Question

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.7 inches, and standard deviation of 5.2 inches. What is the probability that the height of a randomly chosen child is between 49.5 and 67.2 inches? Do not round until you get your your final answer, and then round to 3 decimal places.

Answer 1

`P(49.5<X<67.2)=P((49.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(67.2-\mu)/(\sigma))=P((49.5-55.7)/(5.2)<Z<(67.2-55.7)/(5.2))=P(-1.192<z<2.212)`

`P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)`

`P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)=0.987-0.117=0.870`

Explanation:

We define X the random variable that represent the heights of a population for ten year old children, and for this case we know the distribution for X is given by:

`X \sim N(55.7,5.2)`

Where
`\mu=55.7` and
`\sigma=5.2`

We want to find this probability:

`P(49.5<X<67.2)`

We can use the z score formula to solve this problem given by:

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

Using this formula we have:

`P(49.5<X<67.2)=P((49.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(67.2-\mu)/(\sigma))=P((49.5-55.7)/(5.2)<Z<(67.2-55.7)/(5.2))=P(-1.192<z<2.212)`

And we can find this probability with this difference

`P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)`

We can use tables for the normal standard distribution, excel or a calculator and we got this

`P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)=0.987-0.117=0.870`