| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Multi-period repeated application |
| Difficulty | Moderate -0.3 This is a straightforward S2 Poisson question requiring standard techniques: calculating probabilities from tables, adding independent Poisson distributions (X+Y ~ Po(6.5)), and using binomial distribution for repeated trials. Part (c) involves basic mean/variance calculations and the standard Poisson property check (mean ≈ variance). All steps are routine applications of learned procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04c Calculate binomial probabilities5.01a Permutations and combinations: evaluate probabilities5.02i 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 |
| \(X \sim Po(5.0)\), \(P(X < 4) = P(X \leq 3) = 0.265\) | B2 | \(0.440\) to \(0.441\) for B1; CAO |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim Po(1.5)\), \(P(Y=4) = \dfrac{e^{-1.5} \times (1.5)^4}{4!}\) | M1 | |
| \(= 0.0471\) | A1 | \(0.047\) to \(0.0471\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = X + Y \sim Po(6.5)\) | B1 | |
| \(P(T > 5) = 1 - P(T \leq 5)\) | B1 | \((1 - 0.2237)\) or \((1 - 0.5265)\) |
| \(= 1 - 0.369 = 0.631\) | B1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = \,^8C_7\,(0.631)^7(0.369) + (0.631)^8\) | M1ft | ft on their \(p\) from (b)(i); Either part attempted |
| \(p = 0.11758 + 0.02513 = 0.143\) | A1ft, A1 | Both parts correct; AWFW \(1.142\) to \(0.143\) (CAO) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mean \(= 8\) | B1 | CAO |
| Variance \(= s^2 = 16.9\) (sample variance \(= 15.2\)) | B1 | AWRT |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Poisson not a good model for data | B1dep | |
| Mean \(\neq\) Variance | B1 | |
| Total | 2 |
# Question 2:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(5.0)$, $P(X < 4) = P(X \leq 3) = 0.265$ | B2 | $0.440$ to $0.441$ for B1; CAO |
| **Total** | **2** | |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim Po(1.5)$, $P(Y=4) = \dfrac{e^{-1.5} \times (1.5)^4}{4!}$ | M1 | |
| $= 0.0471$ | A1 | $0.047$ to $0.0471$ |
| **Total** | **2** | |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = X + Y \sim Po(6.5)$ | B1 | |
| $P(T > 5) = 1 - P(T \leq 5)$ | B1 | $(1 - 0.2237)$ or $(1 - 0.5265)$ |
| $= 1 - 0.369 = 0.631$ | B1 | |
| **Total** | **3** | |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = \,^8C_7\,(0.631)^7(0.369) + (0.631)^8$ | M1ft | ft on their $p$ from (b)(i); Either part attempted |
| $p = 0.11758 + 0.02513 = 0.143$ | A1ft, A1 | Both parts correct; AWFW $1.142$ to $0.143$ (CAO) |
| **Total** | **3** | |
## Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= 8$ | B1 | CAO |
| Variance $= s^2 = 16.9$ (sample variance $= 15.2$) | B1 | AWRT |
| **Total** | **2** | |
## Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Poisson not a good model for data | B1dep | |
| Mean $\neq$ Variance | B1 | |
| **Total** | **2** | |
---2 John works from home. The number of business letters, $X$, that he receives on a weekday may be modelled by a Poisson distribution with mean 5.0.
The number of private letters, $Y$, that he receives on a weekday may be modelled by a Poisson distribution with mean 1.5.
\begin{enumerate}[label=(\alph*)]
\item Find, for a given weekday:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 4 )$;
\item $\quad \mathrm { P } ( Y = 4 )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Assuming that $X$ and $Y$ are independent random variables, determine the probability that, on a given weekday, John receives a total of more than 5 business and private letters.
\item Hence calculate the probability that John receives a total of more than 5 business and private letters on at least 7 out of 8 given weekdays.
\end{enumerate}\item The numbers of letters received by John's neighbour, Brenda, on 10 consecutive weekdays are
$$\begin{array} { l l l l l l l l l l }
15 & 8 & 14 & 7 & 6 & 8 & 2 & 8 & 9 & 3
\end{array}$$
\begin{enumerate}[label=(\roman*)]
\item Calculate the mean and the variance of these data.
\item State, giving a reason based on your answers to part (c)(i), whether or not a Poisson distribution might provide a suitable model for the number of letters received by Brenda on a weekday.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2009 Q2 [14]}}