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

Answer:

$P(X>8)=P(\frac{X-\mu}{\sigma}>\frac{8-\mu}{\sigma})=P(Z>\frac{8-10}{2})=P(Z>-1)$  

And we can find this probability with the complement rule:  

$P(Z>-1)=1-P(Z<-1)=1-0.159=0.841$  

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 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:  

$z=\frac{x-\mu}{\sigma}$  

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

$P(X>8)=P(\frac{X-\mu}{\sigma}>\frac{8-\mu}{\sigma})=P(Z>\frac{8-10}{2})=P(Z>-1)$  

And we can find this probability with the complement rule:  

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