| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard results for expectation and variance of linear combinations of independent random variables. Part (i) requires simple arithmetic with E(X)=Var(X)=5 and E(Y)=Var(Y)=4. Part (ii) is a direct calculation showing equality. Part (iii) tests basic understanding that Poisson distributions require non-negative integer coefficients. No problem-solving or novel insight required—purely routine manipulation of formulas covered in S3. |
| Spec | 5.02i Poisson distribution: random events model5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(S) = 22\) | B1 | |
| \(\text{Var}(S)=E(S)\) | B1 | 2 |
| \(E(T) = \frac{1}{2}x5 - \frac{1}{4}x4=1.5\) | B1 | |
| \(\text{Var}(T)=\frac{1}{4}x5 + \frac{1}{16}x4\) | M1 | |
| \(= 1.5 = E(T)\) AG | A1 | 3 |
| Answer | Marks |
|---|---|
| Using \(\text{Var}(aX+bY)\) CWO |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\) only does not have a Poisson distribution. Some values of \(T\) are EITHER negative OR fractional | B1 | Unless wrong reason |
| \(B1\) | 2(7) |
**1(i)**
$E(S) = 22$ | B1 |
$\text{Var}(S)=E(S)$ | B1 | 2
$E(T) = \frac{1}{2}x5 - \frac{1}{4}x4=1.5$ | B1 |
$\text{Var}(T)=\frac{1}{4}x5 + \frac{1}{16}x4$ | M1 |
$= 1.5 = E(T)$ AG | A1 | 3
**1(ii)**
Using $\text{Var}(aX+bY)$ CWO | |
**1(iii)**
$T$ only does not have a Poisson distribution. Some values of $T$ are EITHER negative OR fractional | B1 | Unless wrong reason
$B1$ | 2(7)
---1 The random variables $X$ and $Y$ are independent with $X \sim \operatorname { Po } ( 5 )$ and $Y \sim \operatorname { Po } ( 4 )$. $S$ denotes the sum of 2 observations of $X$ and 3 observations of $Y$.\\
(i) Find $\mathrm { E } ( S )$ and $\operatorname { Var } ( S )$.\\
(ii) The random variable $T$ is defined by $\frac { 1 } { 2 } X - \frac { 1 } { 4 } Y$. Show that $\mathrm { E } ( T ) = \operatorname { Var } ( T )$.\\
(iii) State which of $S$ and $T$ (if either) does not have a Poisson distribution, giving a reason for your answer.
\hfill \mbox{\textit{OCR S3 2011 Q1 [7]}}