Notes · StatisticsUK · A-Levels

Exponential and Poisson distributions

This topic treats the Poisson distribution as a model for the number of events occurring at a constant average rate, and the exponential distribution as the model for the waiting time between such events. It covers their probabilities, means and variances, the additivity of independent Poisson variables, the Poisson approximation to the binomial, and the memoryless property that links the two models.

5 sections~12 min reading time3 competenciesLevel Standard 2 · Advanced 3

T·0777 / 18

Exam profile

AO1 · Calculate Poisson and exponential probabilities, means and variancesAO2 · Justify the modelling conditions and the Poisson approximation to the binomialAO3 · Model counts and waiting times in context and interpret the results

Operators:calculatefinddeterminemodelshow thatapproximate

basic level

The AS-level treatment of these models is limited; the Poisson and exponential distributions are principally full A-Level (Paper 2) content.

higher level

The full A-Level expects Poisson and exponential probabilities, the additivity and approximation results, and the memoryless property.

Depth

Text

Contents · 5 sections

Exponential and Poisson distributions

The Poisson model and its conditions#

StandardPearson Edexcel 9ST0, Topic 10 (Paper 2)

The Poisson distribution models the number of events occurring in a fixed interval of time or space when the events happen singly, at a constant average rate, and independently of one another. Typical contexts are the number of calls to a switchboard in an hour, flaws per metre of cable, or accidents at a junction per month. We write

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

, where

\lambda

is the mean number of events in the interval.

The conditions are that events occur independently, at a constant mean rate, and cannot occur simultaneously (two events do not coincide). Where the rate varies over the interval — rush-hour traffic, seasonal demand — a single Poisson model is inappropriate, and saying so with a reason is an examined skill just as it is for the binomial.

A defining and examinable feature of the Poisson is that its mean and variance are equal: both are

\lambda

. This gives a quick diagnostic — a count whose sample mean and sample variance are close is consistent with a Poisson model, whereas a variance far larger than the mean (overdispersion) suggests the events are clustered rather than independent.

The Poisson parameter is additive over intervals: if events occur at rate

\lambda

per unit interval, then over

t

units the count is

\text{Po}(\lambda t)

. This scaling is used constantly — a rate of 3 per hour gives

\text{Po}(1.5)

over half an hour and

\text{Po}(6)

over two hours — and must be applied before any probability is computed.

X \sim \text{Po}(\lambda): \quad E(X) = \lambda, \quad \text{Var}(X) = \lambda

The Poisson model

Events occur singly, at constant rate, independently; the mean and variance are both $\lambda$ .

Worked example

Scaling the Poisson rate

Flaws occur in a cable at a mean rate of 0.6 per metre, modelled by a Poisson distribution. State the distribution of the number of flaws in a 5-metre length and give its mean and variance.

Scale the rate
Over 5 metres the mean number of flaws is $0.6 \times 5 = 3$ .
State the distribution
The number of flaws in 5 metres is $\text{Po}(3)$ .
Mean and variance
Both equal $\lambda = 3$ .

Result: The number of flaws in 5 metres is $\text{Po}(3)$ , with mean 3 and variance 3.

Typical mistakes

Failing to scale $\lambda$ to the required interval before computing a probability.
Applying the Poisson where the rate is not constant over the interval.

Active revision

Calls arrive at a helpline at a mean rate of 4 per hour. State the distribution of the number of calls in a 15-minute period.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Calculating Poisson probabilities, mean and variance#

StandardPearson Edexcel 9ST0, Topic 10 (Paper 2)

A Poisson distribution

Fig. 1Po(3) has mean and variance 3; probabilities for x = 0 to 8 (the tail beyond 8 is negligible).

