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Sagot :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 33 years, thus [tex]\mu = 33[/tex].
- Standard deviation of 2 years, thus [tex]\sigma = 2[/tex].
Item 1:
This probability is the p-value of Z when X = 36, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{36 - 33}{2}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] 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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.253 = \frac{X - 33}{2}[/tex]
[tex]X - 33 = -0.253(2)[/tex]
[tex]X = 32.5[/tex]
40% of NFL players will retire by an age of 32.5 years.
Item 3:
Sample of 36, thus [tex]n = 36, s = \frac{2}{\sqrt{36}} = 0.3333[/tex]
This probability is the p-value of Z when X = 32, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{32 - 33}{0.3333}[/tex]
[tex]Z = -3[/tex]
[tex]Z = -3[/tex] 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
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