Topics

The Poisson distribution

What it models

The Poisson distribution counts events occurring at random over an interval of time, space, or area — where there is no fixed number of trials.
Write it as:

$X \sim \text{Poisson}(\lambda)$

where $\lambda$ (lambda) is the mean number of events per interval.
Typical situations: calls arriving at a helpline per hour, faults per kilometre of cable, earthquakes above a magnitude per year, weeds per square metre, customers entering a shop per ten minutes.

The conditions

Condition	What it means	Fails when
Events occur at random	No pattern or schedule	Events are scheduled or triggered
Constant average rate	$\lambda$ is the same throughout the interval	The rate varies by time of day or season
Independence	One event does not make another more or less likely	Events cluster or trigger each other
No simultaneous events	Events happen one at a time	Events arrive in groups

The rate condition and the independence condition are the two that fail in practice, and questions are built around them. Customers arriving at a café do not arrive at a constant rate across the day; aftershocks are triggered by earthquakes.

The formula and the calculator

$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$

In practice use poissonpdf for $P(X = x)$ and poissoncdf for $P(X \leqslant x)$ , with the same inequality conversions as for the binomial.
Unlike the binomial, $X$ has no upper limit — any whole number is possible, though large values become vanishingly unlikely.

Mean and standard deviation

$\mu = \lambda \qquad \sigma = \sqrt{\lambda}$

The mean and the variance are both $\lambda$ . This is a distinctive property, and it gives you a quick way to check whether a Poisson model fits a data set: if the observed variance is much larger than the mean, the events are clustering and the Poisson model is wrong.

Changing the interval — the crucial skill

$\lambda$ is tied to a specific interval, and must be rescaled when the interval changes.
If faults occur at 3.2 per kilometre, then over 5 km:

$\lambda_{\text{new}} = 3.2 \times 5 = 16$

If calls arrive at 12 per hour, then in 20 minutes:

$\lambda_{\text{new}} = 12 \times \tfrac{20}{60} = 4$

Rescale $\lambda$ before doing anything else. Forgetting to is the single most common Poisson error, and it produces an answer that is wrong by a large factor.

Shape

Small $\lambda$ → strongly right-skewed, with the peak at 0 or 1.
Large $\lambda$ → increasingly symmetric, approaching a normal shape.

Binomial or Poisson?

The distinction is examined constantly. The test is simple:

Ask	Binomial	Poisson
Is there a fixed number of trials?	Yes — $n$ is given	No
Can you count non-events?	Yes — $n - x$ failures	No — you cannot count the calls that did not arrive
What is given?	$n$ and $p$	A rate per interval

The clearest test is the second one. "Of 40 seeds, how many germinate?" — you can count the 8 that did not, so it is binomial. "How many texts arrive in an hour?" — there is no way to count the texts that did not arrive, so it is Poisson.

Tips and Tricks

Rescale $\lambda$ to the interval in the question before calculating. Write the new $\lambda$ down explicitly: "over 15 minutes, $\lambda = 12 \times 0.25 = 3$ ".

State the distribution: " $X$ = number of faults in 5 km, $X \sim \text{Poisson}(16)$ ."

Use $\sigma = \sqrt{\lambda}$ as a check, and remember mean = variance. If a data set has a variance far above its mean, say so — it is evidence the events are clustering and the Poisson model is inappropriate.

To decide between binomial and Poisson, ask whether you could count the failures. If yes, binomial; if no, Poisson.

Two Poisson variables over the same kind of interval add: independent processes with rates $\lambda_1$ and $\lambda_2$ combine to $\text{Poisson}(\lambda_1 + \lambda_2)$ . This is often the neat route through a question about two sources of events.

Worked ExamplePoisson with a rescaled interval

A regional emergency call centre receives calls at an average rate of 7.2 per hour during the night shift, and calls arrive at random.

(a) State the distribution for the number of calls in one hour, and find the probability of exactly 5 calls. (b) Find the probability of more than 10 calls in one hour. (c) Find the probability of no calls in a 20-minute period. (d) The centre has two operators, each able to handle 6 calls per hour. Find the probability that a given hour brings more calls than the pair can handle, and comment. (e) Records over 120 night-shift hours give a mean of 7.2 calls per hour and a variance of 19.4. Comment on the Poisson model.