The Poisson probability function is

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

for

x = 0, 1, 2, \ldots

, defined for every non-negative integer with no upper limit. As with the binomial, cumulative probabilities

P(X \leq x)

are read from tables or a calculator, and the upper tail is found by complement,

P(X \geq x) = 1 - P(X \leq x - 1)

The distribution

\text{Po}(3)

below shows the characteristic Poisson shape: a peak near the mean (here the mode is at

x = 2

and

x = 3

), a short left tail bounded at zero, and a right tail that decays but never quite ends. For small

\lambda

the distribution is strongly right-skewed; as

\lambda

grows it becomes more symmetric and, for large

\lambda

, approximately normal.

Computations follow the formula directly. For

\text{Po}(3)

P(X = 5) = \dfrac{e^{-3}3^{5}}{5!} = \dfrac{0.0498 \times 243}{120} = 0.1008

(4 d.p.). Keeping

e^{-\lambda}

to enough figures matters, because it multiplies every term; rounding it early corrupts the whole calculation.

The equality of mean and variance is used both as a property and as a solving tool. Given a Poisson count with a known variance, the rate

\lambda

is immediate, and the standard deviation is

\sqrt{\lambda}

. These summaries feed directly into the Poisson hypothesis tests and goodness-of-fit work later in the course.

P(X = x) = \frac{e^{-\lambda}\lambda^{x}}{x!}, \quad x = 0, 1, 2, \ldots

The Poisson probability

Defined for every non-negative integer; $e^{-\lambda}$ normalises the distribution to sum to 1.

Worked example

A Poisson probability

The number of emails arriving in an hour follows $\text{Po}(3)$ . Find the probability that exactly 5 arrive in a given hour.

Apply the formula
$P(X = 5) = \dfrac{e^{-3}3^{5}}{5!}$ .
Evaluate the parts
$e^{-3} = 0.049787$ , $3^5 = 243$ , $5! = 120$ .
Combine
$\dfrac{0.049787 \times 243}{120} = 0.1008$ (4 d.p.).
$P(X = 5) = \frac{0.049787 \times 243}{120} = 0.1008$

Result: $P(X = 5) = 0.1008$ (4 d.p.).

Typical mistakes

Rounding $e^{-\lambda}$ too early, corrupting every term.
Forgetting the factorial $x!$ in the denominator.

Active revision

For $X \sim \text{Po}(3)$ , find $P(X = 5)$ and $P(X \leq 2)$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Sums of Poisson variables and the Poisson approximation to the binomial#

AdvancedPearson Edexcel 9ST0, Topic 10 (Paper 2)

Binomial and its Poisson approximation

Fig. 2B(100, 0.02) and its Poisson approximation Po(2) are nearly identical over x = 0 to 6.

Independent Poisson variables add: if

X \sim \text{Po}(\lambda_1)

and

Y \sim \text{Po}(\lambda_2)

are independent, then

X + Y \sim \text{Po}(\lambda_1 + \lambda_2)

. This is the interval-additivity property seen from another angle, and it lets counts from separate sources or periods be pooled into a single Poisson model before any probability is computed.

The Poisson is the limiting form of the binomial when the number of trials is large and the probability of success is small. Formally, if

X \sim B(n, p)

with large

n

and small

p

, then

X

is approximately

\text{Po}(\lambda)

with

\lambda = np

. The rule of thumb is

n

large (say

n > 50

) and

p

small (say

p < 0.1

), so that

np

is moderate.

The comparison below shows

B(100, 0.02)

against its Poisson approximation

\text{Po}(2)

: the bars are almost indistinguishable, confirming the approximation. It is genuinely useful because a Poisson probability with a single parameter is far quicker to evaluate than a binomial with large

n

, and cumulative Poisson tables are widely available.

The approximation works because, for small

p

, the binomial variance

np(1 - p) \approx np

matches the Poisson variance

\lambda = np

, and the two distributions share the same mean. When

p

is not small, the binomial variance

np(1 - p)

is appreciably below

np

and the Poisson overstates the spread, so the approximation should not be used.

X \sim B(n, p),\; n \text{ large},\; p \text{ small} \;\Rightarrow\; X \approx \text{Po}(np)

Poisson approximation to the binomial

Valid when $np$ is moderate while $n$ is large and $p$ small; independent Poissons also add: $\text{Po}(\lambda_1) + \text{Po}(\lambda_2) = \text{Po}(\lambda_1 + \lambda_2)$ .

