Question

Assume that military aircraft use ejection seats designed for men weighing between 133.8 lb and 208.0 lb. If women’s weights are normally distributed with a mean of 172.6 lb and a standard deviation of 42.4 lb, what percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)? Enter your answer as a percent rounded to one decimal place (do not add a "%"); add a trailing zeros as needed. The percentage of women with weights between 133.8 and 208.0 lb is [EjectPct] percent.

Answer 1

Answer:

61.8

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 172.6, \sigma = 42.4$

What percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)?

We have to find the pvalue of Z when X = 208 subtracted by the pvalue of Z when X = 133.8 for the proportion. Then we multiply by 100 to find the percentage.

X = 208

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

$Z = \frac{208 - 172.6}{42.4}$

$Z = 0.835$

$Z = 0.835$ has a pvalue of 0.798

X = 133.8

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

$Z = \frac{133.8 - 172.6}{42.4}$

$Z = -0.915$

$Z = -0.915$ has a pvalue of 0.180

0.798 - 0.18 = 0.618

0.618*100 = 61.8%

Without the %, the answer is 61.8.