The last chapter discussed one of the most important and widely used discrete random variables in probability and statistics; the Binomial Distribution. In this chapter we will discuss what we can expect to occur during an experiment that incorporates the binomial distribution as well as how its outcomes are dispersed. What we can expect to happen during a binomial distribution, or the mean, has been calculated as $\mathrm{E(X)}=np$ where $n$ is the number of times the experiment has occurred and $p$ is the probability of a success during each experiment. How these outcomes are dispersed is referred to as the variance and is calculated as:

$\mathrm{Var(X)}=npq$ where $n$ is the number of trials, $p$ is the probability of a success, and $q$ is the probability of a failure. When this is the case we say that $\mathrm{X\sim B}\in(n,p)$. That is, “$\rm X$ is distributed binomially with $n$ trials and probability of success $p$.”

Example 1: A city wide police survey shows that $85\%$ of all violent crimes are solved each year. Suppose that in your city, $150$ such crimes are committed each year and they are each deemed independent of each other. How many of these crimes can be expected to be solved?

Explanation: The expected value, or mean, of solving $85\%$ of these $150$ crimes is:

$\mathrm{E(X)}=np$
$\rm E(X)=150\left(\dfrac{85}{100}\right)=127.5$.

In other words, if there are $150$ violent crimes committed in this city this year, then you can expect that about $127$ or $128$ of these crimes will be solved.

Example 2: Determine the variance and standard deviation of the previous statistic and interpret the results.

Explanation: As was discussed in the previous chapter, the standard deviation is the square root of the variance and describes on average how far a data value is away from the average. Since the variance of a binomial distribution is $\mathrm Var (x)=npq$ then

$\mathrm{Var(X)}=150\left(\dfrac{85}{100}\right)\left(\dfrac{15}{100}\right)$
$\mathrm{Var(X)}=19.1$.

Thus the standard deviation is $\sqrt{19.1}=4.37$. Therefore, if this city would continue to have $150$ violent crimes each year for the foreseeable future, then on average, the city can expect to solve anywhere from:

$127.5−4.37\approx 123$ and
$127.5+4.37\approx 132$ violent crimes each year.