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Module 9/Normal approximation & the CLT

Normal vs. Poisson: choosing the approximation

Two approximations, two regimes. Moderate pp with large np(1−p)np(1-p) → normal. Tiny pp with small np2np^2 → Poisson. Picking the wrong one can be badly off.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

X∼Bin(10,0.1)X\sim\mathrm{Bin}(10,0.1). Compute the exact P(X=0)=(0.9)10P(X=0)=(0.9)^{10}.

X∼Bin(10,0.1)X\sim\mathrm{Bin}(10,0.1). Compute the exact P(X≤1)=0.910+10(0.1)(0.9)9P(X\le 1)=0.9^{10}+10(0.1)(0.9)^9.

What you’ll be able to do

  • Match a binomial to the right approximation from its parameters.
  • Compute exact binomial probabilities and compare to each approximation.
  • Explain why the wrong approximation fails (via np(1−p)np(1-p) and np2np^2).

In your course

· MATH2015 · Linear Algebra & Probability
§14.3 Normal vs. Poisson approximations to the binomial
  • Example 14.4Bin(10,0.1)\mathrm{Bin}(10,0.1): Poisson beats normal
  • Example 14.5Bin(40,0.5)\mathrm{Bin}(40,0.5): normal beats Poisson
  • Theorem 14.1 / Prop. 14.1Diagnostics np(1−p)np(1-p) and np2np^2
1

Two regimes

The normal approximation is useful when nn is large and pp is moderate (rule: np(1−p)>10np(1-p)>10), especially for interval probabilities. The Poisson approximation is useful when nn is large and pp is small (so np2np^2 is tiny). They often give quite different numbers — context tells you which applies.

2

A quick diagnostic

Compute two quantities: np(1−p)np(1-p) and np2np^2. If np(1−p)>10np(1-p)>10, trust the normal. If np2np^2 is small (and λ=np\lambda=np moderate), trust the Poisson. With p=12p=\tfrac12 successes are not rare, so Poisson is inappropriate; with tiny pp the distribution is too skewed for the normal.

Guidance (Theorem 14.1 & Proposition 14.1)

Use the normal approximation when np(1−p)>10np(1-p)>10 (reliable for intervals). Use the Poisson approximation when np2np^2 is small. The two conditions rarely hold at once, so the parameters usually point clearly to one choice.

Intuition. np(1−p)np(1-p) measures spread (normal needs enough of it); np2np^2 measures rarity (Poisson needs it tiny).

Worked examples

Example 1

X∼Bin(10,0.1)X\sim\mathrm{Bin}(10,0.1). Compare P(X≤1)P(X\le 1): exact vs Poisson vs normal.

  1. 1

    Exact: 0.910+10(0.1)(0.9)9≈0.73610.9^{10}+10(0.1)(0.9)^9\approx 0.7361.

  2. 2

    λ=np=1\lambda=np=1, so Poisson(1)(1): e−1+e−1≈0.7358e^{-1}+e^{-1}\approx 0.7358 — essentially exact.

  3. 3

    Normal: μ=1, σ2=0.9\mu=1,\ \sigma^2=0.9; with continuity correction z=(1.5−1)/0.9≈0.53z=(1.5-1)/\sqrt{0.9}\approx 0.53, giving ≈0.70\approx 0.70 — noticeably worse.

  4. 4

    Here pp is small and np2=0.1np^2=0.1 is tiny, so Poisson wins.

Answer. Poisson (≈0.736\approx 0.736) beats normal (≈0.70\approx 0.70); exact ≈0.7361\approx 0.7361.
Example 2

X∼Bin(40,0.5)X\sim\mathrm{Bin}(40,0.5). Compare P(X=20)P(X=20): exact vs normal vs Poisson.