Question

The weight of National Football League (NFL) players has increased steadily, gaining up to 1.5 lb. per year since 1942. According to ESPN, the average weight of a NFL player is now 252.8 lb. Assume the population standard deviation is 25 lb. If a random sample of 50 players is selected, what is the probability that the sample mean will be more than 262 lb.

Answer 1

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable X can be defined as the weight of National Football League (NFL) players now.

The mean weight is, μ = 252.8 lb.

The standard deviation of the weights is, σ = 25 lb.

A random sample of n = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 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,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The sample of players selected is quite large, i.e. n = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

$\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)$

Compute the probability that the sample mean weight will be more than 262 lb as follows:

$P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z$

*Use a z-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.