What is a Poisson Distribution and how is it used in Edexcel International A Levels Statistics 2?

A Poisson Distribution is a probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space. It is used in Edexcel International A Levels Statistics 2 to model scenarios where events occur independently and at a constant average rate, such as the number of emails received in an hour or the number of cars passing through a toll booth.

How do you calculate the probability of an event using a Poisson Distribution?

To calculate the probability of observing k events in a Poisson Distribution, you use the formula P(X = k) = (λ^k * e^(-λ)) / k!, where λ is the average rate of occurrence, k is the number of events, and e is the base of the natural logarithm. This formula is essential for solving problems in the Statistics 2 module of the Edexcel International A Levels curriculum.

What are the key characteristics of a Poisson Distribution?

The key characteristics of a Poisson Distribution include: events occur independently, the average rate (λ) is constant, and two events cannot occur at the exact same instant. These properties make it suitable for modeling random events in a fixed interval, a concept frequently explored in Edexcel International A Levels Mathematics.

When should a Poisson Distribution be used instead of a Binomial Distribution?

A Poisson Distribution is typically used instead of a Binomial Distribution when the number of trials is very large, the probability of success is very small, and the events occur independently. This is because the Poisson Distribution is a limiting case of the Binomial Distribution under these conditions, a concept taught in the Statistics 2 section of the Edexcel International A Levels.

How can you approximate a Binomial Distribution using a Poisson Distribution?

A Binomial Distribution can be approximated by a Poisson Distribution when the number of trials (n) is large and the probability of success (p) is small, such that the product np (mean of the distribution) is moderate. This approximation is useful in simplifying calculations and is a technique covered in the Edexcel International A Levels Statistics 2 curriculum.

Why is "Writing $P(X \ge c) = 1 - P(X \le c)$ instead of $1 - P(X \le c - 1)$" a common mistake in Poisson Distributions?

Why it happens: Same off-by-one error as binomial. How to avoid it: Always expand: P(X 8) = P(X = 8) + P(X = 9) + = 1 - P(X 7). Source: WST02/01 examiner reports — annual

Why is "Forgetting to rescale $\lambda$ for a different time interval" a common mistake in Poisson Distributions?

Why it happens: Students treat as fixed. How to avoid it: Always identify the question's interval and multiply by the ratio. 'Rate 2 per hour, 30 minutes: = 1.' Source: WST02/01 examiner reports

Why is "Adding Poisson parameters when events are NOT independent" a common mistake in Poisson Distributions?

Why it happens: Students apply the rule without checking. How to avoid it: Independence MUST be stated (or evident from context). Two correlated streams (e.g. flu cases at two adjacent clinics) cannot simply be added.

Why is "Confusing $\lambda$ with an integer count" a common mistake in Poisson Distributions?

Why it happens: Students expect to look like a count. How to avoid it: is a MEAN, so non-integer values like Po(2.4) or Po(0.5) are normal and correct.

Why is "Stating mean $\ne$ variance for a Poisson" a common mistake in Poisson Distributions?

Why it happens: Habit from other distributions. How to avoid it: MEMORISE: E(X) = Var(X) = . This is the diagnostic feature of the Poisson.

Statistics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Poisson Distributions

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Poisson Distributions Study Notes — Edexcel IAL Mathematics S2 (WST02/01, 2018 spec — 2026 onwards)

$X \sim \text{Po}(\lambda)$ counts independent events at a constant rate over a fixed interval. PMF $e^{-\lambda} \lambda^{r}/r!$ . Signature property $E(X) = \text{Var}(X) = \lambda$ . Scaling and additivity make Poisson a powerful modelling tool for arrivals, accidents, defects, and rare events generally. Foundation for the Poisson hypothesis test and the binomial-to-Poisson approximation.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

S2 2.1 — Understand the Poisson distribution as a model for rare events.
S2 2.2 — Calculate Poisson probabilities (exact, cumulative, tail) using PMF / tables / calculator.
S2 2.3 — Apply $E(X) = \text{Var}(X) = \lambda$ .
S2 2.4 — Rescale $\lambda$ for different intervals.
S2 2.5 — Combine independent Poissons by adding parameters.
S2 2.6 — Judge whether Poisson conditions hold in context.

Setup — when Poisson is appropriate

Four conditions: independence, constant rate, single occurrences, rare events.

