Mean and variance of the binomial distribution

The notation X ~ Bin(n, p), the mean and variance of a Bernoulli distribution, the mean np and variance np(1 − p) of a binomial distribution, and finding n and p from a given mean and variance.

Notation

We write $X \sim \text{Bin}(n, p)$ when $X$ counts the successes in $n$ independent trials with probability of success $p$. The probabilities add to $1$, because by the binomial theorem $\displaystyle\sum_{k=0}^{n}\binom{n}{k}p^k q^{n-k} = (p + q)^n = 1$.

The Bernoulli distribution

A single trial gives $X = 1$ with probability $p$ and $X = 0$ with probability $q$. So $E(X) = 0 \times q + 1 \times p = p$, and $\text{Var}(X) = E(X^2) - p^2 = p - p^2 = p(1 - p)$.

Variance of a Bernoulli trial

p(1 − p) is greatest when p = 0.5, when the outcome is least predictable.

Try this

Why is the variance $0$ when $p = 0$ or $p = 1$?

The binomial distribution

Mean and variance

$$X \sim \text{Bin}(n, p): \quad E(X) = np, \qquad \text{Var}(X) = np(1 - p)$$

This follows because $X$ is the sum of $n$ independent Bernoulli variables, and means and variances of independent variables add.

Example 1●●

$X \sim \text{Bin}(20, 0.3)$. Find the mean, variance and standard deviation.

$E(X) = 20 \times 0.3 = 6$ and $\text{Var}(X) = 6 \times 0.7 = 4.2$.

$\sigma = \sqrt{4.2} \approx 2.049$.

Example 2●●●

A binomial variable has mean $12$ and variance $3$. Find $n$ and $p$.

$np = 12$ and $np(1 - p) = 3$. Dividing, $1 - p = \tfrac14$, so $p = \tfrac34$.

$n = \dfrac{12}{0.75} = 16$.

Flashcards

$X sim ext{Bin}(n, p)$ mean

$np$

$X sim ext{Bin}(n, p)$ variance

$np(1 - p)$

Bernoulli mean and variance

$p$ and $p(1 - p)$

Practice

$X \sim \text{Bin}(20, 0.3)$. Find $E(X)$.
$X \sim \text{Bin}(20, 0.3)$. Find $\text{Var}(X)$.
$X \sim \text{Bin}(20, 0.3)$. Find the standard deviation, correct to three decimal places.
A binomial variable has mean $12$ and variance $3$. Find $n$.
For the same variable, find $p$.
Find the variance of a Bernoulli trial with $p = 0.2$.
The variance of a Bernoulli trial is greatest when $p$ equals:
$\displaystyle\sum_{k=0}^{n}\binom{n}{k}p^k(1 - p)^{n-k}$ equals: