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%

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8