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How do you use the Poisson distribution to model the number of random events in a fixed interval?

Use the Poisson distribution: identify when it applies, calculate probabilities of r events given a mean rate, and use its mean and variance.

A CCEA GCSE Further Mathematics answer on the Poisson distribution, covering the conditions for its use, the probability formula for r events given a mean rate, the equality of mean and variance, and combining intervals in the Statistics unit.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
When the Poisson applies
The probability formula
Scaling the mean to the interval
Mean equals variance
Why this matters

What this dot point is asking

The Poisson distribution models the number of random events occurring in a fixed interval of time or space, and CCEA GCSE Further Mathematics introduces it as a second discrete distribution. You must recognise when a situation is Poisson, calculate the probability of exactly $r$ events given a mean rate, scale the mean to the interval in the question, use the complement for "at least", and know that the mean and variance are equal. It complements the binomial distribution for counting rare events.

When the Poisson applies

The Poisson distribution suits counts of events where there is no fixed number of trials, only an average rate. The defining conditions concern randomness, independence and a steady rate.

Typical examples are phone calls per minute, flaws per metre of cable, or accidents per week, where events happen at random but at a known average rate.

The probability formula

The formula uses the exponential constant $e$ and depends only on the mean $\lambda$ and the number of events $r$ .

Because the whole distribution is fixed by the single parameter $\lambda$ , getting the mean right for the stated interval is the most important step.

Scaling the mean to the interval

The mean rate is given per unit interval, so for a different interval you must scale it in proportion before using the formula. This is the most common place marks are won or lost.

Scale the mean and find a probability

Cars pass a point at an average of $6$ per minute, following a Poisson distribution. Find the probability of exactly $3$ cars in a $30$ second period.

step 1 Scale the mean to the interval

$30$ seconds is half a minute, so the mean is $\lambda = 6 \times \dfrac{1}{2} = 3$ cars.

step 2 Write the probability

We want $P(X = 3)$ with $\lambda = 3$ : $P(X = 3) = \dfrac{e^{-3} \times 3^3}{3!}$ .

step 3 Compute the parts

$e^{-3} = 0.0498$ , $3^3 = 27$ , and $3! = 6$ , so $P(X = 3) = \dfrac{0.0498 \times 27}{6} = \dfrac{1.344}{6}$ .

step 4 Evaluate

$P(X = 3) = 0.224$ to 3 significant figures. Scaling to half a minute was essential before applying the formula.

Mean equals variance

A distinctive feature of the Poisson distribution is that its mean and variance are equal, both being $\lambda$ . This is a useful check and a way to spot whether real data might be Poisson: if a data set has a mean and a variance that are close, a Poisson model may be appropriate. It also contrasts with the binomial, whose variance $np(1-p)$ is always less than its mean $np$ .

Why this matters

The Poisson distribution extends the unit's toolkit to events with no fixed number of trials, which the binomial cannot model. It reinforces the complement technique for "at least" probabilities, introduces the exponential constant from the Pure unit in a statistical setting, and the equal mean and variance is a memorable property that aids both calculation and model selection. It rounds out the discrete-distribution content alongside the binomial.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 3 (style)4 marksCalls arrive at a switchboard at an average rate of

3

per minute, following a Poisson distribution. Find the probability of exactly

2

calls in a given minute.

Show worked answer →

This is Poisson with mean $\lambda = 3$ per minute, and we want $r = 2$ .

Use $P(X = r) = \dfrac{e^{-\lambda}\lambda^r}{r!}$ : $P(X = 2) = \dfrac{e^{-3} \times 3^2}{2!} = \dfrac{0.0498 \times 9}{2}$ .

Compute: $\dfrac{0.4481}{2} = 0.224$ to 3 significant figures.

Marks are for identifying $\lambda$ , the formula, and the value. Forgetting the $e^{-\lambda}$ factor or the factorial is the common error.

CCEA Unit 3 (style)5 marksFlaws occur in a cable at an average of

0.5

per metre, following a Poisson distribution. Find the probability of at least one flaw in a

4

metre length.

Show worked answer →

Scale the mean to the interval: for $4$ metres, $\lambda = 0.5 \times 4 = 2$ flaws.

"At least one" is best via the complement: $P(X \ge 1) = 1 - P(X = 0)$ .

$P(X = 0) = \dfrac{e^{-2} \times 2^0}{0!} = e^{-2} = 0.1353$ , so $P(X \ge 1) = 1 - 0.1353 = 0.865$ .

Marks are for scaling the mean to $4$ metres, the complement, and the value. Using $\lambda = 0.5$ for the whole length is the usual mistake.

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Sources & how we know this

CCEA GCSE Further Mathematics specification (2330) — CCEA (2017)