Definition. $X \sim \text{Po}(\lambda)$ models the count of events over a fixed interval (time, space, volume) when:

Events occur INDEPENDENTLY — one event does not affect the probability of another.
Events occur at a CONSTANT AVERAGE RATE $\lambda$ per unit interval.
Events occur SINGLY — two events do not happen at the same instant.
Events are RARE relative to opportunities (small probability per tiny sub-interval).

Examples that fit.

Phone calls arriving at a switchboard.
Customers entering a shop.
Radioactive decay events per second.
Typos per page in a long manuscript.
Goals scored in football matches (approximately).

Examples that DO NOT fit.

Heavy clustering (events tend to come in bursts) — independence fails.
Rate varying with time (e.g. rush hour) — constant rate fails.
Fixed maximum count (e.g. dice rolls) — Poisson allows arbitrarily large counts.

Connection to binomial. $B(n, p)$ approaches $\text{Po}(np)$ when $n$ is large and $p$ is small (covered in the Approximations topic). Hence Poisson is the natural model for 'binomial with $n \to \infty$ , $p \to 0$ , $np = \lambda$ '.

Four conditions: independent, constant rate, singly, rare.
Used for counts of events over intervals.
Parameter $\lambda$ is the mean count per interval.

PMF and cumulative probabilities

$P(X = r) = e^{-\lambda} \lambda^{r}/r!$ . Cumulative from tables / calculator.

PMF.

$P(X = r) = e^{-\lambda} \dfrac{\lambda^{r}}{r!}, \quad r = 0, 1, 2, \ldots$

Support is all non-negative integers (no upper limit).

Worked PMF example. $\text{Po}(4)$ , $P(X = 3)$ :

$P(X = 3) = e^{-4} \dfrac{64}{6} \approx 0.01832 \times 10.667 \approx 0.1954.$

Cumulative.

$P(X \le c) = \sum_{r=0}^{c} e^{-\lambda} \dfrac{\lambda^{r}}{r!}.$

The Edexcel booklet tabulates cumulative Poisson for selected $\lambda$ values ( $0.5, 1.0, 1.5, \ldots, 10.0$ ). For other $\lambda$ , use calculator PoissonCD $(c, \lambda)$ .

Upper tail.

$P(X \ge c) = 1 - P(X \le c - 1).$

Same off-by-one trap as binomial — note the $c - 1$ .

Interval.

$P(a \le X \le b) = P(X \le b) - P(X \le a - 1).$

Quick check. For small $\lambda$ , $P(X = 0) = e^{-\lambda}$ is often a useful sanity check.

PMF: $e^{-\lambda} \lambda^{r}/r!$ .
$P(X = 0) = e^{-\lambda}$ .
$P(X \le c)$ : tables or PoissonCD.
$P(X \ge c) = 1 - P(X \le c - 1)$ .

Mean and variance equal $\lambda$

$E(X) = \text{Var}(X) = \lambda$ — diagnostic for Poisson data.

Formulae (in the booklet).

$E(X) = \lambda, \quad \text{Var}(X) = \lambda.$

Diagnostic use. When checking whether real data could be Poisson-distributed:

Compute sample mean $\bar x$ .
Compute sample variance $s^{2}$ .
If $\bar x \approx s^{2}$ , Poisson is a reasonable model.
If $s^{2} > \bar x$ , the data is OVERDISPERSED (more variable than Poisson) — clustering suspected.
If $s^{2} < \bar x$ , the data is UNDERDISPERSED — events more regular than random.

Standard deviation. $\sigma = \sqrt{\lambda}$ .

Why mean = variance. A Poisson is the limit of $B(n, p)$ as $n \to \infty, p \to 0, np = \lambda$ . Then $np = \lambda$ (mean) and $np(1-p) \to \lambda$ (variance, since $1 - p \to 1$ ). So both equal $\lambda$ .

$E(X) = \lambda$ and $\text{Var}(X) = \lambda$ — both!
SD: $\sqrt{\lambda}$ .
Mean $\approx$ variance test for fit.
Over- / underdispersion red-flags non-Poisson data.

Scaling — different intervals

Rate $\lambda$ per unit $\to \text{Po}(\lambda t)$ over $t$ units.

Key fact. A Poisson process has counts proportional to interval length:

$X(t) \sim \text{Po}(\lambda t)$

where $\lambda$ is the rate per unit interval and $t$ is the length of the observation window.

Worked example. Customers arrive at $2.4$ per hour.

