| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Expectation and variance from distribution |
| Difficulty | Challenging +1.2 This is a multi-part S4 question requiring systematic application of standard techniques: computing E(X^n) by integration, using variance formulas, and constructing unbiased estimators. While lengthy (7 parts), each step follows directly from the previous one with no novel insights required. The integration is straightforward, and estimator construction is a standard S4 topic. Slightly above average difficulty due to length and the final variance calculation, but well within expected S4 scope. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^{2\theta} \frac{x^{n+1}}{2\theta^2}\,dx = \left[\frac{x^{n+2}}{2(n+2)\theta^2}\right]\) | M1 | Correct integral |
| \(= 2^{n+1}\theta^n/(n+2)\) | A1 | AEF |
| \(E(X) = 4\theta/3\) | B1ft [3] | B0 if not 'deduced' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X) = 2\theta^2 - (4\theta/3)^2 = 2\theta^2/9\) | M1A1ft | ft (i) with no \(n\) |
| \(\text{Var}(X^2) = E(X^4) - (E(X^2))^2 = 16\theta^4/3 - 4\theta^4 = 4\theta^4/3\) | M1A1ft [4] | ft (i) with no \(n\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(\sum X_i) = 3 \times 4\theta/3 = 4\theta\) | M1, A1ft | ft with no \(n\) |
| \(T_1 = \frac{1}{4}\sum X_i\) | A1ft | ft with no \(n\) or \(\theta\) |
| \(E(\sum X_i^2) = 3 \times 2\theta^2 = 6\theta^2\) | M1, A1ft | ft with no \(n\) |
| \(T_2 = (\sum X_i^2)/27\) | A1ft [6] | ft with no \(n\) or \(\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(T_2) = 1/27^2 \times 3 \times \text{Var}(X^2) = 4\theta^4/729\) | M1, A1 [2] | CAO |
## Question 7:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{2\theta} \frac{x^{n+1}}{2\theta^2}\,dx = \left[\frac{x^{n+2}}{2(n+2)\theta^2}\right]$ | M1 | Correct integral |
| $= 2^{n+1}\theta^n/(n+2)$ | A1 | AEF |
| $E(X) = 4\theta/3$ | B1ft **[3]** | B0 if not 'deduced' |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = 2\theta^2 - (4\theta/3)^2 = 2\theta^2/9$ | M1A1ft | ft (i) with no $n$ |
| $\text{Var}(X^2) = E(X^4) - (E(X^2))^2 = 16\theta^4/3 - 4\theta^4 = 4\theta^4/3$ | M1A1ft **[4]** | ft (i) with no $n$ |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(\sum X_i) = 3 \times 4\theta/3 = 4\theta$ | M1, A1ft | ft with no $n$ |
| $T_1 = \frac{1}{4}\sum X_i$ | A1ft | ft with no $n$ or $\theta$ |
| $E(\sum X_i^2) = 3 \times 2\theta^2 = 6\theta^2$ | M1, A1ft | ft with no $n$ |
| $T_2 = (\sum X_i^2)/27$ | A1ft **[6]** | ft with no $n$ or $\theta$ |
### Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(T_2) = 1/27^2 \times 3 \times \text{Var}(X^2) = 4\theta^4/729$ | M1, A1 **[2]** | CAO |
---7 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { x } { 2 \theta ^ { 2 } } & 0 \leqslant x \leqslant 2 \theta \\ 0 & \text { otherwise } \end{cases}$$
where $\theta$ is an unknown positive constant.\\
(i) Find $\mathrm { E } \left( X ^ { n } \right)$, where $n \neq - 2$, and hence write down the value of $\mathrm { E } ( X )$.\\
(ii) Find
\begin{enumerate}[label=(\alph*)]
\item $\operatorname { Var } ( X )$,
\item $\operatorname { Var } \left( X ^ { 2 } \right)$.\\
(iii) Find $\mathrm { E } \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \right)$ and $\mathrm { E } \left( X _ { 1 } ^ { 2 } + X _ { 2 } ^ { 2 } + X _ { 3 } ^ { 2 } \right)$, where $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$ are independent observations of $X$. Hence construct unbiased estimators, $T _ { 1 }$ and $T _ { 2 }$, of $\theta$ and $\operatorname { Var } ( X )$ respectively, which are based on $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$.\\
(iv) Find $\operatorname { Var } \left( T _ { 2 } \right)$.
\end{enumerate}
\hfill \mbox{\textit{OCR S4 2010 Q7 [15]}}