Answer:
The probability that the sample mean weight will be more than 262 lb is 0.0047.
Step-by-step explanation:
The random variable <em>X</em> can be defined as the weight of National Football League (NFL) players now.
The mean weight is, <em>μ</em> = 252.8 lb.
The standard deviation of the weights is, <em>σ</em> = 25 lb.
A random sample of <em>n</em> = 50 NFL players are selected.
According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the sample means is given by,
![\mu_{\bar x}=\mu](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%7D%3D%5Cmu)
And the standard deviation of the sample means is given by,
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The sample of players selected is quite large, i.e. <em>n</em> = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.
![\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)](https://tex.z-dn.net/?f=%5Cbar%20X%5Csim%20N%28%5Cmu_%7B%5Cbar%20x%7D%3D252.8%2C%5C%20%5Csigma_%7B%5Cbar%20x%7D%3D3.536%29)
Compute the probability that the sample mean weight will be more than 262 lb as follows:
![P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Cbar%20X%3E262%29%3DP%28%5Cfrac%7B%5Cbar%20X-%5Cmu_%7B%5Cbar%20x%7D%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%3E%5Cfrac%7B262-252.8%7D%7B3.536%7D%29%5C%5C%5C%5C%3DP%28Z%3E2.60%29%5C%5C%5C%5C%3D1-P%28Z%3C2.60%29%5C%5C%5C%5C%3D1-0.99534%5C%5C%5C%5C%3D0.00466%5C%5C%5C%5C%5Capprox%200.0047)
*Use a <em>z</em>-table for the probability.
Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.