Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randomly selected, find the probability that they have a mean height
Greater than 63.0 inches.
A)0.9918
B)0.0082
C)0.2881
D)0.8989

Question

Quizplus · Accepted Answer

To find the probability that the mean height of 100 randomly selected women is greater than 63.0 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, $\sigma = 2.5$ inches and $n = 100$, so $SE = \frac{2.5}{\sqrt{100}} = \frac{2.5}{10} = 0.25$ inches. We then calculate the z-score for a mean height of 63.0 inches: $z = \frac{X - \mu}{SE} = \frac{63.0 - 63.6}{0.25} = -2.4$. Looking up the z-score of -2.4 in the standard normal distribution table or using a calculator, we find the probability of having a z-score less than -2.4. Since we want the probability of the mean height being greater than 63.0 inches, we subtract this value from 1. The probability corresponding to a z-score of -2.4 is approximately 0.9918, indicating a high likelihood that the average height of 100 randomly selected women will be greater than 63.0 inches.

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