# 10.2 - Is X Binomial?

10.2 - Is X Binomial?
Binomial Random Variable

A discrete random variable $$X$$is a binomial random variable if:

• An experiment, or trial, is performed in exactly the same way $$n$$ times.
• Each of the $$n$$trials has only two possible outcomes. One of the outcomes is called a "success," while the other is called a "failure." Such a trial is called a Bernoulli trial.
• The $$n$$ trials are independent.
• The probability of success, denoted $$p$$, is the same for each trial. The probability of failure is $$q=1-p$$.
• The random variable $$X=$$the number of successes in the $$n$$ trials.

## Example 10-2

A coin is weighted in such a way so that there is a 70% chance of getting a head on any particular toss. Toss the coin, in exactly the same way, 100 times. Let $$X$$equal the number of heads tossed. Is $$X$$a binomial random variable?

Yes, $$X$$ is a binomial random variable, because:

1. The coin is tossed in exactly the same way 100 times.
2. Each toss results in either a head (success) or a tail (failure).
3. One toss doesn't affect the outcome of another toss. The trials are independent.
4. The probability of getting a head is 0.70 for each toss of the coin.
5. $$X$$ equals the number of heads (successes).

## Example 10-3

A college administrator randomly samples students until he finds four that have volunteered to work for a local organization. Let $$X$$ equal the number of students sampled. Is $$X$$ a binomial random variable?

No, $$X$$ is not a binomial random variable, because the number of trials $$n$$was not fixed in advance, and $$X$$ does not equal the number of volunteers in the sample.

## Example 10-4

A Quality Control Inspector (QCI) investigates a lot containing 15 skeins of yarn. The QCI randomly samples (without replacement) 5 skeins of yarn from the lot. Let $$X$$equal the number of skeins with acceptable color. Is $$X$$ a binomial random variable?

No, $$X$$ is not a binomial random variable, because $$p$$, the probability that a randomly selected skein has acceptable color changes from trial to trial. For example, suppose, unknown to the QCI, that 9 of the 15 skeins of yarn in the lot are acceptable. For the first trial, $$p$$equals $$\frac{9}{15}$$. However, for the second trial, $$p$$equals either $$\frac{9}{14}$$ or $$\frac{8}{14}$$depending on whether an acceptable or unacceptable skein was selected in the first trial. Rather than being a binomial random variable, $$X$$ is a hypergeometric random variable. If we continue to assume that 9 of the 15 skeins of yarn in the lot are acceptable, then $$X$$ has the following probability mass function:

$$f(x)=P(X=x)=\dfrac{\dbinom{9}{x} \dbinom{6}{5-x}}{\dbinom{15}{5}}$$ for $$x=0, 1, \ldots, 5$$

## Example 10-5

A Gallup Poll of $$n = 1000$$ random adult Americans is conducted. Let$$X$$equal the number in the sample who own a sport utility vehicle (SUV). Is $$X$$ a binomial random variable?

No, $$X$$ is technically a hypergeometric random variable, not a binomial random variable, because, just as in the previous example, sampling takes place without replacement. Therefore, $$p$$, the probability of selecting an SUV owner, has the potential to change from trial to trial. To make this point concrete, suppose that Americans own a total of $$N=270,000,000$$ cars. Suppose too that half (135,000,000) of the cars are SUVs, while the other half (135,000,000) are not. Then, on the first trial, $$p$$equals $$\frac{1}{2}$$ (from 135,000,000 divided by 270,000,000). Suppose an SUV owner was selected on the first trial. Then, on the second trial, $$p$$ equals 134,999,999 divided by 269,999,999, which equals.... punching into a calculator... 0.499999... Hmmmmm! Isn't that 0.499999... close enough to $$\frac{1}{2}$$ to just call it $$\frac{1}{2}$$?Yes...that's what we do!
In general, when the sample size $$n$$is small in relation to the population size $$N$$, we assume a random variable $$X$$, whose value is determined by sampling without replacement, follows (approximately) a binomial distribution. On the other hand, if the sample size $$n$$is close to the population size $$N$$, then we assume the random variable $$X$$ follows a hypergeometric distribution.