Expected Value and SD of a Binomial Random Variable

Suppose that last quarter, TV World sold 300 television sets. If there is .20 likelihood of selling an extended warranty with each television sold, the number of extended warranties sold last quarter should be around 60 , give or take 7 or so (we will compute this later). The first number is called the expected value of the number of warranties sold; the second number is the standard deviation. Recall that X, the number of warranties sold, is a Binomial random variable. The expected value , denoted E(X)  , of a Binomial random variable X with parameters n and p  is computed as:

 

E(X) = np (4.5)

The expected value E(X)  is also called the average or mean  of X, and denoted $\mu$ . The standard deviation  of a Binomial random variable is computed as 

$$\mbox{SD}(X) = \sigma_X = \sqrt{np(1-p)}$$ (4.6)

Returning to the example, since n=300 and p=.20, we have E(X)= 300 (.20) =60.0, and $\mbox{SD}(X)= \sqrt{ 300 (.20)(.80)} = 6.93$, or approximately $60 \pm 7$.

The SD for random variables is interpreted similarly to the SD for a sample. If the store sells 300 TV sets every quarter, they won't sell exactly 60 extended warranties every time; sometimes they will sell more, sometimes they will sell less. By how much more, and how much less? The answer is, ``By 7, on the average''. Similarly, a baseball player with .200 batting average won't necessarily get 60 hits in 300 at bats . We expect him to get 60 hits, give or take 7 hits or so.

Exercise: Suppose that 5% of videos rented at Campus Video incur a late rental fee. If 700 videos were rented last week, the number that will incur a late rental fee should be around __________ give or take __________ or so.