| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Expectation and variance of Poisson-related expressions |
| Difficulty | Moderate -0.8 This question tests basic understanding of linear transformations of random variables (mean and variance scale by 2 and 4 respectively) and the defining property that Poisson distributions require mean equals variance. Both parts are direct application of standard results with no problem-solving required, making it easier than average. |
| Spec | 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02m Poisson: mean = variance = lambda5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean \(= 2.6\) | B1 | |
| \(\text{Var} = 4 \times 1.3\) | M1 | M1 for either \(4\times\), or for \(\text{Var}(X) = 1.3\) implied |
| \(= 5.2\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var} \neq \text{mean}\) or \(2X\) does not take all integer values | B1 [1] | \(X\) and \(X\) are not independent oe |
## Question 1:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean $= 2.6$ | B1 | |
| $\text{Var} = 4 \times 1.3$ | M1 | M1 for either $4\times$, or for $\text{Var}(X) = 1.3$ implied |
| $= 5.2$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var} \neq \text{mean}$ or $2X$ does not take all integer values | B1 [1] | $X$ and $X$ are not independent oe |
---1 The random variable $X$ has the distribution $\operatorname { Po } ( 1.3 )$. The random variable $Y$ is defined by $Y = 2 X$.\\
(i) Find the mean and variance of $Y$.\\
(ii) Give a reason why the variable $Y$ does not have a Poisson distribution.
\hfill \mbox{\textit{CAIE S2 2011 Q1 [4]}}