HookThe flying bombs were not aiming at anything

In 1944, V-1 flying bombs were falling on London, and south Londoners were convinced the hits clustered — that certain streets were being targeted. After the war, the statistician R. D. Clarke divided \(144\ \text{km}^2\) of south London into \(576\) equal squares and counted the \(537\) recorded hits. If the bombs fell at random, the number per square should follow a Poisson distribution with mean \(\lambda = \frac{537}{576} \approx 0.932\). The model predicted about \(227\) squares with no hits; the actual count was \(229\). Predicted with exactly one hit: about \(211\); observed: \(211\). Square by square, the 'targeting' dissolved into the exact signature of pure randomness — clusters included, because genuine randomness clumps far more than intuition expects.

That is the Poisson distribution's job: counting events that occur singly, independently and at a constant average rate — bomb strikes per grid square, calls per hour, flaws per metre of cable, typos per page. One parameter \(\lambda\) controls everything: it is simultaneously the mean and the variance, a coincidence so restrictive it doubles as a diagnostic test for whether the model fits at all. This section covers the conditions that justify the model, the probability calculations, the additivity that lets you combine independent Poisson streams, and the exam's favourite finale — an exact hypothesis test on a Poisson mean from a single observation.

ModelWhen Poisson applies — the three conditions

Write \(X \sim \mathrm{Po}(\lambda)\) when \(X\) counts events in a fixed interval of time or space and three conditions hold: events occur singly (no simultaneous double-events), independently (one occurrence neither triggers nor suppresses another), and at a constant average rate (the same \(\lambda\) throughout the interval, with occurrences uniformly likely across it). All three are examinable in context, and AQA's favourite mark is the condition that fails: arrivals at a clinic during an epidemic are not independent (one infection causes others); customers entering a café are not at constant rate across a day (lunchtime spike); a machine producing flawed items in bursts violates both.

The rate scales with the interval, and this is a load-bearing fact rather than a footnote: if emails arrive at \(\lambda = 6\) per hour, then the count in \(20\) minutes is \(\mathrm{Po}(2)\) and the count in a working day of \(8\) hours is \(\mathrm{Po}(48)\). Every mixed-units question in this topic begins by rescaling \(\lambda\) to match the interval actually asked about — do it first, write it down, and the rest of the question is routine.

ModelCalculating the probabilities

The probability function is \[\mathrm{P}(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}, \qquad x = 0, 1, 2, \dots\] — an infinite tail of possible counts, weighted so the total is \(1\) (the sum is \(e^{-\lambda}\sum \frac{\lambda^x}{x!} = e^{-\lambda}e^{\lambda}\), the Maclaurin series of \(e^{\lambda}\) doing statistics work). Single values come from the formula or a calculator's Poisson pdf; cumulative probabilities \(\mathrm{P}(X \le x)\) from the calculator's cdf. The translation drills matter more than the formula: 'fewer than 3' is \(\mathrm{P}(X \le 2)\); 'at least 2' is \(1 - \mathrm{P}(X \le 1)\); 'more than 4' is \(1 - \mathrm{P}(X \le 4)\). Because the distribution has no upper limit, every 'at least' question goes through the complement.

For small cases the formula is often faster than the calculator and shows the method: \(\mathrm{P}(X = 0) = e^{-\lambda}\) and \(\mathrm{P}(X \le 1) = e^{-\lambda}(1 + \lambda)\) are worth knowing on sight — the second is the engine of most one-tailed hypothesis tests later in this section.

Worked example

A bakery sells sourdough loaves at a mean rate of \(3.2\) per hour, modelled as Poisson. Probability of exactly \(5\) sales in an hour: \[\mathrm{P}(X = 5) = \frac{e^{-3.2} \times 3.2^5}{5!} = \frac{e^{-3.2} \times 335.54\ldots}{120} \approx 0.114.\] Probability of at least one sale in a \(15\)-minute window: rescale first — \(\lambda = 3.2 \div 4 = 0.8\) — then use the complement: \(\mathrm{P}(X \ge 1) = 1 - \mathrm{P}(X = 0) = 1 - e^{-0.8} \approx 0.551.\) The rescaling line is a mark on its own; skipping straight to a calculator with the wrong \(\lambda\) is the most common route to losing every mark in the part.

ModelMean equals variance — one parameter, two jobs, one diagnostic

For \(X \sim \mathrm{Po}(\lambda)\): \[\mathrm{E}(X) = \lambda \qquad \text{and} \qquad \mathrm{Var}(X) = \lambda.\] The standard deviation is \(\sqrt{\lambda}\), not \(\lambda\) — a distinction exam questions probe deliberately. No other standard distribution ties its mean and variance together this tightly, and the equality cuts both ways. Forwards: given the rate, you know the spread for free. Backwards: given data, comparing the sample mean with the sample variance is a legitimate, examinable check on whether Poisson is a plausible model at all.

