Binomial Mean and Variance

The mean is np

Toss a fair coin 5 times. The number of heads X has the distribution $B\left(5,\frac{1}{2}\right)$B(5, 1/2), and on average half the tosses are heads: $5\times \frac{1}{2}=2.5$5 × 1/2 = 2.5. For any binomial, X ~ B(n, p), the mean is E(X) = np.

The reason is that X is a sum. Each trial on its own is worth 1 for a success and 0 for a failure, and its mean is p. X adds up n of these, so its mean is p + p + … + p = np.

The long sum agrees. For $B\left(5,\frac{1}{2}\right)$B(5, 1/2), with each probability over 32: $E\left(X\right)=\frac{0\times 1+1\times 5+2\times 10+3\times 10+4\times 5+5\times 1}{32}=\frac{0+5+20+30+20+5}{32}=\frac{80}{32}=2.5$E(X) = (0 × 1 + 1 × 5 + 2 × 10 + 3 × 10 + 4 × 5 + 5 × 1)/32 = (0 + 5 + 20 + 30 + 20 + 5)/32 = 80/32 = 2.5.

X ~ $B\left(5,\frac{1}{2}\right)$B(5, 1/2), with each probability in 32nds. The bars are symmetric about the gap between 2 and 3, and the mean, np = 2.5, is at the center of that gap, under the peak.

The variance is np(1 − p)

One trial has variance p(1 − p). The n trials are independent, and the variances of independent variables add, so the variance of X is n times p(1 − p): Var(X) = np(1 − p).

For the 5 tosses, Var$\left(X\right)=5\times \frac{1}{2}\times \frac{1}{2}=1.25$(X) = 5 × 1/2 × 1/2 = 1.25, and the standard deviation is $\sqrt{1.25}\approx 1.12$√1.25 ≈ 1.12 heads. Twenty tosses have mean $20\times \frac{1}{2}=10$20 × 1/2 = 10 and variance $20\times \frac{1}{2}\times \frac{1}{2}=5$20 × 1/2 × 1/2 = 5.

A check with the long sums

Take X ~ $B\left(4,\frac{1}{6}\right)$B(4, 1/6), the number of sixes in 4 rolls. Its probabilities, in 1296ths, are 625, 500, 150, 20 and 1. The long sum for the mean is $\frac{0\times 625+1\times 500+2\times 150+3\times 20+4\times 1}{1296}=\frac{864}{1296}=\frac{2}{3}$(0 × 625 + 1 × 500 + 2 × 150 + 3 × 20 + 4 × 1)/1296 = 864/1296 = 2/3, and np $=4\times \frac{1}{6}=\frac{2}{3}$= 4 × 1/6 = 2/3.

For the variance, $E\left(X^2\right)=\frac{1\times 500+4\times 150+9\times 20+16\times 1}{1296}=\frac{1296}{1296}=1$E(X²) = (1 × 500 + 4 × 150 + 9 × 20 + 16 × 1)/1296 = 1296/1296 = 1, so Var$\left(X\right)=1-{\left(\frac{2}{3}\right)}^2=1-\frac{4}{9}=\frac{5}{9}$(X) = 1 − (2/3)² = 1 − 4/9 = 5/9. The formula gives np$\left(1-p\right)=4\times \frac{1}{6}\times \frac{5}{6}=\frac{20}{36}=\frac{5}{9}$(1 − p) = 4 × 1/6 × 5/6 = 20/36 = 5/9.

The variance depends on p

For a fixed n, the variance np(1 − p) depends on p through the product p(1 − p). At p = 0.1 it is 0.1 × 0.9 = 0.09; at p = 0.2 it is 0.16; at p = 0.3 it is 0.21; at p = 0.5 it is 0.25. Past $\frac{1}{2}$1/2 it falls again: at p = 0.8 it is 0.16, the same as at 0.2.

