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6.1 The Poisson distribution

Syllabus
9709–2028–2029
Topic
6.1
Level
A2

Learning objectives

6.1.1Poisson probabilities• use formulae to calculate probabilities for the distribution Po m^h6.1.2The fact that if X Po+• use the fact that if X Po+ m^h then the mean and variance of X are each equal to m Proofs are not required.6.1.3Poisson distribution• understand the relevance of the Poisson distribution to the distribution of random events, and use the Poisson distribution as a model6.1.4Binomial distribution• use the Poisson distribution as an approximation to the binomial distribution where appropriate The conditions that n is large and p is small should be known; n > 50 and np < 5, approximately.6.1.5Normal distribution• use the normal distribution, with continuity correction, as an approximation to the Poisson distribution where appropriate. The condition that m is large should be known; 152m, approximately.

A Poisson model counts events in a fixed interval at a constant average rate

If events occur independently at constant mean rate λ per interval, X~Po(λ) and P(X=r)=e^{−λ}λ^r/r!.

Match λ to the interval length, check that events are countable and rare enough for the model, and use complements for “at least one” questions.

If a call centre averages 3 calls per minute, P(2 calls in one minute)=e^{−3}3²/2.

Changing the interval changes λ proportionally; it is not a universal parameter for every time window.

A Poisson model’s mean and variance both equal its parameter

For X~Po(λ), E(X)=λ and Var(X)=λ. For a time or area interval scaled by k, the mean becomes kλ under a constant-rate model.

Use the equality as a model check, not as a statement that every sample has equal mean and variance. Estimate λ from appropriate exposure.

If the observed average is 4 events per hour, a two-hour interval has Po(8), not Po(4).

Sample variance need not equal sample mean exactly; the equality describes the distributional parameter.

Use Poisson only for stable independent random-event counts

Check Poisson-model requirement
response a count 0,1,2,…0,1,2,\ldots in fixed time, length, area or volume
rate constant average rate over the exposure
occurrence events occur independently
simultaneity in a very small exposure, more than one event is negligible

If the average is $r$ events per unit exposure, then an exposure of size $t$ usesX\sim Po(rt).

If flaws occur randomly at an average of 0.8 per metre, the number in 5 metres can be modelled by Po(4)Po(4), provided the rate is stable and flaws do not cluster.

Question the model when the rate varies systematically, one event triggers another, observations are capped by a fixed number of trials, or the exposure itself is unclear. State the modelling assumption when context makes it relevant.

A count variable is not automatically Poisson. The theorem that independent Poisson variables add belongs to linear combinations and does not replace the model-suitability checks here.

Approximate rare binomial successes with $Po(np)$

For $X\sim B(n,p)$, the syllabus guide isn>50\quad\text{and}\quad np<5(approximately).Whensuitable,use(approximately). When suitable, useX\approx Y,\qquad Y\sim Po(\lambda),\quad\lambda=np.

Step Action
1 check large nn and small pp using the stated guide
2 calculate λ=np\lambda=np
3 keep the same integer event (=,≤,≥=,\le,\ge)
4 calculate with the Poisson formula/table; no continuity correction

If X∼B(200,0.01)X\sim B(200,0.01), then n>50n>50 and np=2<5np=2<5, so use Y∼Po(2)Y\sim Po(2). Thus P(X≥2)≈P(Y≥2)=1−P(Y=0)−P(Y=1)=1−e−2(1+2).P(X\ge2)\approx P(Y\ge2)=1-P(Y=0)-P(Y=1)=1-e^{-2}(1+2).

The approximation replaces many rare independent success opportunities by a random-event count with the same expected value npnp.

Small pp alone is insufficient. Do not use a half-unit continuity correction: both binomial and Poisson variables are discrete.

Approximate a large-mean Poisson count with $N(\lambda,\lambda)$

For $X\sim Po(\lambda)$, when $\lambda$ is large (the syllabus guide is approximately $\lambda>15$), useY\sim N(\lambda,\lambda).Hence the normal standard deviation is $\sqrt{\lambda}$.

Poisson event Corrected normal event
X≤kX\le k Y<k+0.5Y<k+0.5
X≥kX\ge k Y>k−0.5Y>k-0.5
a≤X≤ba\le X\le b a−0.5<Y<b+0.5a-0.5<Y<b+0.5

If X∼Po(25)X\sim Po(25), the guide condition holds and Y∼N(25,25)Y\sim N(25,25). Therefore P(X≥30)≈P(Y>29.5)=P(Z>29.5−255)=P(Z>0.9).P(X\ge30)\approx P(Y>29.5)=P\left(Z>\frac{29.5-25}{5}\right)=P(Z>0.9).

State the large-mean check, write the approximating normal distribution, correct the integer boundary, then standardise with λ\sqrt{\lambda}.

Do not mix in binomial conditions. For a Poisson variable both normal mean and variance are λ\lambda, but the denominator in the zz-score is λ\sqrt{\lambda}.