| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Explain or apply conditions in context |
| Difficulty | Moderate -0.3 This is a straightforward Poisson distribution application requiring scaling the parameter (λ=2 for 3 acres, λ=2/3 for 1 acre) and calculating P(X≥4) and P(X=2) using standard formulas. Part (ii) requires basic understanding of Poisson assumptions. Slightly easier than average due to direct application of learned techniques with minimal problem-solving. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\text{Po}(2): 1 - \text{P}(\leq 3) = 0.1429\) | Marks: M1 A1 | Guidance: Po(2) tables, "1 –" used. Answer, a.r.t. 0.143 |
| Answer: \(\text{Po}(2/3): e^{-2/3}\frac{(2/3)^2}{2!} = 0.114\) | Marks: M1 M1 A1 | Guidance: Parameter 2/3. Poisson formula correct, \(r = 2\), any \(\mu\). Answer, a.r.t. 0.114 |
| Answer: Foxes may congregate so not independent | Marks: B1 B1 | Guidance: Independent/not constant rate/singly used. Any valid relevant application in context |
**Answer:** $\text{Po}(2): 1 - \text{P}(\leq 3) = 0.1429$ | **Marks:** M1 A1 | **Guidance:** Po(2) tables, "1 –" used. Answer, a.r.t. 0.143
**Answer:** $\text{Po}(2/3): e^{-2/3}\frac{(2/3)^2}{2!} = 0.114$ | **Marks:** M1 M1 A1 | **Guidance:** Parameter 2/3. Poisson formula correct, $r = 2$, any $\mu$. Answer, a.r.t. 0.114
**Answer:** Foxes may congregate so not independent | **Marks:** B1 B1 | **Guidance:** Independent/not constant rate/singly used. Any valid relevant application in context1 In a study of urban foxes it is found that on average there are 2 foxes in every 3 acres.\\
(i) Use a Poisson distribution to find the probability that, at a given moment,
\begin{enumerate}[label=(\alph*)]
\item in a randomly chosen area of 3 acres there are at least 4 foxes,
\item in a randomly chosen area of 1 acre there are exactly 2 foxes.\\
(ii) Explain briefly why a Poisson distribution might not be a suitable model.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 Q1 [7]}}