| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson approximation justification or comparison |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson approximation to binomial with standard calculations. Part (a) requires recognizing n=12500, p=1/5000 gives λ=2.5 and using tables/calculator. Part (b) involves computing E(X)=np and Var(X)=np(1-p), then noting they're approximately equal (2.5 vs 2.4995) to justify Poisson. Slightly above average due to the approximation justification, but still routine S2 material with no novel problem-solving required. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim \text{Po}(2.5)\) | B1 | SOI. |
| \(e^{-2.5}(1 + 2.5 + \frac{2.5^2}{2} + \frac{2.5^3}{3!})\) | M1 | Any \(\lambda\). Allow one end error. |
| \(= 0.758\) (3 sf) | A1 | SC use of binomial B1 for 0.758. SC when no working is shown, \(X \sim \text{Po}(2.5)\) seen scores B1, 0.758 seen also scores B1. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = \frac{5}{2}\) or \(2.5\), \(\text{Var}(X) = \frac{4999}{2000}\) or \(2.4995\) | *B1 | Just an answer of 2.5 for the variance is not sufficient. However, 2.4995 is sufficient. |
| These are almost equal | DB1 | Condone 'equal'. |
| 2 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim \text{Po}(2.5)$ | B1 | SOI. |
| $e^{-2.5}(1 + 2.5 + \frac{2.5^2}{2} + \frac{2.5^3}{3!})$ | M1 | Any $\lambda$. Allow one end error. |
| $= 0.758$ (3 sf) | A1 | SC use of binomial B1 for 0.758. SC when no working is shown, $X \sim \text{Po}(2.5)$ seen scores B1, 0.758 seen also scores B1. |
| | **3** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = \frac{5}{2}$ or $2.5$, $\text{Var}(X) = \frac{4999}{2000}$ or $2.4995$ | *B1 | Just an answer of 2.5 for the variance is not sufficient. However, 2.4995 is sufficient. |
| These are almost equal | DB1 | Condone 'equal'. |
| | **2** | |6 It is known that 1 in 5000 people in Atalia have a certain condition. A random sample of 12500 people from Atalia is chosen for a medical trial. The number having the condition is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item Use an appropriate approximating distribution to find $\mathrm { P } ( X \leqslant 3 )$.
\item Find the values of $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$, and explain how your answers suggest that the approximating distribution used in (a) is likely to be appropriate.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q6 [5]}}