| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Binomial of Poisson approximations |
| Difficulty | Standard +0.8 Part (i)(a) is routine Poisson calculation, but part (i)(b) requires recognizing that daily counts follow Poisson, then the number of 'zero days' follows binomial, which must be approximated by normal distribution—a multi-layered conceptual leap. Part (ii) requires critical thinking about model assumptions. This exceeds standard S2 questions in conceptual sophistication. |
| Spec | 2.04d Normal approximation to binomial5.02i 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 |
|---|---|---|
| Hence \(\text{P}(\text{exactly 2}) = 0.9856 - 0.9098 = 0.0758\) | B1 | For use of correct Poisson mean |
| M1 | For relevant use of tables (or formula) | |
| A1 | For correct answer 0.0758 | |
| (i)(b) No. of days with no cars ~ \(\text{B}(365, 0.6065)\) | M1 | For relevant Poisson probability of P(0) |
| Normal approximation is \(\text{N}(221.3725, 87.11)\) | A1 | For identifying correct binomial distribution |
| A1 | For correct use of \(np\) and \(npq\) | |
| \(\text{P}(<205) = \text{P}\left(Z < \frac{204.5 - 221.3725}{\sqrt{87.11}}\right)\) | M1 | For standardising (with or without c.c. here) |
| \(= \Phi(-1.808) = 0.0353\) | A1 | For completely correct expression |
| A1 | For correct answer 0.0353 | |
| (ii) Events (cars running out of petrol) must occur at a constant average rate. This seems unlikely, given that there will be different volumes of traffic on different days of the week (e.g. weekdays and weekends) | B1 | For correct statement of the condition |
| B1 | For a correct explanation |
(i)(a) For one day, the distribution is $\text{Po}(0.5)$
Hence $\text{P}(\text{exactly 2}) = 0.9856 - 0.9098 = 0.0758$ | B1 | For use of correct Poisson mean
| M1 | For relevant use of tables (or formula)
| A1 | For correct answer 0.0758
(i)(b) No. of days with no cars ~ $\text{B}(365, 0.6065)$ | M1 | For relevant Poisson probability of P(0)
Normal approximation is $\text{N}(221.3725, 87.11)$ | A1 | For identifying correct binomial distribution
| A1 | For correct use of $np$ and $npq$
$\text{P}(<205) = \text{P}\left(Z < \frac{204.5 - 221.3725}{\sqrt{87.11}}\right)$ | M1 | For standardising (with or without c.c. here)
$= \Phi(-1.808) = 0.0353$ | A1 | For completely correct expression
| A1 | For correct answer 0.0353
(ii) Events (cars running out of petrol) must occur at a constant average rate. This seems unlikely, given that there will be different volumes of traffic on different days of the week (e.g. weekdays and weekends) | B1 | For correct statement of the condition
| B1 | For a correct explanation
**Total: 11 marks**
---6 On average a motorway police force records one car that has run out of petrol every two days.
\begin{enumerate}[label=(\roman*)]
\item (a) Using a Poisson distribution, calculate the probability that, in one randomly chosen day, the police force records exactly two cars that have run out of petrol.\\
(b) Using a Poisson distribution and a suitable approximation to the binomial distribution, calculate the probability that, in one year of 365 days, there are fewer than 205 days on which the police force records no cars that have run out of petrol.
\item State an assumption needed for the Poisson distribution to be appropriate in part (i), and explain why this assumption is unlikely to be valid.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 Q6 [11]}}