| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Expectation and variance of Poisson-related expressions |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for expectation and variance of linear combinations of independent random variables, requiring only recall of E(aX+bY) and Var(aX+bY) formulas plus recognition that Z cannot be Poisson since it takes negative values. Slightly above average difficulty due to being Further Maths content, but mechanically routine. |
| Spec | 5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(Z) = 6\) | B1 | |
| \(\text{Var}(Z) = \frac{1}{4}(16) + 2\) | M1 | |
| \(= 6\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No | B1 | Unless accompanied by a spurious reason. Allow \(Z \neq X+Y\) |
| Difference between Poisson distributions is not Poisson, or \(Z\) may be fractional or negative. | B1 | SC Allow B1 for 'no, you cannot subtract Poisson distributions'. |
| [2] |
## Question 1:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(Z) = 6$ | B1 | |
| $\text{Var}(Z) = \frac{1}{4}(16) + 2$ | M1 | |
| $= 6$ | A1 | |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No | B1 | Unless accompanied by a spurious reason. Allow $Z \neq X+Y$ |
| Difference between Poisson distributions is not Poisson, or $Z$ may be fractional or negative. | B1 | SC Allow B1 for 'no, you cannot subtract Poisson distributions'. |
| **[2]** | | |
---1 The independent random variables $X$ and $Y$ have Poisson distributions with parameters 16 and 2 respectively, and $Z = \frac { 1 } { 2 } X - Y$.\\
(i) Find $\mathrm { E } ( Z )$ and $\operatorname { Var } ( Z )$.\\
(ii) State whether $Z$ has a Poisson distribution, giving a reason for your answer.
\hfill \mbox{\textit{OCR S3 2014 Q1 [5]}}