Poisson Distribution: When to Approximate from Binomial and When to Approximate to Normal (CIE 9709)

Poisson Distribution: When to Approximate from Binomial and When to Approximate to Normal (CIE 9709).

When can you approximate a Binomial distribution as Poisson?

You can approximate X~B(n,p) as X~Po(λ) when n is large (n > 50) and p is small enough that np < 5. You then set λ = np and use the Poisson formula instead of the Binomial one. This matters on Cambridge International (CIE) 9709 papers because Binomial probabilities with n in the thousands are impossible to calculate directly by hand or calculator. The Poisson approximation is what makes them tractable at all.

The Poisson distribution itself models the number of times an event happens in a fixed interval, when events occur independently at a constant average rate. If X~Po(λ), then:

P(X = x) = e^(−λ)λ^x / x!, for x = 0, 1, 2, 3, …

Here λ (lambda) is the mean number of occurrences in the interval. A key property examiners test constantly: for a Poisson distribution, the mean and the variance are equal. E(X) = Var(X) = λ, so the standard deviation is √λ.

Why does λ change when the interval changes?

Because λ is a rate scaled to an interval, not a fixed constant. Double the interval and you double λ. If a photocopier breaks down on average 8 times a week, λ = 8 for a whole week, but λ = 8/5 = 1.6 for a single working day, and λ = 16 for a fortnight. The same logic applies to any Poisson scenario: if planes land at a constant average rate of 1 every 10 minutes, λ = 6 for a one-hour window but λ = 1.6 for a 16-minute window. Always rescale λ to match the time (or space) interval the question actually asks about before you calculate anything.

Why do you need a continuity correction going from Poisson to Normal?

You need a continuity correction because Poisson is a discrete distribution (it only takes whole-number values) and the Normal distribution you're approximating it with is continuous. Without adjusting by ±0.5, you'd be comparing a "step" probability to a smooth curve at the wrong point, and get a noticeably wrong answer for it. This approximation, X~Po(λ) ≈ N(λ, λ), is valid when λ > 15.

Binomial-to-Poisson vs Poisson-to-Normal: the two rules side by side

 

Binomial → Poisson

Poisson → Normal

Condition

n > 50 and np < 5

λ > 15

Approximating distribution

X~Po(λ)

X~N(λ, λ)

Parameters used

λ = np

mean = λ, variance = λ

Continuity correction needed?

No

Yes, ±0.5 (discrete → continuous)

Typical exam trigger

Very large n, rare event (misprints, faulty items, rare gene)

Poisson scenario extended over a long time/space interval, so λ grows past 15

Worked example: Binomial approximated to Poisson (CIE 9709/2020/O/N/P62)

A gene is carried by 1 in 50,000 people. In a sample of 150,000 people, find the probability that more than 2 people carry the gene.

This is Binomial: X~B(150,000, 1/50,000). Check the approximation conditions: n = 150,000 > 50, and np = 150,000 × (1/50,000) = 3, which is less than 5. Both conditions hold, so approximate X~Po(3), with λ = np = 3.

"More than 2" means P(X > 2) = 1 − P(X ≤ 2). Using the Poisson cumulative probabilities for λ = 3:

P(X ≤ 2) = P(0) + P(1) + P(2) = e^(−3)(1 + 3 + 4.5) = 0.4232

P(X > 2) = 1 − 0.4232 = 0.577 (3 s.f.)

Worked example: Poisson approximated to Normal with continuity correction (CIE 9709/2018/M/J/P71)

E-reader sales at a shop follow a Poisson distribution with mean 5.1 per 10 days. Find the probability that fewer than 140 are sold in 300 days.

First, rescale λ to the new interval. 300 days is 30 lots of 10 days, so λ = 5.1 × 30 = 153. Since λ = 153 > 15, approximate X~Po(153) as X~N(153, 153).

"Fewer than 140" is a discrete condition (X ≤ 139), so apply the continuity correction: use X < 139.5 in the Normal model.

Z = (139.5 − 153) / √153 = −13.5 / 12.37 = −1.0914

P(Z < −1.0914) = 0.138 (3 s.f.)

Miss the continuity correction here and you'd calculate Z using 140 instead of 139.5. That's a small shift, and exactly the kind of thing that costs marks on CIE mark schemes, which award a specific method mark for applying it correctly.

How do you know which approximation to use in an exam question?

Check what you're given first. If the question describes a Binomial setup (fixed number of trials, two outcomes, probability p) with large n and small np, approximate to Poisson. If the question already gives you a Poisson rate and asks about a stretched-out interval that pushes λ above 15, approximate to Normal and always apply the ±0.5 continuity correction. Never apply both approximations to the same variable at once, and never skip stating λ > 15 or np < 5 as your justification. CIE examiners specifically credit that check, even when the arithmetic that follows it is correct.

For the underlying Binomial groundwork these approximations build on, see SortMyPrep's guide to the binomial distribution.

FAQ

What is the Poisson distribution formula? For X~Po(λ), P(X = x) = e^(−λ)λ^x / x!, where x = 0, 1, 2, 3, … and λ is the mean number of occurrences in the given interval. The mean and variance of a Poisson distribution are always equal, both equal to λ.

When do you approximate Binomial as Poisson in CIE 9709? When n > 50 and np < 5. Set λ = np and use X~Po(λ) instead of the Binomial formula, which becomes impractical to compute directly once n is very large.

When do you approximate Poisson as Normal, and why is a continuity correction needed? When λ > 15, approximate X~Po(λ) as X~N(λ, λ). A continuity correction of ±0.5 is required because Poisson is discrete and Normal is continuous. Without it, the boundary probability gets calculated at the wrong point.

Does λ stay the same if the time interval in the question changes? No. λ scales directly with the interval. If the average rate is given per week, halve it for half a week, or multiply it up for a fortnight. Always match λ to the exact interval the question asks about before substituting into any formula.

Why are the mean and variance the same for a Poisson distribution? It's a defining property of the distribution: for X~Po(λ), E(X) = Var(X) = λ, so the standard deviation is √λ. This is different from the Binomial distribution, where mean (np) and variance (np(1−p)) are not equal unless p = 0.