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Mathematics (AS & A Level) 9709 · Chapter 28

The Poisson distribution

Paper 6. Poisson probabilities, the mean and variance, when the model fits, and approximations using the Poisson for the binomial and the normal for the Poisson.

Written for the 2026–2027 syllabus · 4 min read

6.1 The Poisson distribution

When to use it

X∼Po(λ)X \sim \mathrm{Po}(\lambda) counts how many events happen in a fixed interval of time or space, such as calls per hour or flaws per metre of cloth.

The model fits when events occur:

  • at random
  • independently of each other
  • singly, not at exactly the same moment
  • at a constant average rate

Probabilities

P(X=2)=e−3322=0.224\mathrm{P}(X = 2) = \mathrm{e}^{-3}\frac{3^2}{2} = 0.224

P(X≤1)=e−3(1+3)=0.199P(X≥2)=1−0.199=0.801\mathrm{P}(X \leq 1) = \mathrm{e}^{-3}(1 + 3) = 0.199 \qquad \mathrm{P}(X \geq 2) = 1 - 0.199 = 0.801

Mean and variance

So a Poisson model is plausible only when the mean and variance of the data are close. A big difference between them suggests that it does not fit.

They are close, so this supports using Po(3.1)\mathrm{Po}(3.1).

Scaling the interval

The mean grows in proportion to the size of the interval.

For 30 minutes, λ=1\lambda = 1, so P(0)=e−1=0.368\mathrm{P}(0) = \mathrm{e}^{-1} = 0.368.

For 3 hours, λ=6\lambda = 6, so P(5)=e−6655!=0.161\mathrm{P}(5) = \mathrm{e}^{-6}\dfrac{6^5}{5!} = 0.161.

Using the Poisson for a binomial

For a large number of trials with a small probability, the two distributions are close.

The working is much shorter this way, since the binomial needs very large and very small numbers.

X∼B(200,0.01)X \sim \mathrm{B}(200, 0.01), with n>50n > 50 and np=2<5np = 2 < 5, so use Po(2)\mathrm{Po}(2).

P(X≤1)=e−2(1+2)=0.406\mathrm{P}(X \leq 1) = \mathrm{e}^{-2}(1 + 2) = 0.406

Using the normal for a Poisson

For a large mean, the Poisson distribution becomes bell-shaped.

Use Y∼N(20,20)Y \sim \mathrm{N}(20, 20), and X≥25X \geq 25 becomes Y>24.5Y > 24.5:

z=24.5−2020=1.0061−Φ(1.006)=0.157z = \frac{24.5 - 20}{\sqrt{20}} = 1.006 \qquad 1 - \Phi(1.006) = 0.157

Choosing an approximation

Starting fromConditionsUse
B(n,p)\mathrm{B}(n, p)n>50n > 50 and np<5np < 5Po(np)\mathrm{Po}(np)
B(n,p)\mathrm{B}(n, p)np>5np > 5 and n(1−p)>5n(1 - p) > 5N(np,np(1−p))\mathrm{N}(np, np(1 - p)), with a continuity correction
Po(λ)\mathrm{Po}(\lambda)λ>15\lambda > 15N(λ,λ)\mathrm{N}(\lambda, \lambda), with a continuity correction

Finding a value

Some questions give a probability and ask for λ\lambda.

e−λ=0.2\mathrm{e}^{-\lambda} = 0.2, so λ=−ln⁡0.2=1.61\lambda = -\ln 0.2 = 1.61.

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