Mathematics, 18.03.2020 02:25, monsterking4549

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?

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answered: Ilcienne9736

$P(X8)=P(\frac{X-\mu}{\sigma}\frac{8-\mu}{\sigma})=P(Z\frac{8-10}{2})=P(Z-1)$

And we can find this probability with the complement rule:

P(Z-1)=1-P(Z

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(X8)

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:

$P(X8)=P(\frac{X-\mu}{\sigma}\frac{8-\mu}{\sigma})=P(Z\frac{8-10}{2})=P(Z-1)$

And we can find this probability with the complement rule:

P(Z-1)=1-P(Z

answered: Guest

eve and marks ratios are equivalent

step-by-step explanation:

answered: Guest

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step-by-step explanation:

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