# 10.5 - The Mean and Variance

10.5 - The Mean and Variance## Theorem

If \(X\) is a binomial random variable, then the **mean of *** X* is:

\(\mu=np\)

### Proof

## Theorem

If \(X\) is a binomial random variable, then the **variance of \(X\)** is:

\(\sigma^2=np(1-p)\)

and the **standard deviation of \(X\)** is:

\(\sigma=\sqrt{np(1-p)}\)

The proof of this theorem is quite extensive, so we will break it up into three parts:

### Proof

#### Part 1

#### Part 2

The definition of the expected value of a function gives us:

\(E[X(X-1)]=\sum\limits_{x=0}^n x(x-1)\times f(x)=\sum\limits_{x=0}^n x(x-1)\times \dfrac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\)

The first two terms of the summation equal zero when \(x=0\) and \(x=1\). Therefore, the bottom index on the summation can be changed from \(x=0\) to \(x=2\), as it is here:

\(E[X(X-1)]=\sum\limits_{x=2}^n x(x-1)\times \dfrac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\)

Now, let's see how we can simplify that summation:

And, here's the final part that ties all of our previous work together:

#### Part 3

## Example 10-8

The probability that a planted radish seed germinates is 0.80. A gardener plants nine seeds. Let \(X\) denote the number of radish seeds that successfully germinate? What is the average number of seeds the gardener could expect to germinate?

#### Solution

Because \(X\) is a binomial random variable, the mean of \(X\) is \(np\). Therefore, the gardener could expect, on average, \(9\times 0.80=0.72\) seeds to germinate.

What does it mean that the average is 7.2 seeds? Obviously, a seed either germinates or not. You can't have two-tenths of a seed germinating. Recall that the mean is a long-run (population) average. What the 7.2 means is... if the gardener conducted this experiment... that is, planting nine radish seeds and observing the number that germinated... over and over and over again, the average number of seeds that would germinate would be 7.2. The number observed for any particular experiment would be an integer (that is, whole seeds), but when you take the average of all of the integers from the repeated experiments, you need not obtain an integer, as is the case here. In general, the average of a discrete random variable need not be an integer.

What is the variance and standard deviation of \(X\)?

#### Solution

The variance of \(X\) is:

\(np(1-p)=9\times 0.80\times 0.2=1.44\)

Therefore, the standard deviation of \(X\) is the square root of 1.44, or 1.20.