| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| 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 Poisson distribution question requiring standard techniques: identifying the distribution from context, calculating probabilities using tables/calculator, finding the mode algebraically (simple inequality manipulation), and interpreting parameter constraints. Part (d) requires understanding that for Poisson mode = 10, we need 9 < λ ≤ 10, which is slightly beyond routine but still standard Further Stats content. Overall slightly easier than average A-level due to the structured guidance through each part. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Po}(14.4)\) | B1 [1] | Stated or implied. Allow just "Poisson" if \(14.4\) used in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - P(\leq 20)\) | M1 | \(0.094(0)\) or \(0.0372\): M1A0. \(0.06\) or \(0.060\): M1A0 |
| \(= 0.0604\ (0.060351\ldots)\) | A1 [2] | In range \([0.0603, 0.0604]\). NB: be aware that \(0.94\) is M0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-14.4}\frac{14.4^r}{r!} > e^{-14.4}\frac{14.4^{r+1}}{(r+1)!}\) | M1 | Correct inequality |
| \(\Rightarrow r+1 > 14.4\) so \(r > 13.4\) AG | A1 [2] | Correctly derive AG, allow \(=\) throughout until conclusion. One further intermediate line needed, condone omissions of brackets |
| Answer | Marks | Guidance |
|---|---|---|
| \(14\) | B1 [1] | \(14\) only |
| Answer | Marks | Guidance |
|---|---|---|
| \(14\) (or \(14.4\)) is some way away from \(10\) (and \(10\) is based on a large sample) so a Poisson distribution (or \(\lambda = 14.4\)) is unlikely. | B1ft [1] | OE. FT on their mode. Not too definite – not just "invalid". \(P(R=10) = 0.0589\) irrelevant. Allow "the population parameter may not be close to \(14\)", but *not* "the mode may not be \(10\)". Not \(10 \neq 14\). B0 if other wrong or irrelevant statements seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(10) > P(11)\) (and \(> P(9)\)) \(\Rightarrow \lambda < 11\) | M1 | Either \(\lambda < 11\) or \(\lambda > 10\) (or \(\lambda \geq 10\)) |
| \(10 < \lambda < 11\) | A1 [2] | This or exact equivalent only, but allow \(10 \leq \lambda < 11\), allow \(\mu\). Allow M1A1 if both \(\lambda < 11\) and \(\lambda > 10\) seen but wrongly combined at end |
# Question 5:
## Part (a):
$\text{Po}(14.4)$ | **B1 [1]** | Stated or implied. Allow just "Poisson" if $14.4$ used in (b)
## Part (b)(i):
$1 - P(\leq 20)$ | **M1** | $0.094(0)$ or $0.0372$: M1A0. $0.06$ or $0.060$: M1A0
$= 0.0604\ (0.060351\ldots)$ | **A1 [2]** | In range $[0.0603, 0.0604]$. NB: be aware that $0.94$ is M0
## Part (b)(ii):
$e^{-14.4}\frac{14.4^r}{r!} > e^{-14.4}\frac{14.4^{r+1}}{(r+1)!}$ | **M1** | Correct inequality
$\Rightarrow r+1 > 14.4$ so $r > 13.4$ **AG** | **A1 [2]** | Correctly derive AG, allow $=$ throughout until conclusion. One further intermediate line needed, condone omissions of brackets
## Part (b)(iii):
$14$ | **B1 [1]** | $14$ only
## Part (c):
$14$ (or $14.4$) is some way away from $10$ (and $10$ is based on a large sample) so a Poisson distribution (or $\lambda = 14.4$) is unlikely. | **B1ft [1]** | OE. FT on their mode. Not too definite – not just "invalid". $P(R=10) = 0.0589$ irrelevant. Allow "the population parameter may not be close to $14$", but *not* "the mode may not be $10$". Not $10 \neq 14$. B0 if other wrong or irrelevant statements seen
## Part (d):
$P(10) > P(11)$ (and $> P(9)$) $\Rightarrow \lambda < 11$ | **M1** | Either $\lambda < 11$ or $\lambda > 10$ (or $\lambda \geq 10$)
$10 < \lambda < 11$ | **A1 [2]** | This or exact equivalent only, but allow $10 \leq \lambda < 11$, allow $\mu$. Allow M1A1 if both $\lambda < 11$ and $\lambda > 10$ seen but wrongly combined at end
---5 The manager of an emergency response hotline believes that calls are made to the hotline independently and at constant average rate throughout the day. From a small random sample of the population, the manager finds that the mean number of calls made in a 1-hour period is 14.4.
Let $R$ denote the number of calls made in a randomly chosen 1-hour period.
\begin{enumerate}[label=(\alph*)]
\item Using evidence from the small sample, state a suitable distribution with which to model $R$. You should give the value(s) of any parameter(s).
\item In this part of the question, use the distribution and value(s) of the parameter(s) from your answer to part (a).
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { P } ( R > 20 )$.
\item Given that $\mathrm { P } ( \mathrm { R } = \mathrm { r } ) > \mathrm { P } ( \mathrm { R } = \mathrm { r } + 1 )$, show algebraically that $r > 13.4$.
\item Hence write down the mode of the distribution.
The manager also finds, from records over many years, that the modal value of $R$ is 10 .
\end{enumerate}\item Use this result to comment on the validity of the distribution used in part (b).
\item Assume now that the type of distribution used in part (b) is valid. Find the range(s) of values of the parameter(s) of this distribution that would correspond to the modal value of $R$ being 10.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2022 Q5 [9]}}