Question

Assume that military aircraft use ejection seats designed for men weighing between 132.4132.4 lb and 217217 lb. if​ women's weights are normally distributed with a mean of 168.7168.7 lb and a standard deviation of 48.848.8 ​lb, what percentage of women have weights that are within those​ limits

Answer 1

60.93% fall within those limits.

We will use z-scores to answer this.  The formula for a z-score is:

$z=\frac{X-\mu}{\sigma}$

Using the information for the lower end, we have:
$z=\frac{132.4-168.7}{48.8}=\frac{-36.3}{48.8}=-0.74$

For the upper bound we have:
$z=\frac{217-168.7}{48.8}=\frac{48.3}{48.8}=0.99$

Using a z-table (http://www.z-table.com), we see that the area under the curve to the left of the lower limit is 0.2296.  The area under the curve to the left of the upper limit is 0.8389.  We want the middle portion of this, so we subtract the two:

0.8389-0.2296 = 0.6093 = 60.93%