In $1$ hour: $\text{Po}(2.4)$ .
In $2$ hours: $\text{Po}(4.8)$ .
In $30$ minutes: $\text{Po}(1.2)$ .
In $15$ minutes: $\text{Po}(0.6)$ .

Practical tip. Always identify the rate's natural interval (per hour, per page, per shift) and multiply by the question's time factor. Write 'so $\lambda_{\text{new}} = \ldots$ ' explicitly.

Disjoint intervals are independent. Counts in disjoint intervals are independent Poisson — useful for compound questions ('probability of $\ge 5$ in the first hour AND $\le 2$ in the second hour').

Rescale $\lambda$ by the time / space factor.
$\lambda$ over $t$ units = $\lambda t$ .
Disjoint intervals are independent.
Always state the new $\lambda$ explicitly.

Sum of independent Poissons

Independent $\text{Po}(\lambda_1) + \text{Po}(\lambda_2) = \text{Po}(\lambda_1 + \lambda_2)$ .

Additivity property. If $X_1 \sim \text{Po}(\lambda_1)$ and $X_2 \sim \text{Po}(\lambda_2)$ are independent, then

$X_1 + X_2 \sim \text{Po}(\lambda_1 + \lambda_2).$

Extends to any number of independent Poissons.

Why. Independent events of two types can be re-classified as 'event of any type'. The combined process retains the Poisson properties: independence, constant rate $\lambda_1 + \lambda_2$ , singleness, rarity.

Worked example. Email queries $\text{Po}(3)$ per hour, phone queries $\text{Po}(2)$ per hour, independent. Total queries in an hour: $\text{Po}(5)$ .

Critical caveat. Independence is essential. If the two streams are correlated (e.g. flu cases at two clinics in the same town), the sum is NOT Poisson.

Splitting (thinning). The reverse process: if $\text{Po}(\lambda)$ events are independently classified as 'type A' with probability $q$ , the type-A counts are $\text{Po}(\lambda q)$ (not directly tested at S2 but useful).

$\text{Po}(\lambda_1) + \text{Po}(\lambda_2) = \text{Po}(\lambda_1 + \lambda_2)$ for independent.
INDEPENDENCE IS ESSENTIAL.
Generalises to $k$ streams.
Useful for combined-event questions.

Modelling judgements

Check the four conditions against context. Independence and constant rate are usually the weak ones.

Standard exam prompt. 'State the conditions for the Poisson model and comment on each in context.'

Four conditions (memorise).

Events occur independently.
Events occur at a constant average rate.
Events occur singly (no simultaneous events).
Events are rare relative to opportunities.

Approach.

List the four conditions.
For each, evaluate against the specific scenario.
Identify weak ones — usually independence (clustering) or constant rate (peak/off-peak).
Verdict: 'reasonable' / 'reasonable first approximation' / 'unreasonable'.

Worked example. Typing errors per page, $X \sim \text{Po}(2.5)$ .

Independence — plausible per error, less so if fatigue accumulates.
Constant rate — plausible if conditions uniform; less so over a long session.
Singly — yes, errors are individual.
Rare — yes, an error per thousands of keystrokes is rare.

Verdict: 'Reasonable, with caveats about fatigue.'

What examiners reward.

Specific scenario references.
A clear verdict.
Identifying the WEAKEST condition.

Four conditions: independent, constant rate, singly, rare.
Link each to context — not generic.
End with a verdict.

How it’s examined

Poisson appears on every WST02/01 paper — typically one short cumulative-probability question ( $4$ - $5$ marks), one scaling / additivity question ( $5$ - $6$ marks), and the Poisson hypothesis test (covered in Hypothesis Tests topic, $8$ - $10$ marks). Mean-equals-variance is the diagnostic identity that examiners test directly ('find $\lambda$ given $\text{Var}(X) = 4$ '). Off-by-one errors in upper tails are the most common slip. Calculator use is allowed throughout; PoissonPD and PoissonCD on a Casio FX-991 give all probabilities directly — but always write the PMF or cumulative formula to secure M-marks.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Poisson Distributions

Step-by-step solutions to past-paper-style questions on poisson distributions, written exactly the way a tutor would explain them at the board.

