Question

The average homicide rate for the cities and towns in a state is 10 per 100,000 population with a standard deviation of 2. If the variable is normally distributed, what is the probability that a randomly selected town will have a homicide rate greater than 8?

Answer 1

`P(X>8)=P((X-\mu)/(\sigma)>(8-\mu)/(\sigma))=P(Z>(8-10)/(2))=P(Z>-1)`

And we can find this probability with the complement rule:

`P(Z>-1)=1-P(Z<-1)=1-0.159=0.841`

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 average homicide rate for the cities of a population, and for this case we know the distribution for X is given by:

`X \sim N(10,2)`

Where
`\mu=10` and
`\sigma=2`

We are interested on this probability

`P(X>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(X>8)=P((X-\mu)/(\sigma)>(8-\mu)/(\sigma))=P(Z>(8-10)/(2))=P(Z>-1)`

And we can find this probability with the complement rule:

`P(Z>-1)=1-P(Z<-1)=1-0.159=0.841`