| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Explain or apply conditions in context |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard Poisson conditions and basic calculations. Part (a) requires textbook knowledge of Poisson assumptions with contextual explanation. Part (b)(i-ii) involves routine probability calculations using tables/calculator. Part (b)(iii) requires algebraic manipulation of Poisson probabilities but follows a predictable pattern showing the sum of two independent Poisson distributions—a standard result that stronger students would recognize immediately. |
| 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 |
| Several calls may all refer to the same incident | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Calls occur at constant average rate | B1 | This condition only, allow "average" omitted, *not* "constant *probability*", *not* "random" |
| E.g. No, because incidents are less/more common at night | B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - \left(1 + 2.74 + \frac{2.74^2}{2!}\right)e^{-2.74}\) | M1, M1 | Formula for any one correct Poisson probability for \(r \geq 1\); correct overall formula, allow 1 error |
| \(= \mathbf{0.516(1)}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((e^{-2}\times1)(e^{-3}\times3) + (e^{-2}\times2)(e^{-3}\times1)\) | M1, A1 | Two correct terms multiplied, or all 4 bits seen |
| \(= \mathbf{0.0337}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((e^{-\lambda}\times1)(e^{-\mu}\times\mu) + (e^{-\lambda}\times\lambda)(e^{-\mu}\times1)\) | M1 | Correct algebraic expression [*Ignore 1! throughout*] |
| \(= e^{-\lambda}\times e^{-\mu}(\lambda + \mu)\) | M1, A1 | Take out factor of \(e^{-\lambda}\times e^{-\mu}\); correctly obtain exact answer \([e^{-\lambda-\mu}(\lambda+\mu)]\) |
| \(= e^{-(\lambda+\mu)}(\lambda+\mu) = P(T=1)\) | A1 | 4 |
# Question 8:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Several calls may all refer to the same incident | B1 | **1** | Any reason showing correct understanding of "independent". Not just "singly" or equivalent. "Fires might spread": B0 |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Calls occur at constant average rate | B1 | This condition only, allow "average" omitted, *not* "constant *probability*", *not* "random" |
| E.g. No, because incidents are less/more common at night | B1 | **2** | Any comment (with either yes or no) showing correct understanding. "Fires might not occur at constant average rate" is not enough (gets B1 B0) |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - \left(1 + 2.74 + \frac{2.74^2}{2!}\right)e^{-2.74}$ | M1, M1 | Formula for any one correct Poisson probability for $r \geq 1$; correct overall formula, allow 1 error |
| $= \mathbf{0.516(1)}$ | A1 | **3** | Answer a.r.t. 0.516 |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(e^{-2}\times1)(e^{-3}\times3) + (e^{-2}\times2)(e^{-3}\times1)$ | M1, A1 | Two correct terms multiplied, or all 4 bits seen |
| $= \mathbf{0.0337}$ | A1 | **3** | Answer a.r.t. 0.0337 |
## Part (b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(e^{-\lambda}\times1)(e^{-\mu}\times\mu) + (e^{-\lambda}\times\lambda)(e^{-\mu}\times1)$ | M1 | Correct algebraic expression [*Ignore 1! throughout*] |
| $= e^{-\lambda}\times e^{-\mu}(\lambda + \mu)$ | M1, A1 | Take out factor of $e^{-\lambda}\times e^{-\mu}$; correctly obtain exact answer $[e^{-\lambda-\mu}(\lambda+\mu)]$ |
| $= e^{-(\lambda+\mu)}(\lambda+\mu) = P(T=1)$ | A1 | **4** | All correct, and write down correct formula for $P(T=1)$ |8
\begin{enumerate}[label=(\alph*)]
\item A group of students is discussing the conditions that are needed if a Poisson distribution is to be a good model for the number of telephone calls received by a fire brigade on a working day.
\begin{enumerate}[label=(\roman*)]
\item Alice says "Events must be independent". Explain why this condition may not hold in this context.
\item State a different condition that is needed. Explain whether it is likely to hold in this context.
\end{enumerate}\item The random variables $R , S$ and $T$ have independent Poisson distributions with means $\lambda , \mu$ and $\lambda + \mu$ respectively.
\begin{enumerate}[label=(\roman*)]
\item In the case $\lambda = 2.74$, find $\mathrm { P } ( R > 2 )$.
\item In the case $\lambda = 2$ and $\mu = 3$, find $\mathrm { P } ( R = 0$ and $S = 1 ) + \mathrm { P } ( R = 1$ and $S = 0 )$. Give your answer correct to 4 decimal places.
\item In the general case, show algebraically that
$$\mathrm { P } ( R = 0 \text { and } S = 1 ) + \mathrm { P } ( R = 1 \text { and } S = 0 ) = \mathrm { P } ( T = 1 ) .$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR S2 2011 Q8 [13]}}