1Single-point Poisson probability (4 marks, S2)
Core• Adapted from WST02/01 January 2024 Q2• poisson, PMF, calculator
Question
Telephone calls arrive at a switchboard at a mean rate of $4$ calls per minute. The number of calls in a one-minute period is modelled by $X \sim \text{Po}(4)$ . Find (a) $P(X = 3)$ , (b) $P(X = 0)$ .
Step-by-step solution
1. Step 1
  Use the PMF.
  $P(X = r) = e^{-\lambda} \dfrac{\lambda^{r}}{r!}$
2. Step 2
  (a) $r = 3$ :
  $P(X = 3) = e^{-4} \dfrac{4^{3}}{3!} = e^{-4} \dfrac{64}{6}$
3. Step 3
  $e^{-4} \approx 0.01832$ , so
  $P(X = 3) \approx 0.01832 \times 10.667 \approx 0.1954$
4. Step 4
  (b) $r = 0$ : $0! = 1$ and $4^{0} = 1$ , so $P(X = 0) = e^{-4} \approx 0.0183$ .
Answer
(a) $P(X = 3) \approx 0.1954$ . (b) $P(X = 0) \approx 0.0183$ .
Examiner tip
(a) M1 for the PMF formula, A1 for $0.1954$ . (b) B1 for $e^{-4} = 0.0183$ . PoissonPD $(r, \lambda)$ on a Casio gives both directly; still write the formula for the M-mark.
2Cumulative Poisson and upper tail (5 marks, S2)
Core• Adapted from WST02/01 June 2024 Q3• poisson, cumulative, tail
Question
$X \sim \text{Po}(6)$ . Find (a) $P(X \le 4)$ , (b) $P(X \ge 8)$ , (c) $P(5 \le X \le 9)$ .
Step-by-step solution
1. Step 1
  (a) From Edexcel cumulative Poisson tables (or calculator):
  $P(X \le 4) = 0.2851$
2. Step 2
  (b) Upper tail. Use complement.
  $P(X \ge 8) = 1 - P(X \le 7)$
3. Step 3
  From table: $P(X \le 7) = 0.7440$ . So
  $P(X \ge 8) = 1 - 0.7440 = 0.2560$
4. Step 4
  (c) Interval:
  $P(5 \le X \le 9) = P(X \le 9) - P(X \le 4)$
5. Step 5
  $P(X \le 9) = 0.9161$ , $P(X \le 4) = 0.2851$ .
  $P(5 \le X \le 9) = 0.9161 - 0.2851 = 0.6310$
Answer
(a) $0.2851$ . (b) $0.2560$ . (c) $0.6310$ .
Examiner tip
(a) B1 for $0.2851$ . (b) M1 for $1 - P(X \le 7)$ , A1 for $0.2560$ . (c) M1 for $P(X \le 9) - P(X \le 4)$ , A1 for $0.6310$ . Off-by-one is the killer error: $P(X \ge 8) \ne 1 - P(X \le 8)$ .
3Time scaling — different intervals (5 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q5• poisson, scaling, interval
Question
Customers arrive at a shop at an average rate of $2.4$ per hour, modelled as a Poisson process. Find the probability that (a) exactly $5$ arrive in a $2$ -hour period, (b) more than $3$ arrive in a $30$ -minute period.
Step-by-step solution
1. Step 1
  (a) Over $t = 2$ hours, the rate scales: $\lambda_{2h} = 2.4 \times 2 = 4.8$ . So $Y \sim \text{Po}(4.8)$ .
2. Step 2
  $P(Y = 5)$ :
  $P(Y = 5) = e^{-4.8} \dfrac{4.8^{5}}{5!} \approx 0.1747$
3. Step 3
  (b) Over $30$ minutes ( $t = 0.5$ hours), $\lambda_{30m} = 2.4 \times 0.5 = 1.2$ . So $W \sim \text{Po}(1.2)$ .
4. Step 4
  $P(W > 3) = 1 - P(W \le 3)$ .
  $P(W \le 3) = e^{-1.2}\left(1 + 1.2 + \dfrac{1.2^{2}}{2} + \dfrac{1.2^{3}}{6}\right)$
5. Step 5
  $e^{-1.2} \approx 0.3012$ ; sum in brackets $= 1 + 1.2 + 0.72 + 0.288 = 3.208$ .
  $P(W \le 3) \approx 0.3012 \times 3.208 \approx 0.9662$
6. Step 6
  Therefore
  $P(W > 3) \approx 1 - 0.9662 = 0.0338$
Answer
(a) $\approx 0.1747$ . (b) $\approx 0.0338$ .
Examiner tip
(a) B1 for $\lambda = 4.8$ , M1 for PMF, A1 for $0.1747$ . (b) B1 for $\lambda = 1.2$ , M1 for $1 - P(W \le 3)$ , A1 for $0.0338$ . The most common error: forgetting to rescale $\lambda$ for the new time interval.
4Sum of independent Poissons (5 marks, S2)
Extended• Adapted from WST02/01 June 2024 Q4• poisson, sum, additivity
Question
A help-desk receives email queries at a rate of $\text{Po}(3)$ per hour and phone queries at $\text{Po}(2)$ per hour, independently. Find the probability that the help-desk receives at most $4$ queries in total in a one-hour period.
Step-by-step solution
1. Step 1
  Independent Poissons add: total $T = E + P$ where $E \sim \text{Po}(3)$ , $P \sim \text{Po}(2)$ .
  $T \sim \text{Po}(3 + 2) = \text{Po}(5)$
2. Step 2
  Find $P(T \le 4)$ from the cumulative table for $\lambda = 5$ :
  $P(T \le 4) = 0.4405$
Answer
$P(T \le 4) = 0.4405$ .
Examiner tip
M1 for adding the parameters $3 + 2 = 5$ . A1 for stating $T \sim \text{Po}(5)$ . M1 for cumulative lookup. A1 for $0.4405$ . Independence is essential — without it, the sum is NOT Poisson.
5Poisson modelling validity (5 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q8• poisson, modelling, context
Question
The number of typing errors per page in a long manuscript is modelled as $X \sim \text{Po}(2.5)$ . State the conditions for the Poisson model and comment on each in this context.
Step-by-step solution
1. Step 1
  Condition 1 — Events occur INDEPENDENTLY. Each typing error is independent of others. Plausible for a careful typist, less so if fatigue accumulates (later pages have more errors).
2. Step 2
  Condition 2 — Events occur at a CONSTANT AVERAGE RATE. Plausible if typing conditions are uniform; less plausible if quality varies (e.g. fatigue at end of session).
3. Step 3
  Condition 3 — Events occur SINGLY (not in clusters). For typing errors this is reasonable; two errors are unlikely to be 'simultaneous' in a meaningful sense.
4. Step 4
  Condition 4 — Events are RARE relative to opportunities. A typist makes thousands of keystrokes per page; an error rate of $2.5$ per page is very small per keystroke, so 'rare' is satisfied.
5. Step 5
  Verdict: the Poisson model is reasonable, though the constant-rate and independence assumptions may not hold if the typist tires.
Answer
Four conditions: independent events, constant rate, single occurrences, rare events. All are approximately satisfied for typing errors, though fatigue may violate constant-rate / independence.
Examiner tip
B1 each for the four conditions (4 marks). B1 for the verdict comment (1 mark). Mark schemes give credit for any TWO sensible contextual remarks linked to the scenario. 'Independent events' alone is insufficient — must reference the specific context.