Worked example

Poisson approximation to a binomial

A machine produces items with a 2% defect rate, independently. In a sample of 100, use a Poisson approximation to find the probability of exactly 3 defectives, and compare with the exact binomial value 0.1822.

Set up the approximation
$X \sim B(100, 0.02)$ with $n$ large and $p$ small, so $X \approx \text{Po}(\lambda)$ with $\lambda = np = 2$ .
Poisson probability
$P(X = 3) = \dfrac{e^{-2}2^{3}}{3!} = \dfrac{0.135335 \times 8}{6} = 0.1804$ (4 d.p.).
$P(X = 3) \approx \frac{e^{-2}2^{3}}{3!} = 0.1804$
Compare
The exact binomial value is 0.1822, so the approximation is accurate to within 0.002.

Result: The Poisson estimate is 0.1804, agreeing closely with the exact binomial value 0.1822.

Typical mistakes

Adding Poisson variables that are not independent.
Using the Poisson approximation when $p$ is not small, so the variances disagree.

Active revision

A rare fault affects 1.5% of items. In a batch of 200, use a Poisson approximation to estimate the probability of at most 2 faulty items.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

The exponential distribution for waiting times#

AdvancedPearson Edexcel 9ST0, Topic 10 (Paper 2)

The exponential density with a tail probability

Fig. 3Exponential density with rate 0.5: mean at x = 2, and the shaded tail P(X > 3) = e^(-1.5) = 0.223.

The exponential distribution is a continuous model for the waiting time until the next event in a Poisson process. Its probability density function is

f(x) = \lambda e^{-\lambda x}

for

x \geq 0

(and 0 otherwise), where

\lambda

is the same rate as in the associated Poisson model. The density starts at

\lambda

when

x = 0

and decays exponentially, as the curve below for

\lambda = 0.5

shows.

Probabilities are areas under the density. The cumulative distribution function is

F(x) = 1 - e^{-\lambda x}

, so the probability of waiting longer than

x

is the clean upper-tail formula

P(X > x) = e^{-\lambda x}

. This is the workhorse result: the shaded region in the figure is

P(X > 3) = e^{-0.5 \times 3} = e^{-1.5} = 0.223

(3 s.f.).

The mean waiting time is

E(X) = \dfrac{1}{\lambda}

and the variance is

\text{Var}(X) = \dfrac{1}{\lambda^2}

, so the standard deviation equals the mean. For

\lambda = 0.5

per minute the mean wait is 2 minutes, marked on the curve; a higher event rate

\lambda

shortens the mean wait, as intuition demands.

The exponential and the Poisson describe the same process from two viewpoints: the Poisson counts how many events occur in a fixed time, while the exponential measures the time between consecutive events. If events occur as

\text{Po}(\lambda)

per unit time, the gaps between them are

\text{Exp}(\lambda)

, and this correspondence is the reason the two distributions are studied together.

f(x) = \lambda e^{-\lambda x},\; x \geq 0; \qquad P(X > x) = e^{-\lambda x}; \qquad E(X) = \frac{1}{\lambda},\; \text{Var}(X) = \frac{1}{\lambda^2}

The exponential distribution

The waiting-time model of a Poisson process; the upper tail is a clean exponential.

Worked example

An exponential tail probability

The time in minutes between customer arrivals is exponential with $\lambda = 0.5$ . Find the probability that the wait exceeds 3 minutes and state the mean wait.

Use the tail formula
$P(X > 3) = e^{-\lambda x} = e^{-0.5 \times 3} = e^{-1.5}$ .
Evaluate
$e^{-1.5} = 0.2231 = 0.223$ (3 s.f.).
$P(X > 3) = e^{-1.5} = 0.223$
Mean wait
$E(X) = \dfrac{1}{\lambda} = \dfrac{1}{0.5} = 2$ minutes.

Result: $P(X > 3) = 0.223$ (3 s.f.) and the mean waiting time is 2 minutes.

