Question

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?

Answer 1

Answer:

a) $P(58<X<80)=P(\frac{58-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{58-63.4}{2.4}<Z<\frac{80-63.4}{2.4})=P(-2.25<z<6.917)$

And we can find this probability on this way:

$P(-2.25<z<6.917)=P(z<6.917)-P(z<-2.25)$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-2.15<z<6.917)=P(z<6.917)-P(z<-2.15)=1-0.0158=0.984$

And that represent 98.4%

b) $z=2.05<\frac{a-63.4}{2.4}$

And if we solve for a we got

$a=63.4 +2.05*2.4=68.32$

So the value of height that separates the bottom 98% of data from the top 2% is 68.32.  

$z=-2.33<\frac{a-63.4}{2.4}$

And if we solve for a we got

$a=63.4 -2.33*2.4=57.81$

So the value of height that separates the bottom 1% of data from the top 99% is 57.81.  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(63.4,2.4)$  

Where $\mu=63.4$ and $\sigma=2.4$

We are interested on this probability

$P(58<X<80)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(58<X<80)=P(\frac{58-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{80-\mu}{\sigma})=P(\frac{58-63.4}{2.4}<Z<\frac{80-63.4}{2.4})=P(-2.25<z<6.917)$

And we can find this probability on this way:

$P(-2.25<z<6.917)=P(z<6.917)-P(z<-2.25)$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-2.15<z<6.917)=P(z<6.917)-P(z<-2.15)=1-0.0158=0.984$

And that represent 98.4%

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.02$   (a)

$P(X<a)=0.98$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.98 of the area on the left and 0.02 of the area on the right it's z=2.05. On this case P(Z<2.05)=0.98 and P(z>2.05)=0.02

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.98$  

$P(z<\frac{a-\mu}{\sigma})=0.98$

But we know which value of z satisfy the previous equation so then we can do this:

$z=2.05<\frac{a-63.4}{2.4}$

And if we solve for a we got

$a=63.4 +2.05*2.4=68.32$

So the value of height that separates the bottom 98% of data from the top 2% is 68.32.  

For the other value we need

$P(X>a)=0.99$   (a)

$P(X<a)=0.01$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.01$  

$P(z<\frac{a-\mu}{\sigma})=0.01$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-2.33<\frac{a-63.4}{2.4}$

And if we solve for a we got

$a=63.4 -2.33*2.4=57.81$

So the value of height that separates the bottom 1% of data from the top 99% is 57.81.  

Answer 2

Correct Answer: 98.78

A. 98. 78

A) Part 2:  ​No, because only a small percentage of women are not allowed to join this branch of the military because of their height.

B. 57.8 ,    68.3