(a) One hour

$X \sim \text{Poisson}(\lambda = 7.2)$

$P(X = 5) = \frac{e^{-7.2}(7.2)^5}{5!} = \text{poissonpdf}(7.2, 5) = 0.12042...$

(b) More than 10 calls

Step 1 — Convert the inequality. "More than 10" excludes 10:

$P(X > 10) = 1 - P(X \leqslant 10)$

Step 2 — Compute.

$P(X \leqslant 10) = \text{poissoncdf}(7.2, 10) = 0.88664...$

$P(X > 10) = 1 - 0.88664... = 0.11336...$

(c) A 20-minute period — rescale first

Step 1 — Rescale $\lambda$ . Twenty minutes is one third of an hour:

$\lambda_{20} = 7.2 \times \frac{20}{60} = 7.2 \times \frac{1}{3} = 2.4$

Step 2 — Compute.

$P(X = 0) = \frac{e^{-2.4}(2.4)^0}{0!} = e^{-2.4} = 0.09071...$

Note what happens if you forget to rescale: using $\lambda = 7.2$ gives $e^{-7.2} = 0.00075$ , which is wrong by a factor of more than 100.

(d) Capacity

Step 1 — Find the capacity. Two operators at 6 calls per hour each handle 12 calls per hour.

Step 2 — Find the probability of exceeding it.

$P(X > 12) = 1 - P(X \leqslant 12) = 1 - \text{poissoncdf}(7.2, 12) = 1 - 0.96376... = 0.03624$

Step 3 — Interpret for the centre.

(e) Assessing the Poisson model against the data

Step 1 — Apply the defining property. For a Poisson distribution, the variance equals the mean:

$\text{predicted variance} = \lambda = 7.2$

Step 2 — Compare with the observed variance.

$\text{observed variance} = 19.4 \quad \text{against} \quad 7.2$

The observed variance is about 2.7 times the mean.

Step 3 — Conclude.

Step 4 — Explain the mechanism, since that is what earns the marks.

Over-dispersion of this kind means the constant rate or independence condition has failed, and for an emergency call centre both are plausible:

The rate is not constant across the night. Calls are far more frequent at 11pm than at 4am, so pooling the whole shift mixes a high-rate period with a low-rate one. A mixture of Poisson processes with different rates has greater variance than its mean, which produces exactly this signature.
Calls are not independent. A single incident — a crash, a storm, a power outage — generates multiple calls at once, which violates both the independence condition and the "no simultaneous events" condition.

Consequence for part (d). The 3.6% figure was computed under a model that understates the variability, so the true probability of demand exceeding capacity is higher than 3.6% — and the overloads will be concentrated in the busy early-evening hours rather than spread evenly.

What to do instead: model each part of the shift separately with its own rate, so that the constant-rate condition holds within each block, and staff to the peak block rather than to the shift average.

The Poisson distribution

Covers AS91586

What it models

The Poisson distribution counts events occurring at random over an interval of time, space, or area — where there is no fixed number of trials.
Write it as:

$X \sim \text{Poisson}(\lambda)$

where $\lambda$ (lambda) is the mean number of events per interval.
Typical situations: calls arriving at a helpline per hour, faults per kilometre of cable, earthquakes above a magnitude per year, weeds per square metre, customers entering a shop per ten minutes.

The conditions

Condition	What it means	Fails when
Events occur at random	No pattern or schedule	Events are scheduled or triggered
Constant average rate	$\lambda$ is the same throughout the interval	The rate varies by time of day or season
Independence	One event does not make another more or less likely	Events cluster or trigger each other
No simultaneous events	Events happen one at a time	Events arrive in groups

The rate condition and the independence condition are the two that fail in practice, and questions are built around them. Customers arriving at a café do not arrive at a constant rate across the day; aftershocks are triggered by earthquakes.

The formula and the calculator

$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$

In practice use poissonpdf for $P(X = x)$ and poissoncdf for $P(X \leqslant x)$ , with the same inequality conversions as for the binomial.
Unlike the binomial, $X$ has no upper limit — any whole number is possible, though large values become vanishingly unlikely.

Mean and standard deviation

$\mu = \lambda \qquad \sigma = \sqrt{\lambda}$

The mean and the variance are both $\lambda$ . This is a distinctive property, and it gives you a quick way to check whether a Poisson model fits a data set: if the observed variance is much larger than the mean, the events are clustering and the Poisson model is wrong.

