IB DP Mathematics • 2023 • 7 Marks

Statistics & Probability: Poisson Distribution and Normal Approximation

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Flaws in a textile fabric occur independently and at a constant average rate of $1.4$ flaws per square metre. (a) Find the probability that a randomly chosen $2\text{ m}^2$ piece of fabric contains exactly 3 flaws. [3] (b) A large roll of $100\text{ m}^2$ of this fabric is examined. Using a suitable Normal approximation with continuity correction, find the probability that there are at least 150 flaws. [4]

Verified Solution & Marking Scheme

(a) Model as Poisson with scaled parameter
For $2\text{ m}^2$: $\lambda = 1.4 \times 2 = 2.8$. Let $X \sim \text{Po}(2.8)$: $P(X = 3) = \frac{e^{-2.8} (2.8)^3}{3!} = \frac{e^{-2.8} \times 21.952}{6} \approx 0.2225$
(b) Setup Normal approximation for large λ
For $100\text{ m}^2$: $\lambda = 1.4 \times 100 = 140$. Since $\lambda = 140 > 10$, $Y \sim \text{Po}(140) \approx N(\mu = 140, \sigma^2 = 140)$. $\sigma = \sqrt{140} \approx 11.832$.
Apply continuity correction for P(Y ≥ 150)
With continuity correction: $P(Y \ge 150) \approx P(W \ge 149.5)$ where $W \sim N(140, 140)$: $Z = \frac{149.5 - 140}{\sqrt{140}} = \frac{9.5}{11.83216} \approx 0.8029$ $P(Z \ge 0.8029) = 1 - \Phi(0.8029) \approx 1 - 0.7890 = 0.211$
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