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

Answer:

$P(32$

And we can find this probability with this difference:

$P(-1.905$

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

$P(-1.905$

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".  

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

$P(32$

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:

$P(32$

And we can find this probability with this difference:

$P(-1.905$

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

$P(-1.905$