Question

Question Help Assume that military aircraft use ejection seats designed for men weighing between 143.6143.6 lb and 208208 lb. If​ women's weights are normally distributed with a mean of 178.6178.6 lb and a standard deviation of 48.148.1 ​lb, what percentage of women have weights that are within those​ limits? Are many women excluded with those​ specifications

Answer 1

Answer:

49.95% of women have weights that are within those​ limits. More than half of women are excluded with those specifications, so yes, there are many women excluded.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 178.6, \sigma = 48.1$

Between 143.6 lb and 208lb.

pvalue of Z when X = 208 subtracted by the pvalue of Z when X = 143.6. So

X = 208

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{208 - 178.6}{48.1}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291

X = 143.6

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{143 - 178.6}{48.1}$

$Z = -0.74$

$Z = -0.74$ has a pvalue of 0.2296

0.7291 - 0.2296 = 0.4995

49.95% of women have weights that are within those​ limits. More than half of women are excluded with those specifications, so yes, there are many women excluded.