The percentage of women that have weights between those limits is

Question

Assume that military aircraft use ejection seats designed for men weighing between 131.7 lb and 206 lb. If​ women's weights are normally distributed with a mean of 165.5lb and a standard deviation of 45.2 ​lb, what percentage of women have weights that are within those​ limits? Are many women excluded with those​ specifications?

Answer 1

Going to assume women's weight follows a normal distribution.

That allows using Standard normal (z) curves for probabilities.

A z value is equal to the target number minus the mean value and the difference divided by standard deviation.

For the lower limit z = (131.7 - 165.5) / 45.2 = -0.75

The upper limit z = (206 - 165.5) / 45.2 = 0.90

From Standard normal (z) curves, -0.75 gives a probability of 0.2266 and 0.90 gives 0.8159.

This means that 22.66% are going to be too light and 81.59% will not be too heavy.

We want to know how many will be excluded. 22.66% are too light.

100% - 81.59% are too heavy or 18.41%.

The total excluded is sum of those too light and too heavy. 22.66% + 18.41% = 41.07%