Typical mistakes

Confusing the rate $\lambda$ with the mean; the mean is $\frac{1}{\lambda}$ .
Integrating unnecessarily when the tail formula $P(X > x) = e^{-\lambda x}$ gives the answer directly.

Active revision

The time (minutes) between arrivals at a counter is exponential with rate $\lambda = 0.5$ . Find $P(X > 3)$ and the mean waiting time.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

The memoryless property and the Poisson-exponential link#

AdvancedPearson Edexcel 9ST0, Topic 10 (Paper 2)

The exponential distribution is memoryless: the probability of waiting a further time

t

does not depend on how long you have already waited, so

P(X > s + t \mid X > s) = P(X > t)

. A component that has survived some time is, under this model, as good as new. This follows algebraically because

\dfrac{P(X > s + t)}{P(X > s)} = \dfrac{e^{-\lambda(s + t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t)

The memoryless property is a strong modelling assumption and is not always realistic — many real components wear out, so the chance of failure rises with age, contradicting memorylessness. Recognising when the assumption is reasonable (truly random failures, radioactive decay) and when it is not (ageing, fatigue) is part of judging the model.

The Poisson and exponential are two descriptions of a single Poisson process at rate

\lambda

: the number of events in a fixed interval is

\text{Po}(\lambda t)

, while the time between consecutive events is

\text{Exp}(\lambda)

. In particular, the probability of no event in a time

t

can be found either as the Poisson

P(N = 0) = e^{-\lambda t}

or as the exponential

P(X > t) = e^{-\lambda t}

— the same expression, as it must be.

This equivalence is powerful in problems: a question about 'no calls in the next 10 minutes' can be answered with either model, and a question mixing counts and waiting times can be moved to whichever description is simpler. Understanding that the two distributions are the discrete and continuous faces of one process is the conceptual heart of this topic.

P(X > s + t \mid X > s) = P(X > t) = e^{-\lambda t}

The memoryless property

The remaining wait is independent of the wait already elapsed; equal to the Poisson $P(N = 0)$ over time $t$ .

Worked example

One process, two models

Machine breakdowns occur as a Poisson process at rate 0.5 per hour. Find the probability of no breakdown in the next 2 hours using both the Poisson and the exponential descriptions.

Poisson view
The number in 2 hours is $\text{Po}(0.5 \times 2) = \text{Po}(1)$ , so $P(N = 0) = e^{-1} = 0.368$ .
Exponential view
The time to the first breakdown is $\text{Exp}(0.5)$ , so $P(X > 2) = e^{-0.5 \times 2} = e^{-1} = 0.368$ .
$P(N = 0) = P(X > 2) = e^{-1} = 0.368$
Reconcile
Both give $e^{-1} = 0.368$ (3 s.f.), confirming the two models describe the same process.

Result: The probability of no breakdown in 2 hours is $e^{-1} = 0.368$ , obtained identically from the Poisson and the exponential.

Typical mistakes

Assuming memorylessness for a component that ages (where failure rate increases with time).
Treating the Poisson count and the exponential gap as unrelated rather than two views of one process.

Active revision

Faults occur on a line as a Poisson process at 0.5 per hour. Find the probability of no fault in the next 2 hours, and verify it two ways.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content (Statistics) (Department for Education / Ofqual)

References & sources

Sources

Pearson Edexcel

Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification

Department for Education / Ofqual

GCE AS and A level subject content (Statistics)

Exponential and Poisson distributions

5 sections~12 min reading time3 competenciesLevel Standard 2 · Advanced 3

Exponential and Poisson distributions

Key points

Scaling the Poisson rate

Key points

A Poisson probability

Key points

Poisson approximation to a binomial

Key points

An exponential tail probability

Key points

One process, two models

Sources

Exponential and Poisson distributions

Key points

Scaling the Poisson rate

Key points

A Poisson probability

Key points

Poisson approximation to a binomial

Key points

An exponential tail probability

Key points

One process, two models

Sources