Question

Assume that military aircraft se ejection was designed for men weighing between 1474 1 and 212 1. women's weights are normally distributed with a mean of 1005 Ib and a standard deviation of 48.31, 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 imit is (Round to two decimal places as needed) Are many women excluded with those specification? O A No, the percentage of women who are excluded, which is equal to the probability found previously shows that very few women are excluded OB. No, the percentage of women who are excluded, which is the complement of the probability found previously shows that very few women are excluded O c. Yes, the percentage of women who we excluded, which is the complement of the probability found previously shows that about half of women are excluded. OD. Yes, the percentage of women who are excluded, which is equal to the probability found previously shows that about half of women are excluded.

Answer 1

Normal distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 160.5 lb

Standard deviation = 48.3 lb

P(147.4 < X < 212) = P(X < 212) - P(X < 147.4)

= P(Z < (212 - 160.5)/48.3) - P(Z < (147.4 - 160.5)/48.3)

= P(Z < 1.07) - P(Z < -0.27)

= 0.8577 - 0.3936

= 0.4641

= 46.41%

C. Yes, the percentage of women who are excluded, which is complememt to the probability found previously, shows that about half of women are excluded.