If a data set has mean \(2.1\) and variance \(4.9\), the counts are overdispersed — more variable than Poisson allows — which typically signals clustering or a rate that drifts (events arriving in bursts breach independence or constant rate). Sample variance far below the mean suggests something regularising the events, like a minimum gap between arrivals. The one-sentence verdict AQA wants pairs the numbers with a condition: 'variance \(4.9\) is much larger than mean \(2.1\), so the constant-rate/independence conditions look violated and Poisson is doubtful.' Clarke's bomb data passed exactly this audit — mean \(0.932\), variance close behind — which is why the fit was so eerily good.

MechanismAdding independent Poisson streams

If \(X \sim \mathrm{Po}(\lambda)\) and \(Y \sim \mathrm{Po}(\mu)\) are independent, then \[X + Y \sim \mathrm{Po}(\lambda + \mu).\] Independent Poisson streams merge into a single Poisson stream whose rate is the sum of the rates — calls to two independent phone lines, defects from two machines, arrivals at two doors. The same fact justifies the interval-scaling rule from earlier: the count over three hours is the sum of three independent one-hour counts, so its parameter is \(3\lambda\).

Two warnings guard the marks. First, independence is required — if the two streams are driven by the same underlying cause (two tills in the same shop during the same rush), additivity needs justifying and may fail. Second, the result is about sums of variables, and it does not extend to differences: \(X - Y\) can be negative, so it cannot be Poisson at all. When a question combines streams, name the result, state independence, give the new parameter, and only then compute.

Worked example

A help desk receives phone queries at \(2.1\) per hour and email queries, independently, at \(1.4\) per hour. Total queries in an hour: \(T \sim \mathrm{Po}(3.5)\). Probability of at most \(2\) queries in an hour: \[\mathrm{P}(T \le 2) = e^{-3.5}\left(1 + 3.5 + \frac{3.5^2}{2}\right) = e^{-3.5}(1 + 3.5 + 6.125) = 10.625\,e^{-3.5} \approx 0.321.\] The three-term bracket is the formula applied at \(x = 0, 1, 2\) with the common factor \(e^{-3.5}\) pulled out — quicker and more transparent than three separate calculator calls, and the layout examiners can follow is the layout that collects method marks when a button-press goes wrong.

CaseTesting a Poisson mean — exact probabilities, no approximations

The section's summit is a hypothesis test on \(\lambda\) using a single observed count and exact Poisson probabilities. The machinery is the standard five steps. State \(H_0: \lambda = \lambda_0\) (the historical or claimed rate, rescaled to the observation interval) and \(H_1\) — one-tailed if the claim is 'decreased' or 'increased', two-tailed for 'changed'. Then compute the tail probability of a result at least as extreme as the observation, under \(H_0\): for a 'decrease' claim with observed count \(x\), that is \(\mathrm{P}(X \le x)\). Compare with the significance level, decide, and conclude in context with appropriately hedged language.

The phrase 'at least as extreme' is where tests are won and lost: the \(p\)-value is a tail, never the single value \(\mathrm{P}(X = x)\). For two-tailed tests, compare the observed tail with \(\frac{\alpha}{2}\). The alternative formulation — building the critical region first, the set of counts whose tail probability sits below \(\alpha\) — earns the same marks and is compulsory when the question asks for it explicitly; because the distribution is discrete, the actual significance level of the critical region is usually less than the nominal \(\alpha\), and stating that actual value is a routine follow-up mark.

Worked example

A server historically logs errors at a rate of \(6\) per day. After an upgrade, the first day shows \(1\) error. Test at the \(5\%\) level whether the error rate has fallen. Hypotheses: \(H_0: \lambda = 6\), \(H_1: \lambda < 6\). Under \(H_0\), with \(X\) the daily error count, the \(p\)-value is the probability of a result at least this extreme: \[\mathrm{P}(X \le 1) = e^{-6}(1 + 6) = 7e^{-6} \approx 0.0174.\] Since \(0.0174 < 0.05\), reject \(H_0\): there is significant evidence at the \(5\%\) level that the error rate has fallen. Note the shape of the conclusion — 'significant evidence that', not 'this proves' — and note what a check of \(\mathrm{P}(X \le 2) = 25e^{-6} \approx 0.0620 > 0.05\) reveals: the critical region is \(X \le 1\), so \(1\) error only just clears the bar, and \(2\) errors would not have been significant at all.

