Answer:
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%
