| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Multi-period repeated application |
| Difficulty | Standard +0.3 This is a straightforward S2 Poisson question testing standard results: direct probability calculation, recognizing variance equals mean, sum of independent Poissons, and a simple binomial application. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02j 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 |
| \(P(X=8) = P(X \leq 8) - P(X \leq 7) = 0.8472 - 0.7440\) | M1 | \(P(X=8) = \frac{e^{-6}(6^8)}{8!}\) |
| \(= 0.103\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 9\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(Y>9) = 1 - P(Y \leq 9) = 1 - 0.5874\) | M1 | |
| \(= 0.4126\) | A1ft | 2 marks; AWFW 0.412 to 0.413 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T \sim \text{Po}(15)\) | B1ft | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T \leq 20) = 0.917\) | B1ft | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T \text{ at least } 21) = 0.083\) | B1ft | |
| \(p = 15 \times (0.083)^4 \times (0.917)^2 = 0.000599\) | M1, A1 | 3 marks; B(6,(iii)) used; CAO; AWFW 0.000598 to 0.0006 |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=8) = P(X \leq 8) - P(X \leq 7) = 0.8472 - 0.7440$ | M1 | $P(X=8) = \frac{e^{-6}(6^8)}{8!}$ |
| $= 0.103$ | A1 | 2 marks total |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 9$ | B1 | 1 mark |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(Y>9) = 1 - P(Y \leq 9) = 1 - 0.5874$ | M1 | |
| $= 0.4126$ | A1ft | 2 marks; AWFW 0.412 to 0.413 |
### Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T \sim \text{Po}(15)$ | B1ft | 1 mark |
### Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T \leq 20) = 0.917$ | B1ft | 1 mark |
### Part (c)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T \text{ at least } 21) = 0.083$ | B1ft | |
| $p = 15 \times (0.083)^4 \times (0.917)^2 = 0.000599$ | M1, A1 | 3 marks; B(6,(iii)) used; CAO; AWFW 0.000598 to 0.0006 |
---2
\begin{enumerate}[label=(\alph*)]
\item The number of telephone calls, $X$, received per hour for Dr Able may be modelled by a Poisson distribution with mean 6 .
Determine $\mathrm { P } ( X = 8 )$.
\item The number of telephone calls, $Y$, received per hour for Dr Bracken may be modelled by a Poisson distribution with mean $\lambda$ and standard deviation 3 .
\begin{enumerate}[label=(\roman*)]
\item Write down the value of $\lambda$.
\item Determine $\mathrm { P } ( Y > \lambda )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Assuming that $X$ and $Y$ are independent Poisson variables, write down the distribution of the total number of telephone calls received per hour for Dr Able and Dr Bracken.
\item Determine the probability that a total of at most 20 telephone calls will be received during any one-hour period.
\item The total number of telephone calls received during each of 6 one-hour periods is to be recorded. Calculate the probability that a total of at least 21 telephone calls will be received during exactly 4 of these one-hour periods.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2008 Q2 [10]}}