Question

assume that the heights of men are normally distributed with a mean of 67.6 inches and a standard deviation of 2.8 inches. if 64 men are randomly selected, find the probability that they have a mean height greater than 68.6 inches. a. 0.8188 b. 0.9005 c. 9.9671 d. 0.0021

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 2.8$ inches and $n = 64$, we have $\sigma_{\bar{x}}=\frac{2.8}{\sqrt{64}}=\frac{2.8}{8}=0.35$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x}=68.6$, $\mu = 67.6$, and $\sigma_{\bar{x}} = 0.35$. So, $z=\frac{68.6 - 67.6}{0.35}=\frac{1}{0.35}\approx2.86$.

Step3: Find the probability

We want $P(\bar{X}>68.6)$, which is equivalent to $P(Z > 2.86)$. Using the standard normal table, $P(Z\leq2.86)=0.9979$. Then $P(Z > 2.86)=1 - P(Z\leq2.86)=1 - 0.9979 = 0.0021$.

Answer:

D. 0.0021