Respuesta :
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,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the sample means is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
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.
[tex]\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)[/tex]
Compute the probability that the sample mean weight will be more than 262 lb as follows:
[tex]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<2.60)\\\\=1-0.99534\\\\=0.00466\\\\\approx 0.0047[/tex]
*Use a z-table for the probability.
Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.
