Mean and Variance of a Binomial Distribution

For n trials with success probability p, the number of successes has mean np and variance np(1−p). The standard deviation, its square root, shows how much the outcome varies.

Explanation

For $n$ trials that each succeed with probability $p$ , the number of successes $X$ has these statistics.

E[X] = np \qquad V[X] = np(1-p) \qquad \sigma = \sqrt{np(1-p)}

The mean is exactly what intuition suggests: 100 tosses of a fair coin average 50 heads. The variance says how far from that you should expect to stray.

Example

Toss a coin 100 times, so $n = 100$ and $p = 0.5$ .

E[X] = 100 \times 0.5 = 50

V[X] = 100 \times 0.5 \times 0.5 = 25 \qquad \sigma = \sqrt{25} = 5

The mean is 50 heads with a standard deviation of 5.

Since $n$ is large, the normal approximation applies: about 95% of the time the count lands within $50 \pm 2\sigma$ , that is between 40 and 60 heads.

A result of 70 heads would sit four standard deviations from the mean — good grounds for suspecting the coin.

When variance peaks

The product $p(1-p)$ is largest at $p = 0.5$ , where it equals 0.25. A fifty-fifty chance is the most unpredictable one.

When $p$ is close to 0 or 1 the outcome is nearly certain, so there is little to vary.