VocabularyKey terms the mark scheme pays for

Poisson distribution

The model \(X \sim \mathrm{Po}(\lambda)\) for counts of events occurring singly, independently and at constant average rate \(\lambda\) in a fixed interval: \(\mathrm{P}(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}\).

Rate parameter λ

The mean number of events per interval. Scales with the interval: a rate of \(6\) per hour gives \(\mathrm{Po}(2)\) for \(20\) minutes and \(\mathrm{Po}(48)\) for \(8\) hours.

Mean–variance equality

For Poisson, \(\mathrm{E}(X) = \mathrm{Var}(X) = \lambda\), so the standard deviation is \(\sqrt{\lambda}\). Sample mean far from sample variance is evidence against a Poisson model.

Overdispersion

Sample variance noticeably exceeding the sample mean — counts more variable than Poisson permits, typically caused by clustering or a drifting rate.

Additivity

If \(X \sim \mathrm{Po}(\lambda)\) and \(Y \sim \mathrm{Po}(\mu)\) are independent, \(X + Y \sim \mathrm{Po}(\lambda + \mu)\). Holds for sums only — a difference of Poissons is never Poisson.

p-value (exact test)

The probability, computed under \(H_0\) with exact Poisson probabilities, of an observation at least as extreme as the one obtained — a tail probability, never a single point probability.

Critical region

The set of observed counts that would lead to rejecting \(H_0\) at the stated significance level. Discreteness means its actual significance level is usually below the nominal one.

Poisson conditions

Events occur singly, independently of one another, and at a constant average rate through the interval — the three modelling assumptions to state, check or fault in context.

TrapsMisconceptions that cost marks

“Random events should spread out evenly — clusters prove something is causing them.”

Actually: Genuine randomness clumps. Clarke's \(576\)-square analysis of \(537\) V-1 strikes matched Poisson predictions almost exactly — including squares with \(4\) or more hits — even though the strikes were unaimed. Uniform spacing is what randomness does <em>not</em> look like.

“The standard deviation of Po(λ) is λ.”

Actually: The <em>variance</em> is \(\lambda\); the standard deviation is \(\sqrt{\lambda}\). For \(\lambda = 9\) the standard deviation is \(3\), not \(9\) — conflating the two wrecks any question that asks for values 'within one standard deviation of the mean'.

“The p-value is P(X = observed value).”

Actually: It is the tail probability of results at least as extreme: for a decrease claim with observed count \(1\), \(\mathrm{P}(X \le 1)\), not \(\mathrm{P}(X = 1)\). Point probabilities shrink automatically as counts grow, so testing with them would reject almost anything.

“Doubling the interval doubles every probability.”

Actually: Doubling the interval doubles \(\lambda\), and probabilities respond through \(e^{-\lambda}\lambda^x/x!\) — nonlinearly. With \(\lambda = 0.8\) per quarter-hour, \(\mathrm{P}(X \ge 1) = 0.551\); over a half hour (\(\lambda = 1.6\)) it is \(1 - e^{-1.6} \approx 0.798\), not \(1.102\) — which is just as well, since that is not a probability.

ExamWhat examiners want

Rescale \(\lambda\) to the question's interval before anything else, and show the line — 'per 15 minutes: \(\lambda = 3.2 \div 4 = 0.8\)' is routinely a standalone mark, and using the unscaled rate poisons every number after it. Translate inequalities in writing: 'fewer than 3' becomes \(\mathrm{P}(X \le 2)\), 'at least 2' becomes \(1 - \mathrm{P}(X \le 1)\), and because the distribution is unbounded above, every 'at least' goes through the complement. When conditions are asked for, tie each one to the context rather than reciting the list — 'one email arriving does not make another more likely (independence)' scores; 'events are independent' floating free often does not.

In hypothesis tests, the exam rewards a fixed skeleton: hypotheses in terms of \(\lambda\) with the interval stated, the observed tail probability computed exactly (write the bracket — \(7e^{-6}\) — before the decimal), an explicit comparison with the significance level, and a hedged contextual conclusion: 'significant evidence that the error rate has fallen', never 'this proves the upgrade worked'. If a critical region is requested, build it by checking consecutive cumulative probabilities against \(\alpha\), state the region as an inequality on the count, and quote its actual significance level — the tail probability at the region's boundary — since the discreteness point is usually the final mark. And whenever data is offered as 'suitable for a Poisson model', compare the sample mean with the sample variance in one sentence: the equality of the two is the model's fingerprint, and naming the violated condition when they disagree is what separates the top-band answer.