 Javion Henry

2022-07-19

A survey found that​ women's heights are normally distributed with mean 63.4
in and standard deviation 2.4
in. A branch of the military requires​ women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?
a) The percentage of women who meet the height requirement is:
The website says I am supposed to use $\frac{z-\lambda }{2.4}$
doing so gives me $\frac{16.6}{2.4}=7.086$ This value literally is useless. What am I doing wrong? phravincegrln2

Expert

It is helpful to define some notation.

Let X be a random variable that describes the height in inches of a randomly selected woman from the population. Then, we are told
$X\sim \mathrm{Normal}\left(\mu =63.4,\sigma =2.4\right),$
that is, the mean height is 63.4 inches and the standard deviation of the height is 2.4 inches. Now, we are also told that in order for a woman to meet the military's height requirement, we need
$58\le X\le 80.$
So the probability that a randomly selected woman meets this requirement is written as
$Pr\left[58\le X\le 80\right].$
Recalling that if X is a normal random variable with mean $\mu$ and standard deviation $\sigma$, the transformation
$Z=\frac{X-\mu }{\sigma }$
makes $Z\sim \mathrm{Normal}\left(0,1\right)$; that is to say, the random variable obtained by subtracting 63.4 from the observed height, and dividing by 2.4, is normally distributed with mean 0 and standard deviation 1. So
$Pr\left[58\le X\le 80\right]=Pr\left[\frac{58-63.4}{2.4}\le \frac{X-\mu }{\sigma }\le \frac{80-63.4}{2.4}\right]=Pr\left[-2.25\le Z\le 6.91667\right].$
What we have done here is called standardizing $X$: rather than dealing with a normal distribution with arbitrary mean and standard deviation, we have converted the original probability into an equivalent probability for a standard normal random variable $Z$. Doing this lets us more conveniently compute the probability using a statistical table, or a computer/calculator. To perform the computation, we note that if
$\mathrm{\Phi }\left(z\right)={F}_{Z}\left(z\right)=Pr\left[Z\le z\right]$
is the cumulative distribution function for a standard normal random variable, then
$Pr\left[-2.25\le Z\le 6.91667\right]=Pr\left[Z\le 6.91667\right]-Pr\left[Z<-2.25\right]=\mathrm{\Phi }\left(6.91667\right)-\mathrm{\Phi }\left(-2.25\right).$
The values 6.91667 and −2.25 are what we call $z$-scores. Most statistical tables don't have z-scores as high as 6.91667 because the probability of $Z$ exceeding such a large value is extraordinarily small--equivalently, $Pr\left[Z\le 6.91667\right]\approx 1$. We can use a computer to evaluate it:
$Pr\left[Z\le 6.91667\right]=0.9999999999976880282593536391778042314419249724181\dots .$
So to a precision of about one trillionth (${10}^{-12}$), this probability is effectively 1. What this means in terms of heights of women, the chance of a randomly selected woman being taller than 80 inches is less than 1 in approximately 432 billion, according to the model provided in the question.

However, what's $Pr\left[X<58\right]=Pr\left[Z\le -2.25\right]=\mathrm{\Phi }\left(-2.25\right)$? Well, we can look this up in a table: we get
$\mathrm{\Phi }\left(-2.25\right)\approx 0.0122245,$
which is about 1 in 82 women. So our answer should be
$Pr\left[58\le X\le 80\right]\approx 1-0.0122245=0.987776,$
that is to say, about 98.8% of all women from this population should meet the military guidelines for height.

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