S2.1 - The Binomial and Poisson distributions

Learning objectives

S2.1.1Binomial and Poisson distributionsThe binomial and Poisson Students will be expected to use these distributions to distributions. model a real-world situation and to comment critically on their appropriateness.; Cumulative probabilities by calculation or by reference to tables.; Students will be expected to use the additive property of the Poisson distribution – e.g. if the number of events per minute ∼ Po(λ) then the number of events per 5 minutes ∼ Po(5λ).S2.1.2Mean and variance of binomial and Poisson distributionsThe mean and variance of the No derivations will be required. binomial and Poisson distributions.S2.1.3TheThe use of the Poisson distribution as an approximation to the binomial distribution.

Binomial and Poisson distributions both model counts, but their random processes differ. Choose the model from the conditions before calculating a probability, and keep the count interval and event inequality explicit.

Model	Conditions and parameter
$X\sim B(n,p)$	a fixed number $n$ of independent trials; each trial has success/failure outcomes; success probability $p$ is constant
$X\sim Po(\lambda)$	events occur independently and singly at a constant average rate in a fixed interval; $\lambda$ is the mean count in that interval

$P_{B}(X=x)=\binom nxp^x(1-p)^{n-x},\qquad P_{Po}(X=x)=e^{-\lambda}\frac{\lambda^x}{x!}$

Translate words into integer events before using a formula or cumulative table: 'fewer than 3' is $P(X\le2)$ , 'at least 6' is $1-P(X\le5)$ , and $4<X<10$ is $P(X\le9)-P(X\le4)$ . For $X\sim Po(4.5)$ , the last probability is $0.9829-0.5321=0.4508$ .

Poisson means scale with exposure. If complaints occur at mean 3 per day, then a 7-day total has distribution $Po(21)$ . More generally, independent Poisson counts add: if $X\sim Po(\lambda_1)$ and $Y\sim Po(\lambda_2)$ independently, then $X+Y\sim Po(\lambda_1+\lambda_2)$ .

A calculation is only as sound as the model. Repeated trials with changing success probabilities are not binomial; clustering, dependence or a changing rate undermines Poisson. State assumptions in context—such as failures being independent with constant probability, or events occurring independently at a constant average rate—and comment when the situation contradicts them.

The mean gives the long-run expected count and the variance measures its spread. Binomial and Poisson models have characteristic mean–variance relationships that can also identify unknown parameters or test plausibility.

Distribution	Mean	Variance
$X\sim B(n,p)$	$np$	$np(1-p)$
$X\sim Po(\lambda)$	$\lambda$	$\lambda$

If a binomial variable has mean 90 and variance 36, then $np=90,\qquad np(1-p)=36.$ Dividing the second equation by the first gives $1-p=36/90=0.4$ , so $p=0.6$ and $n=90/0.6=150$ . Check that $n$ is a positive integer and $0\le p\le1$ .

For $Y\sim B(20,0.3)$ , $E(Y)=6$ and $\operatorname{Var}(Y)=4.2$ . For $W\sim Po(4)$ , both the mean and variance are 4. An observed count dataset whose sample mean is far from its sample variance may therefore cast doubt on a Poisson model, although closeness alone does not prove all Poisson assumptions.

Do not interchange variance and standard deviation, or assume every count with equal mean and variance must be Poisson. The specification requires use of these moment formulae but not their derivations.

When $X\sim B(n,p)$ has many trials and a small success probability, the rare-success count can be approximated by a Poisson variable with the same mean.

$X\sim B(n,p)\ \approx\ Y\sim Po(\lambda),\qquad \lambda=np$

First check that the binomial conditions hold and that the context really describes many independent opportunities for a rare event. Then calculate $\lambda=np$ , preserve the original integer event exactly, and evaluate it with the Poisson formula or table.

If $X\sim B(200,0.012)$ , use $Y\sim Po(2.4)$ . Then $P(X\le1)\approx P(Y\le1)=e^{-2.4}(1+2.4)=0.3084.$ For 'at least 2', use the complementary Poisson event $1-P(Y\le1)$ .

This is a discrete-to-discrete approximation, so no continuity correction is used. It becomes less convincing when the success probability is not small or trials are dependent; use the exact binomial distribution when the approximation is not justified. Normal approximations and continuity correction belong to S2.3, not this objective.

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—S2.2 - Continuous random variablesRead

S2.1 - The Binomial and Poisson distributions

Learning objectives

Choose and use Binomial or Poisson

Use Binomial and Poisson moments

Approximate a Binomial count with Poisson

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