The Binomial Distribution

In many cases, it is appropriate to summarize a group of independent observations by the number of observations in the group that represent one of two outcomes. For example, the proportion of individuals in a random sample who support one of two political candidates fits this description. In this case, the statistic is the count X of voters who support the candidate divided by the total number of individuals in the group n . This provides an estimate of the parameter p , the proportion of individuals who support the candidate in the entire population.

The binomial distribution describes the behavior of a count variable X if the following conditions apply: 1: The number of observations n is fixed. 2: Each observation is independent. 3: Each observation represents one of two outcomes ("success" or "failure"). 4: The probability of "success" p is the same for each outcome.
If these conditions are met, then X has a binomial distribution with parameters n and p , abbreviated B(n,p) .

Suppose individuals with a certain gene have a 0.70 probability of eventually contracting a certain disease. If 100 individuals with the gene participate in a lifetime study, then the distribution of the random variable describing the number of individuals who will contract the disease is distributed B(100,0.7) .

Note: The sampling distribution of a count variable is only well-described by the binomial distribution is cases where the population size is significantly larger than the sample size. As a general rule, the binomial distribution should not be applied to observations from a simple random sample (SRS) unless the population size is at least 10 times larger than the sample size.

To find probabilities from a binomial distribution, one may either calculate them directly, use a binomial table, or use a computer. The number of sixes rolled by a single die in 20 rolls has a B(20,1/6) distribution. The probability of rolling more than 2 sixes in 20 rolls, P(X>2) , is equal to 1 - P(X < 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) . Using the MINITAB command "cdf" with subcommand "binomial n=20 p=0.166667" gives the cumulative distribution function as follows:

Binomial with n = 20 and p = 0.166667 x P( X 
The corresponding graphs for the probability density function and cumulative distribution function for the B(20,1/6) distribution are shown below: 

Since the probability of 2 or fewer sixes is equal to 0.3287, the probability of rolling more than 2 sixes = 1 - 0.3287 = 0.6713.
The probability that a random variable X with binomial distribution B(n,p) is equal to the value k , where k = 0, 1. n , is given by , where . 
The latter expression is known as the binomial coefficient , stated as " n choose k ," or the number of possible ways to choose k "successes" from n observations. For example, the number of ways to achieve 2 heads in a set of four tosses is "4 choose 2", or 4!/2!2! = (4*3)/(2*1) = 6. The possibilities are , where "H" represents a head and "T" represents a tail. The binomial coefficient multiplies the probability of one of these possibilities (which is (1/2)²(1/2)² = 1/16 for a fair coin) by the number of ways the outcome may be achieved, for a total probability of 6/16.
Mean and Variance of the Binomial Distribution
The binomial distribution for a random variable X with parameters n and p represents the sum of n independent variables Z which may assume the values 0 or 1. If the probability that each Z variable assumes the value 1 is equal to p , then the mean of each variable is equal to 1*p + 0*(1-p) = p , and the variance is equal to p(1-p). By the addition properties for independent random variables, the mean and variance of the binomial distribution are equal to the sum of the means and variances of the n independent Z variables, so
These definitions are intuitively logical. Imagine, for example, 8 flips of a coin. If the coin is fair, then p = 0.5. One would expect the mean number of heads to be half the flips, or np = 8*0.5 = 4. The variance is equal to np(1-p) = 8*0.5*0.5 = 2.
Sample Proportions
If we know that the count X of "successes" in a group of n observations with sucess probability p has a binomial distribution with mean np and variance np(1-p) , then we are able to derive information about the distribution of the sample proportion , the count of successes X divided by the number of observations n . By the multiplicative properties of the mean, the mean of the distribution of X/n is equal to the mean of X divided by n , or np/n = p . This proves that the sample proportion is an unbiased estimator of the population proportion p . The variance of X/n is equal to the variance of X divided by n² , or (np(1-p))/n² = (p(1-p))/n . This formula indicates that as the size of the sample increases, the variance decreases.
In the example of rolling a six-sided die 20 times, the probability p of rolling a six on any roll is 1/6, and the count X of sixes has a B(20, 1/6) distribution. The mean of this distribution is 20/6 = 3.33, and the variance is 20*1/6*5/6 = 100/36 = 2.78. The mean of the proportion of sixes in the 20 rolls, X/20 , is equal to p = 1/6 = 0.167, and the variance of the proportion is equal to (1/6*5/6)/20 = 0.007.
Normal Approximations for Counts and Proportions
For large values of n , the distributions of the count X and the sample proportion are approximately normal. This result follows from the Central Limit Theorem. The mean and variance for the approximately normal distribution of X are np and np(1-p) , identical to the mean and variance of the binomial( n,p ) distribution. Similarly, the mean and variance for the approximately normal distribution of the sample proportion are p and (p(1-p)/n) .
Note: Because the normal approximation is not accurate for small values of n , a good rule of thumb is to use the normal approximation only if np > 10 and np(1-p) > 10.
For example, consider a population of voters in a given state. The true proportion of voters who favor candidate A is equal to 0.40. Given a sample of 200 voters, what is the probability that more than half of the voters support candidate A?
The count X of voters in the sample of 200 who support candidate A is distributed B(200,0.4) . The mean of the distribution is equal to 200*0.4 = 80, and the variance is equal to 200*0.4*0.6 = 48. The standard deviation is the square root of the variance, 6.93. The probability that more than half of the voters in the sample support candidate A is equal to the probability that X is greater than 100, which is equal to 1- P(X < 100).
To use the normal approximation to calculate this probability, we should first acknowledge that the normal distribution is continuous and apply the continuity correction . This means that the probability for a single discrete value, such as 100, is extended to the probability of the interval (99.5,100.5). Because we are interested in the probability that X is less than or equal to 100, the normal approximation applies to the upper limit of the interval, 100.5. If we were interested in the probability that X is strictly less than 100, then we would apply the normal approximation to the lower end of the interval, 99.5.