Key Formulae — Poisson Distributions

The formulae you need to memorise for poisson distributions on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Poisson PMF
$P(X = r) = e^{-\lambda} \dfrac{\lambda^{r}}{r!}, \quad r = 0, 1, 2, \ldots$
$\lambda$
mean number of events per interval (must be $> 0$ )
$r$
observed count (non-negative integer)
When to use
Exact probability of $r$ events. IS in the IAL formula booklet. PoissonPD on a calculator gives the same value.
Example
$\text{Po}(4)$ , $P(X = 3) = e^{-4} \dfrac{64}{6} \approx 0.1954$ .
Mean and variance
$E(X) = \text{Var}(X) = \lambda$
$\lambda$
Poisson parameter
When to use
Diagnostic check: real data with sample mean $\approx$ sample variance suggests a Poisson model. IS in the IAL formula booklet.
Example
$\text{Po}(6)$ : mean and variance both $= 6$ ; SD $= \sqrt{6} \approx 2.45$ .
Time / space scaling
$X \sim \text{Po}(\lambda) \text{ per unit interval} \Rightarrow Y \sim \text{Po}(\lambda t) \text{ over } t \text{ units}$
$t$
scaling factor (time or space)
When to use
When the question changes the observation window. NOT explicitly in the booklet — follows from the Poisson process definition.
Example
Rate $2.4$ per hour over $30$ minutes: $\lambda = 2.4 \times 0.5 = 1.2$ .
Sum of independent Poissons
$X_1 \sim \text{Po}(\lambda_1), X_2 \sim \text{Po}(\lambda_2) \text{ independent} \Rightarrow X_1 + X_2 \sim \text{Po}(\lambda_1 + \lambda_2)$
$X_1, X_2$
independent Poisson variables
When to use
Combining two types of events into a single count. INDEPENDENCE is essential — without it, the sum is NOT Poisson.
Example
Email $\text{Po}(3)$ + phone $\text{Po}(2)$ = total $\text{Po}(5)$ .
Cumulative Poisson
$P(X \le c) = \sum_{r=0}^{c} e^{-\lambda} \dfrac{\lambda^{r}}{r!}$
$c$
upper limit
When to use
Read from Edexcel cumulative Poisson tables (booklet) or via calculator PoissonCD.
Example
$\text{Po}(6)$ : $P(X \le 4) = 0.2851$ .

