Assume that military aircraft use ejection seats designed for men weighing between 135.2135.2 lb and 206206 lb. If women's weights are normally distributed with a mean of 173.7173.7 lb and a standard deviation of 46.446.4 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications?

P(135.2 < x < 206) = 55.5%

Although, 55.5% is more than half of the women, but a whooping 44.5% of the women will still be left out, hence, a large percentage of women are still out of the range of allowable weights for ejection seats in the aircrafts.

Step-by-step explanation:

Given,

μ = mean for the women's weight = 173.7 lb

σ = standard deviation for the women's weight = 46.4 lb

To find the percentage of women's weight distributed within the limit (135.2 < x < 206)

This is a normal distribution problem

We need to find P(135.2 < x < 206.0)

First of, we normalize/standardize the two weights of interest.

The standardized score for a value is the value minus the mean then divided by the standard deviation.

For 135.2 lb

z = (x - μ)/σ = (135.2 - 173.7)/46.4 = - 0.83

For 206 lb

z = (x - μ)/σ = (206 - 173.7)/46.4 = 0.70

P(135.2 < x < 206) = P(-0.83 < z < 0.70)

We'll use data from the normal probability table for these probabilities

P(135.2 < x < 206) = P(-0.83 < z < 0.70) = P(z < 0.70) - P(z < -0.83)

P(z < 0.7) = 1 - P(z ≥ 0.7) = 1 - P(z ≤ -0.7) = 1 - 0.242 = 0.758

P(z < -0.83) = 1 - P(z ≥ -0.83) = 1 - P(z ≤ 0.83) 1 - 0.797 = 0.203

P(z < 0.70) - P(z > -0.83) = 0.758 - 0.203 = 0.555 = 55.5%

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d