Questions: Assume that military aircraft use ejection seats designed for men weighing between 140.7 lb and 205 lb. If women's weights are normally distributed with a mean of 163.2 lb and a standard deviation of 46.9 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? The percentage of women that have weights between those limits is 49.79%. (Round to two decimal places as needed.) Are many women excluded with those specifications? A. Yes, the percentage of women who are excluded, which is equal to the probability found previously, shows that about half of women are excluded. B. No, the percentage of women who are excluded, which is equal to the probability found previously, shows that very few women are excluded. C. No, the percentage of women who are excluded, which is the complement of the probability found previously, shows that very few women are excluded. D. Yes, the percentage of women who are excluded, which is the complement of the probability found previously, shows that about half of women are excluded.

Solution Steps

To determine the percentage of women whose weights fall within the specified limits of \(140.7 \, \text{lb}\) and \(205 \, \text{lb}\), we first calculate the Z-scores for these bounds using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

where:

• \(X\) is the value,
• \(\mu = 163.2 \, \text{lb}\) (mean weight of women),
• \(\sigma = 46.9 \, \text{lb}\) (standard deviation of women's weight).

Calculating the Z-scores:

• For the lower bound \(140.7 \, \text{lb}\):

\[ Z_{start} = \frac{140.7 - 163.2}{46.9} \approx -0.4797 \]

• For the upper bound \(205 \, \text{lb}\):

\[ Z_{end} = \frac{205 - 163.2}{46.9} \approx 0.8913 \]

Next, we find the probability that a woman's weight is between these two Z-scores using the cumulative distribution function \( \Phi \):

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.8913) - \Phi(-0.4797) \]

Using the standard normal distribution values, we find:

\[ P \approx 0.4979 \]

Thus, the probability that a woman's weight is between \(140.7 \, \text{lb}\) and \(205 \, \text{lb}\) is:

\[ P \times 100 \approx 49.79\% \]

To find the percentage of women excluded from the weight limits, we calculate the complement of the probability:

\[ \text{Percentage Excluded} = 100\% - P \times 100 \approx 100\% - 49.79\% \approx 50.21\% \]

Since approximately \(50.21\%\) of women are excluded based on the weight limits, we conclude that:

• The percentage of women who are excluded shows that about half of women are excluded.

Final Answer