The average retirement age of NFL players is 33 years with a standard deviation of 2 years.
a) Find the probability that a randomly chosen NFL player retired at over 36 years old.
b) What is the probability that a randomly chosen sample of 10NFL players has average retirement age between 34 and 35 years of age?
The Correct Answer and Explanation is :
Part (a) Finding the probability that a randomly chosen NFL player retired over 36 years old
Given:
- Average retirement age (mean), (\mu = 33)
- Standard deviation, (\sigma = 2)
- We are asked to find the probability that a randomly chosen NFL player retired after 36 years old, i.e., (P(X > 36)).
To find this probability, we will use the Z-score formula, which standardizes the given value:
[
Z = \frac{X – \mu}{\sigma}
]
Where:
- (X = 36) (the value we are comparing against)
- (\mu = 33) (the mean)
- (\sigma = 2) (the standard deviation)
Substituting the values:
[
Z = \frac{36 – 33}{2} = \frac{3}{2} = 1.5
]
Now, we look up the Z-score value of 1.5 in the standard normal distribution table, which gives the cumulative probability up to 36 years old:
[
P(Z < 1.5) \approx 0.9332
]
To find the probability that the player retired after 36 years old, we subtract this value from 1:
[
P(X > 36) = 1 – P(Z < 1.5) = 1 – 0.9332 = 0.0668
]
So, the probability that a randomly chosen NFL player retired over 36 years old is approximately 0.0668 or 6.68%.
Part (b) Finding the probability that the average retirement age of 10 randomly chosen NFL players is between 34 and 35 years old
In this case, we are dealing with a sample of 10 NFL players, so we will use the sampling distribution of the sample mean. The mean of the sample mean remains the same as the population mean, but the standard deviation (standard error) of the sample mean is reduced by a factor of the square root of the sample size (n).
Given:
- Sample size (n = 10)
- Population mean (\mu = 33)
- Population standard deviation (\sigma = 2)
The standard error (SE) is:
[
SE = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{10}} \approx 0.6325
]
Next, we want to find the probability that the sample mean falls between 34 and 35 years. We will calculate the Z-scores for both 34 and 35 years, using the formula:
[
Z = \frac{X – \mu}{SE}
]
For (X = 34):
[
Z_{34} = \frac{34 – 33}{0.6325} \approx 1.58
]
For (X = 35):
[
Z_{35} = \frac{35 – 33}{0.6325} \approx 3.16
]
Now, we find the cumulative probabilities for these Z-scores:
- (P(Z < 1.58) \approx 0.9429)
- (P(Z < 3.16) \approx 0.9992)
The probability that the sample mean is between 34 and 35 years old is the difference between these two probabilities:
[
P(34 < \bar{X} < 35) = P(Z < 3.16) – P(Z < 1.58) = 0.9992 – 0.9429 = 0.0563
]
So, the probability that the average retirement age of 10 randomly chosen NFL players is between 34 and 35 years old is approximately 0.0563 or 5.63%.
Conclusion
- Part (a): The probability that a randomly chosen NFL player retired after 36 years old is approximately 6.68%.
- Part (b): The probability that the average retirement age of a sample of 10 NFL players is between 34 and 35 years old is approximately 5.63%.