- 2068 Jestha (old course) · 5 marks
Normal distribution; approximation of Poisson
Short answer
Normal distribution is continuous with pdf f(x)=(1/(σ√(2π)))e^(-(x-μ)²/(2σ²)), symmetric about μ. Poisson(λ) with large λ (≥10) is approximated by normal N(λ,λ) using continuity correction: z=(x±0.5−λ)/√λ.
Define the normal distribution. Describe normal approximation of Poisson's distribution.
Normal distribution: a continuous distribution with pdf
f(x) = (1 / (σ√(2π))) · e^(−(x − μ)² / (2σ²)), −∞ < x < ∞
where μ is the mean and σ the standard deviation. Its curve is bell-shaped and symmetric about μ, with mean = median = mode = μ, total area 1, and about 68.27%, 95.45% and 99.73% of values within μ ± 1σ, 2σ, 3σ.
Normal approximation of the Poisson distribution
If X ~ Poisson(λ), the mean and variance are both λ. When λ is large (generally λ ≥ 10; some books use λ > 5), the Poisson distribution is closely approximated by the normal distribution with
μ = λ and σ² = λ, i.e. X ≈ N(λ, λ)
Because the Poisson is discrete and the normal is continuous, a continuity correction of 0.5 is used:
- P(X = a) ≈ P(a − 0.5 < X < a + 0.5)
- P(X ≤ a) ≈ P(X < a + 0.5)
- P(X ≥ a) ≈ P(X > a − 0.5)
and the standard variable is z = (x ± 0.5 − λ)/√λ.
Example
Calls arrive at a rate λ = 25 per hour. Find P(X ≤ 30).
z = (30.5 − 25)/√25 = 5.5/5 = 1.10
P(X ≤ 30) ≈ P(Z < 1.10) = 0.8643
The exact Poisson value is 0.8633, so the approximation is good.
✓ Answer checked by Claude Sonnet 5.5
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