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.
Assume that military aircraft se ejection was designed for men weighing between 1474 1 and 212...
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