The heights of American women aged 18 to 24 are normally distributed with a mean of 66 inches and a standard deviation of 2.5 inches. In order to serve in the U.S. Army, women must be between 57 inches and 79 inches tall. What percentage of women are ineligible to serve based on their height?

This website will help you do calculations like this.

http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html

I get 99.98%

To determine what percentage of women are ineligible to serve in the U.S. Army based on their height, we need to find the proportion of women whose height falls below 57 inches or above 79 inches in a normal distribution.

First, let's determine the z-scores for both the lower and upper limits. The z-score is calculated using the formula:

z = (x - μ) / σ

where x is the specific height, μ is the mean height, and σ is the standard deviation.

For the lower limit (57 inches):
z_lower = (57 - 66) / 2.5 = -3.6

For the upper limit (79 inches):
z_upper = (79 - 66) / 2.5 = 5.2

Next, we use a standard normal distribution table or a calculator to find the proportions associated with the z-scores.

The proportion below the lower limit is the proportion of the area to the left of the z-score (to the left of -3.6). Since the z-score is very large, the area under the curve is essentially 0. (You can look this up on a z-table.)

The proportion above the upper limit is the proportion of the area to the right of the z-score (to the right of 5.2). Again, since the z-score is very large, the area under the curve is essentially 0.

Therefore, the total proportion of women who are ineligible to serve based on their height is essentially 0.

In terms of percentages, we can say that 0% of women are ineligible to serve in the U.S. Army based on their height in this normal distribution.