Changing the interval — the crucial skill

$\lambda$ is tied to a specific interval, and must be rescaled when the interval changes.
If faults occur at 3.2 per kilometre, then over 5 km:

$\lambda_{\text{new}} = 3.2 \times 5 = 16$

If calls arrive at 12 per hour, then in 20 minutes:

$\lambda_{\text{new}} = 12 \times \tfrac{20}{60} = 4$

Rescale $\lambda$ before doing anything else. Forgetting to is the single most common Poisson error, and it produces an answer that is wrong by a large factor.

Shape

Small $\lambda$ → strongly right-skewed, with the peak at 0 or 1.
Large $\lambda$ → increasingly symmetric, approaching a normal shape.

Binomial or Poisson?

The distinction is examined constantly. The test is simple:

Ask	Binomial	Poisson
Is there a fixed number of trials?	Yes — $n$ is given	No
Can you count non-events?	Yes — $n - x$ failures	No — you cannot count the calls that did not arrive
What is given?	$n$ and $p$	A rate per interval

The clearest test is the second one. "Of 40 seeds, how many germinate?" — you can count the 8 that did not, so it is binomial. "How many texts arrive in an hour?" — there is no way to count the texts that did not arrive, so it is Poisson.

Tips and Tricks

Rescale $\lambda$ to the interval in the question before calculating. Write the new $\lambda$ down explicitly: "over 15 minutes, $\lambda = 12 \times 0.25 = 3$ ".

State the distribution: " $X$ = number of faults in 5 km, $X \sim \text{Poisson}(16)$ ."

To decide between binomial and Poisson, ask whether you could count the failures. If yes, binomial; if no, Poisson.

Worked ExamplePoisson with a rescaled interval

A regional emergency call centre receives calls at an average rate of 7.2 per hour during the night shift, and calls arrive at random.

(a) One hour

$X \sim \text{Poisson}(\lambda = 7.2)$

$P(X = 5) = \frac{e^{-7.2}(7.2)^5}{5!} = \text{poissonpdf}(7.2, 5) = 0.12042...$

(b) More than 10 calls

Step 1 — Convert the inequality. "More than 10" excludes 10:

$P(X > 10) = 1 - P(X \leqslant 10)$

Step 2 — Compute.

$P(X \leqslant 10) = \text{poissoncdf}(7.2, 10) = 0.88664...$

$P(X > 10) = 1 - 0.88664... = 0.11336...$

(c) A 20-minute period — rescale first

Step 1 — Rescale $\lambda$ . Twenty minutes is one third of an hour:

$\lambda_{20} = 7.2 \times \frac{20}{60} = 7.2 \times \frac{1}{3} = 2.4$

Step 2 — Compute.

$P(X = 0) = \frac{e^{-2.4}(2.4)^0}{0!} = e^{-2.4} = 0.09071...$

Note what happens if you forget to rescale: using $\lambda = 7.2$ gives $e^{-7.2} = 0.00075$ , which is wrong by a factor of more than 100.

(d) Capacity

Step 1 — Find the capacity. Two operators at 6 calls per hour each handle 12 calls per hour.

Step 2 — Find the probability of exceeding it.

$P(X > 12) = 1 - P(X \leqslant 12) = 1 - \text{poissoncdf}(7.2, 12) = 1 - 0.96376... = 0.03624$

Step 3 — Interpret for the centre.

(e) Assessing the Poisson model against the data

Step 1 — Apply the defining property. For a Poisson distribution, the variance equals the mean:

$\text{predicted variance} = \lambda = 7.2$

Step 2 — Compare with the observed variance.

$\text{observed variance} = 19.4 \quad \text{against} \quad 7.2$

The observed variance is about 2.7 times the mean.

Step 3 — Conclude.

Step 4 — Explain the mechanism, since that is what earns the marks.

Over-dispersion of this kind means the constant rate or independence condition has failed, and for an emergency call centre both are plausible:

The rate is not constant across the night. Calls are far more frequent at 11pm than at 4am, so pooling the whole shift mixes a high-rate period with a low-rate one. A mixture of Poisson processes with different rates has greater variance than its mean, which produces exactly this signature.
Calls are not independent. A single incident — a crash, a storm, a power outage — generates multiple calls at once, which violates both the independence condition and the "no simultaneous events" condition.