Jet Ejection Seats The U.S. Air Force once used ACES-II ejection seats designed for men weighing between 140 lb and 211 lb. Given that women’s weights are normally distributed with a mean of 171.1 lb and a standard deviation of 46.1 lb (based on data from the National Health Survey), what percentage of women have weights that are within those limits? Were many women excluded with those past specifications?

Women have weights that are normally distributed with a mean of 171.1 lb anda standard deviationof 46.1 lb.

ACES-II ejection seats were designed for men who have weights between 140 lb and 211 lb.

The percentage of women who have weights within the limits of 140 lb and 211 lb is computed using respective z-scores.

For$x_1=211$, the z-score is given by:

$$\begin{gathered} z_1=\frac{x_1-\mu }{\sigma } \\ =\frac{211-171.1}{46.1} \\ =0.87 \end{gathered}$$

For$x_2=140$, the cumulative area is given by:

$$\begin{gathered} z_2=\frac{x-\mu }{\sigma } \\ =\frac{140-171.1}{46.1} \\ =-0.67 \end{gathered}$$

Referring to the standard normal table for positive z score, the cumulative probability of 0.87 is obtained from the cell intersection for rows 0.8 and the column value 0.07,which implies is 0.8078.

Referring to the standard normal table for negative z score, the cumulative probability of -0.67 is obtained from the cell intersection for rows -0.6 and the column value 0.07,which is 0.2514.

Thus,

$$\begin{gathered} P\left(Z<0.87\right)=0.8078 \\ P\left(Z<-0.67\right)=0.2514 \end{gathered}$$

The probability that random women have a weight between 140 and 211 is:

$$\begin{gathered} P\left(-0.67<Z<0.87\right)=\text{Area}\text{less}\text{than}0.87-\text{Area}\text{less}\text{than}-0.67 \\ =P\left(Z<0.87\right)-P\left(Z<-0.68\right) \\ =0.8078-0.2514 \\ =0.5564 \end{gathered}$$

Expressing the result as a percentage, about 55.64%of women have weights that are within those limits.

The probability that the weight of a woman does not lie between the limits is a complementary event of the event that weight lies between the limits.

Thus,

$$\begin{gathered} \text{Probability of women being excluded}=1-\text{Probability of women have weight within those limits} \\ =1-0.5564 \\ =0.4436 \end{gathered}$$

Thus, about 44.36 % chances are that the women's weight is excluded from the past specifications, which is large. Therefore, many women are excluded.