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Mean and variance of the binomial and Poisson distributions

Edexcel IAL Mathematics · WST02 Statistics 2

The binomial and Poisson distributions have simple formulae for their mean and variance. They are given in the formulae booklet and no proofs are needed in S2.

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The formulae

X ~ B(n, p): E(X) = np, Var(X) = np(1 - p).
X ~ Po(λ): E(X) = λ and Var(X) = λ.

The standard deviation is the square root of the variance.

Distribution	Mean	Variance
B(n, p)	np	np(1 - p)
Po(λ)	λ	λ

Working backwards

If a binomial has mean 6 and variance 4.2, then np = 6 and np(1 - p) = 4.2. Divide: 1 - p = 0.7, so p = 0.3 and n = 20.

Mean and variance as evidence

For a Poisson distribution the mean and variance are equal. If a sample of counts has a mean close to its variance, that supports a Poisson model; if the variance is much larger than the mean, a Poisson model is doubtful. For a binomial, the variance is always less than the mean.

Worked example

A binomial random variable X has mean 6 and variance 4.2. Find n and p.

np = 6 and np(1 - p) = 4.2
Divide: 1 - p = 4.2/6 = 0.7
p = 0.3, so n = 6/0.3

Answer: n = 20, p = 0.3

Key idea: B(n, p): mean np, variance np(1 - p). Po(λ): mean = variance = λ. Mean close to variance supports Poisson.

Check yourself

X ~ B(40, 0.15). Find E(X).
Show answer
6. np = 40 × 0.15 = 6.
X ~ B(40, 0.15). Find Var(X).
Show answer
5.1. np(1 - p) = 6 × 0.85 = 5.1.
X ~ Po(9). Find the standard deviation of X.
Show answer
3. Var(X) = λ = 9, so the standard deviation is 3.
A binomial random variable has mean 6 and variance 4.2. Find p.
Show answer
0.3. 1 - p = 4.2/6 = 0.7, so p = 0.3.
A binomial random variable has mean 6 and variance 4.2. Find n.
Show answer
20. p = 0.3, so n = 6/0.3 = 20.
X ~ Po(4). Find the probability that X is greater than its mean. Give your answer to 4 decimal places.
Show answer
0.3712. P(X > 4) = 1 - P(X ≤ 4) = 1 - 0.6288 = 0.3712.
The numbers of calls in 80 one-minute periods give Σx = 192 and Σx² = 652. Find the variance of the sample. Give your answer to 3 significant figures.
Show answer
2.39. Mean = 2.4; variance = 652/80 - 2.4² = 8.15 - 5.76 = 2.39.
The 80 one-minute periods above have mean 2.4 and variance 2.39. What does this suggest?
- A Poisson model may be suitable, because the mean and variance are close
- A Poisson model is unsuitable, because the variance is less than the mean
- A binomial model must be used, because the data are counts
- Nothing, because the mean and variance cannot be compared
Show answer
A Poisson model may be suitable, because the mean and variance are close. For Poisson, mean = variance, so close values support the model.

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Sources

Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit S2

Last updated 8 November 2026. Spotted a mistake? Tell us.