---
title: "Normal distribution; approximation of Poisson"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:normal-approx-poisson-define"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/normal-approx-poisson-define
site: BCT Engineering
---

# Normal distribution; approximation of Poisson

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2068 Jestha (old course), 5 marks
- Video: [Poisson to Normal Approximation](https://www.youtube.com/watch?v=-YfOip1YB3E) — Mr Camilleri
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/normal-approx-poisson-define
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

Define the normal distribution. Describe normal approximation of Poisson's distribution.

## 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.

## Answer

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For large λ, Poisson(λ) is approximated by:
   A. N(λ, λ)
   B. N(λ, λ^2)
   C. N(0, 1)
   D. N(1/λ, 1/λ)
   Answer: A. Mean and variance of the Poisson are both λ. (Key point: Parameters of approximation)

2. Why is a 0.5 continuity correction used?
   A. Poisson is discrete while normal is continuous
   B. To make the mean equal to 0
   C. Because λ is small
   D. To reduce the variance
   Answer: A. Discrete values are treated as intervals of width 1, e.g. P(X = a) ≈ P(a − 0.5 < X < a + 0.5). (Key point: Reason for continuity correction)

3. P(X ≤ 30) is approximated using which boundary?
   A. 29.5
   B. 30
   C. 30.5
   D. 31
   Answer: C. P(X ≤ a) ≈ P(X < a + 0.5), so z = (30.5 − 25)/5 = 1.10. (Key point: Correction applied to P(X ≤ a))

4. For λ = 25, P(X ≤ 30) ≈ P(Z < 1.10) equals:
   A. 0.3643
   B. 0.8643
   C. 0.1357
   D. 0.9
   Answer: B. 0.5 + 0.3643 = 0.8643, close to the exact Poisson value 0.8633. (Key point: Example result)

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