is the count X of voters who supportthe 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 ifthe 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.
- 2: Each observation is independent.
If these conditions are met, then X has a binomial distribution with parameters n and p, abbreviated B(n,p).
Example
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 thedisease is distributed B(100,0.7).
Note: The sampling distribution of a count variable is only well-described by the binomialdistribution 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 20rolls has a B(20,1/6) distribution. The probability of rolling more than 2 sixesin 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 cumulativedistribution function as follows:
Binomial with n = 20 and p = 0.166667 x P( X <= x) 0 0.0261 1 0.1304 2 0.3287 3 0.5665 4 0.7687 5 0.8982 6 0.9629 7 0.9887 8 0.9972 9 0.9994The corresponding graphs for the probability density function and cumulative distribution functionfor 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 than2 sixes = 1 - 0.3287 = 0.6713.
The probability that a random variable X with binomial distribution B(n,p) isequal 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 achieve2 heads in a set of four tosses is "4 choose 2", or 4!/2!2! = (4*3)/(2*1) = 6. The possibilities are {HHTT, HTHT, HTTH, TTHH, THHT, THTH}, where "H" representsa 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 ofn 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 meanof 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 flipsof 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. Thevariance 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 withsucess probability p has a binomial distribution with mean np and variancenp(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 Xdivided 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 varianceof X divided by n², or (np(1-p))/n² = (p(1-p))/n . This formulaindicates 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 rollinga 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 sampleproportion 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 thatmore than half of the voters support candidate A?
The count X of voters in the sample of 200 who support candidate A is distributedB(200,0.4). The mean of the distribution is equal to 200*0.4 = 80, and the variance is equalto 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 tothe 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 thatthe 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 probabilitythat X is less than or equal to 100, the normal approximation applies to the upper limitof the interval, 100.5. If we were interested in the probability that X is strictly lessthan 100, then we would apply the normal approximation to the lower end of the interval, 99.5.
So, applying the continuity correction and standardizing the variable X gives the following:
1 - P(X< 100)
= 1 - P(X< 100.5)
= 1 - P(Z< (100.5 - 80)/6.93)
= 1 - P(Z< 20.5/6.93)
= 1 - P(Z< 2.96) = 1 - (0.9985) = 0.0015. Since the value 100 is nearly threestandard deviations away from the mean 80, the probability of observing a count this high is extremely small.
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