Assume that the heights of men are normally distributed with a mean of 69.5 inches and a standard deviation of 2.1 inches. If 36 men are randomly selected, find the probability that they have a mean height greater than
70.5 inches.
A)0.0021
B)0.8188
C)0.9005
D)9.9671

Question

Quizplus · Accepted Answer

To find the probability that the mean height of 36 randomly selected men is greater than 70.5 inches, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 2.1$ inches and $n = 36$, so $SE = \frac{2.1}{\sqrt{36}} = \frac{2.1}{6} = 0.35$ inches. Next, we calculate the z-score, which is $z = \frac{\bar{x} - \mu}{SE}$, where $\bar{x}$ is the sample mean and $\mu$ is the population mean. Plugging in the values, we get $z = \frac{70.5 - 69.5}{0.35} = \frac{1}{0.35} = 2.857$. Looking up 2.857 in the z-table gives us the area to the left of this z-score. To find the probability of being greater than 70.5 inches, we subtract this area from 1. The area to the left of z = 2.857 is approximately 0.9979, so the probability of being greater than 70.5 inches is $1 - 0.9979 = 0.0021$.

Assume That the Heights of Men Are Normally Distributed with a Mean