| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Session | Specimen |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Moderate -0.3 This is a straightforward application of the Poisson sum property (V+W ~ Po(10)) requiring a single probability calculation P(X>10) = 1-P(X≤10) and recalling that independence is required. The conceptual demand is minimal—recognizing that Poisson distributions add under independence is a standard result taught explicitly in Further Maths Statistics, and the calculation is routine using tables or technology. |
| Spec | 5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| \(V + W\) is \(\text{Po}(10)\) | M1 | Uses Poisson \(\lambda = 10\); PI |
| \(P(S > 10) = 1 - P(S \leq 10) = 0.417\) | A1 | Obtains correct probability |
| Answer | Marks | Guidance |
|---|---|---|
| Purchases of printers at *Verigood* are not independent of those at *Winnerprint* | E1 | States that model requires independence of purchases from store to store |
## Question 4(a):
$V + W$ is $\text{Po}(10)$ | M1 | Uses Poisson $\lambda = 10$; PI
$P(S > 10) = 1 - P(S \leq 10) = 0.417$ | A1 | Obtains correct probability
## Question 4(b):
Purchases of printers at *Verigood* are not independent of those at *Winnerprint* | E1 | States that model requires independence of purchases from store to store4 The number of printers, $V$, bought during one day from the Verigood store can be modelled by a Poisson distribution with mean 4.5
The number of printers, $W$, bought during one day from the Winnerprint store can be modelled by a Poisson distribution with mean 5.5
4
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total number of printers bought during one day from Verigood and Winnerprint stores is greater than 10.\\[0pt]
[2 marks]
4
\item State the circumstance under which the distributional model you used in part (a) would not be valid.\\[0pt]
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics Q4 [3]}}