Question

Adult male height is normally distributed with a mean of 69.4 inches and a standard deviation of 2.32 inches. If an adult male is randomly selected, what is the probability that the adult male has a height between 67.5 and 70.8 inches? Round your final answer to four decimal places.

Answer 1

`P(67.5<X<70.8)=P((67.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(70.8-\mu)/(\sigma))=P((67.5-69.4)/(2.32)<Z<(70.8-69.4)/(2.32))=P(-0.819<z<0.603)`

`P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)`

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

`P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)=0.7267-0.2064=0.5203`

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".

Solution to the problem

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(69.4,2.32)`

Where
`\mu=69.4` and
`\sigma=2.32`

We are interested on this probability

`P(67.5<X<70.8)`

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

`z=(x-\mu)/(\sigma)`

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

`P(67.5<X<70.8)=P((67.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(70.8-\mu)/(\sigma))=P((67.5-69.4)/(2.32)<Z<(70.8-69.4)/(2.32))=P(-0.819<z<0.603)`

And we can find this probability with this difference:

`P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)`

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

`P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)=0.7267-0.2064=0.5203`