Question

Costs for standard veterinary services at a local animal hospital follow a Normal distribution with a mean of $72 and a standard deviation of $21. What is the probability that one bill for veterinary services costs between $32 and $111

Answer 1

`P(32<X<111)=P((32-\mu)/(\sigma)<(X-\mu)/(\sigma)<(111-\mu)/(\sigma))=P((32-72)/(21)<Z<(111-72)/(21))=P(-1.905<z<1.857)`

And we can find this probability with this difference:

`P(-1.905<z<1.857)=P(z<1.857)-P(z<-1.905)`

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

`P(-1.905<z<1.857)=P(z<1.857)-P(z<-1.905)=0.968-0.0284=0.940`

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 costs of services of a population, and for this case we know the distribution for X is given by:

`X \sim N(72,21)`

Where
`\mu=72` and
`\sigma=21`

We are interested on this probability

`P(32<X<111)`

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(32<X<111)=P((32-\mu)/(\sigma)<(X-\mu)/(\sigma)<(111-\mu)/(\sigma))=P((32-72)/(21)<Z<(111-72)/(21))=P(-1.905<z<1.857)`

And we can find this probability with this difference:

`P(-1.905<z<1.857)=P(z<1.857)-P(z<-1.905)`

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

`P(-1.905<z<1.857)=P(z<1.857)-P(z<-1.905)=0.968-0.0284=0.940`