6.1 - Approximating with Poisson and normal models | Cambridge A-Level Mathematics

A rare-event binomial count can often be approximated by a Poisson distribution, while a large-mean Poisson count can often be approximated by a normal curve. Learn to justify the choice, match its parameters and handle the event boundaries correctly.

Rare binomial events

Suppose $X\sim\mathrm B(200,0.01)$ . There are many opportunities for success, but each success is rare and the expected count is only $2$ . Is there a simpler model for its probabilities?

For small $p$ , the binomial variance $np(1-p)$ is close to its mean $np$ . A Poisson model has exactly equal mean and variance, so matching its parameter to $np$ is a natural approximation.

Poisson approximation to a binomial distribution

For $X\sim\mathrm B(n,p)$ , a Poisson approximation is suitable when $n$ is large and $p$ is small. Use the approximate guidance

n>50,\qquad np<5,

and replace the binomial count by $Y\sim\operatorname{Po}(np)$ .
The Poisson probability formula is in the formula booklet. These approximation conditions and the choice $\lambda=np$ are not: learn them.

These are approximate suitability guidelines, not a sharp boundary at which the distributions become equal. Check the actual values when justifying your choice. The underlying experiment must first satisfy the binomial assumptions.

Worked example 1: a rare-event point probability

Question. Let $X\sim\mathrm B(200,0.01)$ . Use a suitable Poisson approximation to find $\mathrm P(X=2)$ , justifying the approximation.

Plan. Check the rare-event conditions, use mean $np$ , then calculate a Poisson point probability.

Here $n=200>50$ , $p=0.01$ is small and $np=2<5$ . Use $Y\sim\operatorname{Po}(2)$ :

\mathrm P(X=2)\approx\mathrm P(Y=2) =\frac{\mathrm e^{-2}2^2}{2!}=0.2706706\ldots.

Answer. Approximately $0.271$ .

Check. The exact binomial expression $\binom{200}{2}(0.01)^2(0.99)^{198}$ is about $0.272$ , close to the approximation. The Poisson mean $2$ matches the binomial mean, while its variance $2$ is close to the binomial variance $1.98$ .

Keep the model and change only the event to more than two successes. The approximation is still discrete, so there is no continuity correction.

Worked example 2: a rare-event tail

Question. For the same binomial variable, use a Poisson approximation to find $\mathrm P(X>2)$ .

Plan. Use $Y\sim\operatorname{Po}(2)$ and subtract counts $0,1,2$ from $1$ .

\begin{aligned} \mathrm P(X>2)&\approx1-\mathrm P(Y\le2)\\ &=1-\mathrm e^{-2}\left(1+2+\frac{2^2}{2!}\right)\\ &=0.3233236\ldots \end{aligned}

Answer. Approximately $0.323$ .

Check. The approximating probability of at most two is $5\mathrm e^{-2}=0.6766764\ldots$ . The two complementary approximations add to $1$ ; the count $2$ is excluded from the requested upper tail.

A large $n$ alone is insufficient. For example, $\mathrm B(200,0.4)$ has mean $80$ , so it does not satisfy these rare-event conditions. Its normal approximation uses binomial variance $np(1-p)$ , not Poisson variance $np$ .

Watch out

In a Poisson approximation to a binomial, use $\lambda=np$ , not $p$ . Keep the integer event unchanged: $X>2$ becomes $Y>2$ , with no shift to $2.5$ .

Approximate a lower tail for a new binomial count.

A large Poisson mean

As the Poisson parameter increases, its distribution becomes less skewed. A normal curve can then give a useful approximation, matching both the Poisson mean and variance to the same parameter.

Normal approximation to a Poisson distribution

For $X\sim\operatorname{Po}(\lambda)$ , use

Y\sim\mathrm N(\lambda,\lambda)

when $\lambda$ is large: $\lambda>15$ , approximately. Use a continuity correction when translating the discrete event to the normal area. Standardise using standard deviation $\sqrt\lambda$ .
The Poisson mean and variance are in the formula booklet. This approximation condition and the continuity correction are not: learn them.

The half-unit correction has the same bar-edge meaning as in the earlier binomial approximation. Count $r$ occupies the interval from $r-0.5$ to $r+0.5$ . Correct the event before standardising, and keep the original discrete variable separate from the normal variable.

Worked example 3: a lower tail

Question. Let $X\sim\operatorname{Po}(25)$ . Use a normal approximation to find $\mathrm P(X\le29)$ .

Plan. Match mean and variance, include the whole bar for $29$ , then use a normal left-tail area.

Since $25>15$ , use $Y\sim\mathrm N(25,25)$ , with standard deviation $5$ :

\begin{aligned} \mathrm P(X\le29)&\approx\mathrm P(Y<29.5)\\ &=\Phi\left(\frac{29.5-25}{5}\right)\\ &=\Phi(0.9)=0.8159 \end{aligned}

Answer. Approximately $0.816$ .

Check. The corrected boundary is $0.9$ standard deviations above the mean, so the left area should exceed $0.5$ . The bar centred on $29$ is included completely.

Now request the complementary upper tail. The boundary stays fixed; the side of the curve changes.

Worked example 4: more than the same count

Question. For $X\sim\operatorname{Po}(25)$ , use a normal approximation to find $\mathrm P(X>29)$ .

Plan. The first permitted count is $30$ , whose left edge is $29.5$ ; find the normal right tail.

\begin{aligned} \mathrm P(X>29)&\approx\mathrm P(Y>29.5)\\ &=1-\Phi\left(\frac{29.5-25}{5}\right)\\ &=1-0.8159=0.1841 \end{aligned}

Answer. Approximately $0.184$ .

Check. This is the complement of the preceding event. The two normal areas add to $1$ , with the same corrected boundary $29.5$ .

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6.1 - Approximating with Poisson and normal models

Rare binomial events

A large Poisson mean

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