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

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:

[tex]X \sim N(72,21)[/tex]

Where [tex]\mu=72[/tex] and [tex]\sigma=21[/tex]

We are interested on this probability

[tex]P(32<X<111)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(32<X<111)=P(\frac{32-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{111-\mu}{\sigma})=P(\frac{32-72}{21}<Z<\frac{111-72}{21})=P(-1.905<z<1.857)[/tex]

And we can find this probability with this difference:

[tex]P(-1.905<z<1.857)=P(z<1.857)-P(z<-1.905)[/tex]

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

[tex]P(-1.905<z<1.857)=P(z<1.857)-P(z<-1.905)=0.968-0.0284=0.940[/tex]