Key Definitions and Keywords — Poisson Distributions

Definitions to memorise and the exact keywords mark schemes credit for poisson distributions answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Poisson distribution
Examiner keyword
Discrete distribution counting events occurring independently at constant average rate $\lambda$ over a fixed interval. Notation $X \sim \text{Po}(\lambda)$ .
Poisson conditions
Examiner keyword
Four required assumptions: (i) events independent; (ii) constant average rate; (iii) events occur singly; (iv) events are rare relative to opportunities.
Poisson process
Examiner keyword
Stochastic process where counts in disjoint intervals are independent and Poisson-distributed, with mean proportional to the interval length.
Parameter $\lambda$
Examiner keyword
Mean number of events per interval. For Poisson, $\lambda$ equals BOTH mean AND variance.
Rate
Events per unit interval. Over $t$ units, the count is $\text{Po}(\lambda t)$ .

Common Mistakes and Misconceptions — Poisson Distributions

The traps other students keep falling into on poisson distributions questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Writing $P(X \ge c) = 1 - P(X \le c)$ instead of $1 - P(X \le c - 1)$
WST02/01 examiner reports — annual
Why it happens
Same off-by-one error as binomial.
How to avoid it
Always expand: $P(X \ge 8) = P(X = 8) + P(X = 9) + \ldots = 1 - P(X \le 7)$ .
✕Forgetting to rescale $\lambda$ for a different time interval
WST02/01 examiner reports
Why it happens
Students treat $\lambda$ as fixed.
How to avoid it
Always identify the question's interval and multiply $\lambda$ by the ratio. 'Rate $2$ per hour, $30$ minutes: $\lambda = 1$ .'
✕Adding Poisson parameters when events are NOT independent
Why it happens
Students apply the rule without checking.
How to avoid it
Independence MUST be stated (or evident from context). Two correlated streams (e.g. flu cases at two adjacent clinics) cannot simply be added.
✕Confusing $\lambda$ with an integer count
Why it happens
Students expect $\lambda$ to look like a count.
How to avoid it
$\lambda$ is a MEAN, so non-integer values like $\text{Po}(2.4)$ or $\text{Po}(0.5)$ are normal and correct.
✕Stating mean $\ne$ variance for a Poisson
Why it happens
Habit from other distributions.
How to avoid it
MEMORISE: $E(X) = \text{Var}(X) = \lambda$ . This is the diagnostic feature of the Poisson.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Poisson Distributions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Poisson Distributions

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Single-point Poisson probability (4 marks, S2)

2Cumulative Poisson and upper tail (5 marks, S2)

3Time scaling — different intervals (5 marks, S2)

4Sum of independent Poissons (5 marks, S2)

5Poisson modelling validity (5 marks, S2)

Poisson PMF

Mean and variance

Time / space scaling

Sum of independent Poissons

Cumulative Poisson

Poisson distribution

Poisson conditions

Poisson process

Parameter λ\lambdaλ

Rate

✕Writing P(X≥c)=1−P(X≤c)P(X \ge c) = 1 - P(X \le c)P(X≥c)=1−P(X≤c) instead of 1−P(X≤c−1)1 - P(X \le c - 1)1−P(X≤c−1)

✕Forgetting to rescale λ\lambdaλ for a different time interval

✕Adding Poisson parameters when events are NOT independent

✕Confusing λ\lambdaλ with an integer count

✕Stating mean ≠\ne= variance for a Poisson

Practice questions

Past paper style quiz

Poisson Distributions — frequently asked questions

Parameter $\lambda$

✕Writing $P(X \ge c) = 1 - P(X \le c)$ instead of $1 - P(X \le c - 1)$

✕Forgetting to rescale $\lambda$ for a different time interval

✕Confusing $\lambda$ with an integer count

✕Stating mean $\ne$ variance for a Poisson