| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Moderate -0.3 This is a straightforward application of the Poisson distribution requiring only standard calculations and stating textbook conditions. Part (i) is pure recall, parts (ii) and (iii) involve routine Poisson probability calculations with clearly stated parameters. The only mild challenge is recognizing that different time periods need combining in part (iii), but this is explicitly guided by the question structure. Overall, slightly easier than average due to minimal problem-solving required. |
| 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 |
|---|---|---|
| 4 | (i) | Receipt of an email is an event which occurs randomly, |
| Answer | Marks |
|---|---|
| and at a uniform average rate. | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.3 | |
| 3.3 | Allow constant average rate | |
| 4 | (ii) | (A) |
| P(X = 10) = 0.0829 | B1 | |
| [1] | 1.1 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (ii) | (B) |
| = 0.2123. | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | E |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (iii) | (A) |
| and outside working hours are independent. | E1 | |
| [1] | M | |
| 3.5b | E.g. sum of independent |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (iii) | (B) |
| P(X ≤ 20) = 0.9863 | I |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | BC |
Question 4:
4 | (i) | Receipt of an email is an event which occurs randomly,
independently
and at a uniform average rate. | E1
E1
[2] | 3.3
3.3 | Allow constant average rate
4 | (ii) | (A) | Poisson mean 7.4
P(X = 10) = 0.0829 | B1
[1] | 1.1 | N
BC
4 | (ii) | (B) | P(X ≥ 10) = 1 – 0.7877
= 0.2123. | M1
A1
[2] | 3.4
1.1 | E
OR P(X ≥ 10) = 1 – P(X ≤ 9)
BC
4 | (iii) | (A) | Numbers of junk emails arriving during working hours
and outside working hours are independent. | E1
[1] | M
3.5b | E.g. sum of independent
Poisson distributions is
Poisson
4 | (iii) | (B) | Mean = 7.4 + 16×0.3 = 12.2
P(X ≤ 20) = 0.9863 | I
C
B1
B1
[2] | 3.3
1.1 | BC\begin{enumerate}[label=(\roman*)]
\item State the conditions under which the Poisson distribution is an appropriate model for the number of emails received by one person in a day. [2]
\end{enumerate}
Jane records the number of junk emails which she receives each day. During working hours ($9$am to $5$pm, Monday to Friday) the mean number of junk emails is $7.4$ per day. Outside working hours ($5$pm to $9$am), the mean number of junk emails is $0.3$ per hour.
For the remainder of this question, you should assume that Poisson models are appropriate for the number of junk emails received during each of "working hours" and "outside working hours".
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the probability that the number of junk emails which she receives between $9$am and $5$pm on a Monday is
\begin{enumerate}[label=(\alph*)]
\item exactly $10$, [1]
\item at least $10$. [2]
\end{enumerate}
\item \begin{enumerate}[label=(\alph*)]
\item What assumption must you make to calculate the probability that the number of junk emails which she receives from $9$am Monday to $9$am Tuesday is at most $20$? [1]
\item Find the probability. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor Q4 [8]}}