| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Mean and variance of linear combination |
| Difficulty | Challenging +1.2 This is a multi-part Poisson distribution question requiring understanding of independent Poisson variables, probability calculations with redistribution logic, and linear combinations. Part (i) requires careful enumeration of overflow scenarios, part (ii) involves finding specific combinations summing to £400, and part (iii) tests knowledge of linear transformations of Poisson variables. While conceptually more demanding than routine single-distribution problems, the techniques are standard for S3 level with no novel insights required beyond systematic case analysis. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A + B \sim Po(12)\) seen | B1 | |
| \(1 - 0.7720\) | M1 | |
| 0.228 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = 0\ \&\ B = 5\) AND \(A = 4\ \&\ B = 2\) identified | B1 | These 2 pairs only. Allow B=5 alone (+A4,B2) for this mark. |
| \(e^{-6.5} \times e^{-5.5}\frac{5.5^5}{5!} + e^{-6.5} \times \frac{6.5^4}{4!} \times e^{-5.5} \times \frac{5.5^2}{2!}\) | B1;B1 | Each product seen (not nec added), or 0.0015x0.1714; 0.11182x0.06181. Or 0.000258; 0.00691. Allow from tables. Eg 0.5289-0.3575=0.1714 |
| 0.00717 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 830\) | B1 | |
| \(\text{Var} = 60^2 \times 6.5 + 80^2 \times 5.5\) | M1 | |
| \(= 58\,600\) | A1 | |
| Var \(\neq\) Mean, so no. | B1ft | Any correct reason. eg Not all integer values possible. |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A + B \sim Po(12)$ seen | B1 | |
| $1 - 0.7720$ | M1 | |
| 0.228 | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = 0\ \&\ B = 5$ AND $A = 4\ \&\ B = 2$ identified | B1 | These 2 pairs only. Allow B=5 alone (+A4,B2) for this mark. |
| $e^{-6.5} \times e^{-5.5}\frac{5.5^5}{5!} + e^{-6.5} \times \frac{6.5^4}{4!} \times e^{-5.5} \times \frac{5.5^2}{2!}$ | B1;B1 | Each product seen (not nec added), or 0.0015x0.1714; 0.11182x0.06181. Or 0.000258; 0.00691. Allow from tables. Eg 0.5289-0.3575=0.1714 |
| 0.00717 | B1 | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 830$ | B1 | |
| $\text{Var} = 60^2 \times 6.5 + 80^2 \times 5.5$ | M1 | |
| $= 58\,600$ | A1 | |
| Var $\neq$ Mean, so no. | B1ft | Any correct reason. eg Not all integer values possible. |
---5 Two guesthouses, the Albion and the Blighty, have 8 and 6 rooms respectively. The demand for rooms at the Albion has a Poisson distribution with mean 6.5 and the demand for rooms at the Blighty has an independent Poisson distribution with mean 5.5. The owners have agreed that if their guesthouse is full, they will re-direct guests to the other.\\
(i) Find the probability that, on any particular night, the two guesthouses together do not have enough rooms to meet demand.\\
(ii) The Albion charges $\pounds 60$ per room per night, and the Blighty $\pounds 80$. Find the probability, that on a particular night, the total income of the two guesthouses is exactly $\pounds 400$.\\
(iii) If $A$ is the number of rooms demanded at the Albion each night, and $B$ the number of rooms demanded at the Blighty each night, find the mean and variance of the variable $C = 60 A + 80 B$. State whether $C$ has a Poisson distribution, giving a reason for your answer.
\hfill \mbox{\textit{OCR S3 2015 Q5 [11]}}