| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics question with three standard parts: (a) routine normal approximation to Poisson sum requiring only formula application, (b) conceptual explanation about model validity requiring understanding but not calculation, and (c) algebraic manipulation of Poisson probability formulas. All parts are textbook exercises with no novel insight required, making it slightly easier than average even for Further Maths. |
| 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 |
|---|---|---|
| \(\text{Po}(59)\) | M1\* | Attempt \(\text{Po}[10(3.2+2.7)]\) |
| \(0.912\quad[0.911542\ldots]\) | M1dep | 0.92965: M1A0 |
| A1 | Answer in range \([0.911, 0.912]\). SC \(\text{Po}(5.9)\) stated or implied: M1 (no credit for answer 1) |
| Answer | Marks | Guidance |
|---|---|---|
| If the mass becomes significantly smaller, the average rate may not be constant | B1 | Comment on constant average rate, in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-\lambda}\dfrac{\lambda^r}{r!} = e^{-\lambda}\dfrac{\lambda^{r+1}}{(r+1)!}\) | M1 | Correct formula used in either equation |
| A1 | One correct equation, needn't be simplified | |
| \(\lambda = r+1\) | A1 | Cancel \(e^{-\lambda}\) and \(r!\) to obtain a correct linear equation |
| \(1.5e^{-\lambda}\dfrac{\lambda^{r-1}}{(r-1)!} = e^{-\lambda}\dfrac{\lambda^r}{r!}\) | A1 | A second correct equation, allow 1.5 on wrong side |
| \(1.5r = \lambda\) | A1 | Second correct linear equation |
| \(1.5r = r+1 \Rightarrow \lambda = 3,\ r = 2\) | M1 | Solve simultaneously to get both values |
| A1 | Both correct, exact only |
## Question 8:
### Part (a):
$\text{Po}(59)$ | **M1\*** | Attempt $\text{Po}[10(3.2+2.7)]$
$0.912\quad[0.911542\ldots]$ | **M1dep** | 0.92965: M1A0
| **A1** | Answer in range $[0.911, 0.912]$. SC $\text{Po}(5.9)$ stated or implied: M1 (no credit for answer 1)
[3]
### Part (b):
If the mass becomes significantly smaller, the average rate may not be constant | **B1** | Comment on constant average rate, in context
[1]
### Part (c):
$e^{-\lambda}\dfrac{\lambda^r}{r!} = e^{-\lambda}\dfrac{\lambda^{r+1}}{(r+1)!}$ | **M1** | Correct formula used in either equation
| **A1** | One correct equation, needn't be simplified
$\lambda = r+1$ | **A1** | Cancel $e^{-\lambda}$ and $r!$ to obtain a correct linear equation
$1.5e^{-\lambda}\dfrac{\lambda^{r-1}}{(r-1)!} = e^{-\lambda}\dfrac{\lambda^r}{r!}$ | **A1** | A second correct equation, allow 1.5 on wrong side
$1.5r = \lambda$ | **A1** | Second correct linear equation
$1.5r = r+1 \Rightarrow \lambda = 3,\ r = 2$ | **M1** | Solve simultaneously to get both values
| **A1** | Both correct, exact only
[7]8
\begin{enumerate}[label=(\alph*)]
\item A substance emits particles randomly at a constant average rate of 3.2 per minute. A second substance emits particles randomly, and independently of the first source, at a constant average rate of 2.7 per minute.
Find the probability that the total number of particles emitted by the two sources in a ten-minute period is less than 70 .
\item The random variable $X$ represents the number of particles emitted by a substance in a fixed time interval $t$ minutes. It may be assumed that particles are emitted randomly and independently of each other.
In general, the rate at which particles are emitted is proportional to the mass of the substance, but each particle emitted reduces the mass of the substance.
Explain why a Poisson distribution may not be a valid model for $X$ if the value of $t$ is very large.
\item The random variable $Y$ has the distribution $\operatorname { Po } ( \lambda )$. It is given that\\
$\mathrm { P } ( \mathrm { Y } = \mathrm { r } ) = \mathrm { P } ( \mathrm { Y } = \mathrm { r } + 1 )$\\
$\mathrm { P } ( \mathrm { Y } = \mathrm { r } ) = 1.5 \times \mathrm { P } ( \mathrm { Y } = \mathrm { r } - 1 )$.
Determine the following, in either order.
\begin{itemize}
\item The value of $r$
\item The value of $\lambda$
\end{itemize}
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2021 Q8 [11]}}