| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2002 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Calculate single probability using Poisson approximation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard Poisson approximation to binomial. Part (i) requires simple algebra (n = E(X)/p = 2.55/0.015 = 170). Part (ii) involves calculating P(X < 3) using Poisson with λ = np = 3.15, which is routine calculator work. No conceptual challenges or problem-solving required—pure procedural application of a standard approximation technique. |
| Spec | 2.04d Normal approximation to binomial5.02d Binomial: mean np and variance np(1-p)5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.015n = 2.55\), \(n = 170\) | M1, A1 (2 marks) | For equation linking \(n\), \(p\) and mean. For correct answer. |
| (ii) mean \(= 210 \times 0.015 = 3.15\) | B1 | For new mean. |
| \(P(0) + P(1) + P(2) = e^{-3.15}\left(1 + 3.15 + \frac{3.15^2}{2}\right) = 0.390\) or \(0.391\) | M1, A1 (3 marks) | For evaluating Poisson \(P(0) + P(1) + P(2) + [P(3)]\). For correct answer. |
(i) $0.015n = 2.55$, $n = 170$ | M1, A1 (2 marks) | For equation linking $n$, $p$ and mean. For correct answer.
(ii) mean $= 210 \times 0.015 = 3.15$ | B1 | For new mean.
$P(0) + P(1) + P(2) = e^{-3.15}\left(1 + 3.15 + \frac{3.15^2}{2}\right) = 0.390$ or $0.391$ | M1, A1 (3 marks) | For evaluating Poisson $P(0) + P(1) + P(2) + [P(3)]$. For correct answer.
SR use of Binomial scores B1 for final correct answer $\geq 0.389$
---1.5% of the population of the UK can be classified as 'very tall'.
\begin{enumerate}[label=(\roman*)]
\item The random variable $X$ denotes the number of people in a sample of $n$ people who are classified as very tall. Given that E$(X) = 2.55$, find $n$. [2]
\item By using the Poisson distribution as an approximation to a binomial distribution, calculate an approximate value for the probability that a sample of size 210 will contain fewer than 3 people who are classified as very tall. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2002 Q2 [5]}}