| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Multi-period repeated application |
| Difficulty | Moderate -0.8 This is a straightforward application of standard Poisson distribution properties. Part (a) involves direct calculation from Poisson probability formulas, part (b)(i) requires knowing that the sum of independent Poisson variables is Poisson with summed means (a key textbook result), and parts (b)(ii)-(iii) are routine cumulative probability calculations. No problem-solving insight or novel reasoning is required—just recall and application of standard S2 techniques. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=3) = \frac{e^{-3.5} \times (3.5)^3}{3!} = 0.216\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(Y \geq 5) = 1 - P(Y \leq 4) = 1 - 0.2851 = 0.715\) | M1, A1 | "used" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T \sim Po(9.5)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(7 \leq T \leq 10) = P(T \leq 10) - P(T \leq 6) = 0.6453 - 0.1649 = 0.480\) | M1, A1, A1 | Accept 0.48 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = (0.4804)^3 = 0.111\) | M1, \(A1\checkmark\) |
# Question 2(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=3) = \frac{e^{-3.5} \times (3.5)^3}{3!} = 0.216$ | M1, A1 | |
# Question 2(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(Y \geq 5) = 1 - P(Y \leq 4) = 1 - 0.2851 = 0.715$ | M1, A1 | "used" |
# Question 2(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T \sim Po(9.5)$ | B1 | |
# Question 2(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(7 \leq T \leq 10) = P(T \leq 10) - P(T \leq 6) = 0.6453 - 0.1649 = 0.480$ | M1, A1, A1 | Accept 0.48 |
# Question 2(b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = (0.4804)^3 = 0.111$ | M1, $A1\checkmark$ | |
---2 The number of telephone calls per day, $X$, received by Candice may be modelled by a Poisson distribution with mean 3.5.
The number of e-mails per day, $Y$, received by Candice may be modelled by a Poisson distribution with mean 6.0.
\begin{enumerate}[label=(\alph*)]
\item For any particular day, find:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X = 3 )$;
\item $\quad \mathrm { P } ( Y \geqslant 5 )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down the distribution of $T$, the total number of telephone calls and e-mails per day received by Candice.
\item Determine $\mathrm { P } ( 7 \leqslant T \leqslant 10 )$.
\item Hence calculate the probability that, on each of three consecutive days, Candice will receive a total of at least 7 but at most 10 telephone calls and e-mails.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2007 Q2 [10]}}