So the spread is largest when $p=\frac{1}{2}$p = 1/2, where success and failure are equally likely and the result is hardest to predict.

The curve y = p(1 − p) for p from 0 to 1. It is 0 at both ends, highest at $p=\frac{1}{2}$p = 1/2, where it is 0.25, and the same at 0.2 and 0.8, where it is 0.16.

A smaller p

Take the same 5 trials with $p=\frac{1}{5}$p = 1/5. The mean falls to np $=5\times \frac{1}{5}=1$= 5 × 1/5 = 1, and the variance to np$\left(1-p\right)=5\times \frac{1}{5}\times \frac{4}{5}=0.8$(1 − p) = 5 × 1/5 × 4/5 = 0.8.

The bars crowd toward 0: $P\left(X=0\right)={\left(\frac{4}{5}\right)}^5=\frac{1024}{3125}$P(X = 0) = (4/5)⁵ = 1024/3125, about 0.33, and $P\left(X=1\right)=5\times \frac{1}{5}\times {\left(\frac{4}{5}\right)}^4=\frac{1280}{3125}$P(X = 1) = 5 × 1/5 × (4/5)⁴ = 1280/3125, about 0.41. Five successes have probability only ${\left(\frac{1}{5}\right)}^5=\frac{1}{3125}$(1/5)⁵ = 1/3125.

X ~ $B\left(5,\frac{1}{5}\right)$B(5, 1/5), with each probability in 3125ths: 1024, 1280, 640, 160, 20 and 1. The tallest bar is at 1, the mean, and the bars at 4 and 5 barely rise above the axis.

When p is 1 or 0

At p = 1 every trial succeeds, so X = n every time. One bar holds the whole probability, nothing varies, and np(1 − p) = n × 1 × 0 = 0. At p = 0 every trial fails, X = 0 every time, and the variance is 0 again.

The standard deviation

The standard deviation is $\sqrt{\text{np}\left(1-p\right)}$√(np(1 − p)). For 100 passengers who each turn up with probability 0.9, the mean is 100 × 0.9 = 90 and the variance is 100 × 0.9 × 0.1 = 9, so the standard deviation is $\sqrt{9}=3$√9 = 3 passengers.

Multiply n by 4 and the mean is multiplied by 4, but the standard deviation only by $\sqrt{4}=2$√4 = 2. For 400 passengers the mean is 360 and the standard deviation is $\sqrt{36}=6$√36 = 6. As n grows, the spread grows more slowly than the mean.

Bars for B(n, 0.15). At n = 20 the mean is np = 3 and the standard deviation is $\sqrt{20\times 0.15\times 0.85}\approx 1.60$√(20 × 0.15 × 0.85) ≈ 1.60. Drag n: the mean moves right in step with n, and the bars widen more slowly, like $\sqrt{n}$√n.

The usual mistakes

Giving n as the mean. $B\left(40,\frac{1}{4}\right)$B(40, 1/4) has mean $40\times \frac{1}{4}=10$40 × 1/4 = 10; a mean of 40 would need every trial to succeed.

Applying p twice. The mean of $B\left(40,\frac{1}{4}\right)$B(40, 1/4) is 10, not $40\times \frac{1}{4}\times \frac{1}{4}=2.5$40 × 1/4 × 1/4 = 2.5.

Giving np as the variance. For $B\left(50,\frac{1}{5}\right)$B(50, 1/5), np = 10 is the mean; the variance is $10\times \frac{4}{5}=8$10 × 4/5 = 8.

Leaving out the p. $n\left(1-p\right)=50\times \frac{4}{5}=40$n(1 − p) = 50 × 4/5 = 40 is not the variance of $B\left(50,\frac{1}{5}\right)$B(50, 1/5); np(1 − p) needs both p and 1 − p.

Adding variances where standard deviations are asked for. The mean plus two standard deviations of B(400, 0.9) is 360 + 2 × 6 = 372, not 360 + 2 × 36.