*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.

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

*, stated as"*

**binomial coefficient***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 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 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 variance

*np(1-p)*, then we are able to derive information about the distribution of the

*, the count of successes*

**sample proportion***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 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 distributed*B(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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