The Poisson distribution

A-Level Maths 9709 · Topic 6.1 revision notes

Everything from the 2 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

The Poisson distribution
Approximations: binomial to Poisson, Poisson to normal

1The Poisson distribution

Syllabus 6.1

If events occur singly, independently and at random at a constant mean rate, the number \(X\) in a fixed interval satisfies \(X \sim \text{Po}(\lambda)\) with \(\text{P}(X = r) = e^{-\lambda}\dfrac{\lambda^{r}}{r!}\) for \(r = 0, 1, 2, \ldots\). The mean and the variance are both \(\lambda\), and \(\lambda\) scales in proportion to the length of the interval.

Fig. 1: the Poisson distribution with mean 3, drawn to scale. The teal bars add to P(X ≤ 2) = 0.423; the bars for r = 2 and r = 3 are equal because P(X = 3) = P(X = 2) × 3/3

Worked example: Is a Poisson model suitable?

In 100 one-minute intervals, the numbers of cars passing a point were recorded. The sample mean was 1.96 and the sample variance was 1.82. A second road, next to a set of traffic lights, gave mean 2.0 and variance 0.4. Comment on each.

For a Poisson distribution the mean equals the variance.
First road: \(1.96\) and \(1.82\) are close, so a Poisson model with \(\lambda \approx 1.96\) is reasonable.
Second road: the variance \(0.4\) is much smaller than the mean \(2.0\).
Traffic lights release cars in bunches at regular times, so the cars do not arrive independently at random.

Answer: Poisson is suitable for the first road but not for the second (variance far below the mean; arrivals are not independent).

Define \(X\), fix \(\lambda\) for the interval, then calculate.

Define: "Let \(X\) be the number of ... in ..." and write \(X \sim \text{Po}(\lambda)\).
Scale: multiply the rate by the length of the interval in the question.
Boundaries: write the event as an inequality (\(X \ge 3\), \(X \le 2\)) before calculating.
Complements: \(\text{P}(X \ge k) = 1 - \text{P}(X \le k - 1)\).
Context: state conditions and conclusions in terms of the situation.
Accuracy: keep full calculator values; give answers to 3 s.f.

Watch out!

Do not just recite "independent and random". Cambridge wants the condition in context: "the calls arrive independently of each other" or "the mean rate of calls is constant during the day". A condition with no reference to the situation usually earns nothing.

"More than 2" means \(X \ge 3\), so it is \(1 - \text{P}(X \le 2)\), not \(1 - \text{P}(X \le 3)\). Write the inequality in symbols before you calculate; most lost marks here are a boundary slipped by one.

In a two-stage question, do not add the rates to \(\lambda = 6\) for the week and find \(\text{P}(X = 0)\): that is the probability of no faults all week, a different event. Identify what is being counted at each stage.

Revisit it: The Poisson distribution lesson · worked practice

2Approximations: binomial to Poisson, Poisson to normal

Syllabus 6.1

If \(X \sim B(n, p)\) with \(n > 50\) and \(np < 5\), use \(X \approx \text{Po}(np)\). If \(X \sim \text{Po}(\lambda)\) with \(\lambda > 15\), use \(X \approx N(\lambda, \lambda)\): the variance equals the mean, and a continuity correction of \(0.5\) is needed because a discrete count is replaced by a continuous variable.

Fig. 1: Po(20) (bars, drawn to scale) and the normal curve N(20, 20). The teal bars are X ≤ 15; the approximating area stops at 15.5, half-way through the next bar

Worked example: A rare defect

1.5% of bolts made by a machine are faulty. A random sample of 200 bolts is taken. Use a suitable approximation to find the probability that at most 2 are faulty.

\(X \sim B(200, 0.015)\). Here \(n = 200 > 50\) and \(np = 3 < 5\), so use \(X \approx \text{Po}(3)\).
\(\text{P}(X \le 2) \approx e^{-3}\left(1 + 3 + \dfrac{3^{2}}{2}\right) = 8.5e^{-3}\).
\(= 0.4232\). (The exact binomial value is \(0.4215\), so the approximation is good.)

Answer: \(0.423\) (3 s.f.)

Check the conditions, choose, then correct.

Binomial, rare successes: \(n > 50\), \(np < 5\): use \(\text{Po}(np)\), no continuity correction.
\(p\) close to 1: count the failures instead.
Poisson, large mean: \(\lambda > 15\): use \(N(\lambda, \lambda)\), standard deviation \(\sqrt{\lambda}\).
Continuity correction: list the integers included, then go half a unit beyond the last one.
Justify: quote the numerical conditions in your answer.

Watch out!

State the conditions with numbers: "\(n = 200 > 50\) and \(np = 3 < 5\)". Writing "\(n\) is large and \(p\) is small" with no values often loses the mark, and using \(\text{Po}(np(1-p))\) is wrong: the Poisson parameter is the mean, \(np\).

The standard deviation of \(N(20, 20)\) is \(\sqrt{20}\), not 20. Dividing by 20 is the commonest error in this topic, closely followed by forgetting the continuity correction altogether.

Do not apply a continuity correction when approximating binomial by Poisson. Both are discrete, so \(\text{P}(X \le 2)\) stays \(\text{P}(X \le 2)\).

Revisit it: Approximations: binomial to Poisson, Poisson to normal lesson · worked practice