p-value of 0.001.

0.001 = 0.1% probability that their mean retirement age will be less than 32 years.

A similar problem is given at https://brainly.com/question/25218719