| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | PDF with multiple constants |
| Difficulty | Challenging +1.2 This is a multi-part S4 question requiring standard techniques: validating a PDF using non-negativity constraints, finding expectation by integration, constructing unbiased estimators using E(X), and comparing estimators via variance. While it has 5 parts and requires careful algebra, each step follows routine procedures for PDF problems with no novel insights needed. Slightly above average difficulty due to the estimator construction in parts (iii)-(v), which is less commonly practiced than basic PDF manipulation. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(2) = \frac{1}{4}(1 + 2a) \geq 0 \Rightarrow a\geq-\frac{1}{2}\) | M1, A1 | Using \(f(x) \geq 0\), Allow omission of ¼ |
| [2] |
| Answer | Marks |
|---|---|
| \(= [\frac{1}{4}(x^2/2 + ax^3/3)] = 4a/3\) | M1, A1 |
| [2] | |
| (iii) \(E(3X/4) = a \Rightarrow T_1 = 3X_1/4\) | M1 A1 |
| [2] |
| Answer | Marks |
|---|---|
| \(= ^3/_8 (^3/_4a + ^3/_4a)=a, \Rightarrow T_2\) unbiased for \(a\) | M1, A1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(T_2) < \text{Var}(T_1) \Rightarrow T_2\) better | M1 A1, M1 | M1 for \(\sigma^2 \not 2\) for either T. |
| [4] |
**(i)** $f(– 2) = \frac{1}{4}(1 – 2a) \geq 0 \Rightarrow a\leq \frac{1}{2}$
$f(2) = \frac{1}{4}(1 + 2a) \geq 0 \Rightarrow a\geq-\frac{1}{2}$ | M1, A1 | Using $f(x) \geq 0$, Allow omission of ¼
| [2] |
**(ii)** $\int_{-\frac{1}{4}}^1 (x + az^x)dx$
$= [\frac{1}{4}(x^2/2 + ax^3/3)] = 4a/3$ | M1, A1 |
| [2] |
**(iii)** $E(3X/4) = a \Rightarrow T_1 = 3X_1/4$ | M1 A1 |
| [2] |
**(iv)** $E(T_2) = ^3/_8 (E(X_1) + E(X_2)).$
$= ^3/_8 (^3/_4a + ^3/_4a)=a, \Rightarrow T_2$ unbiased for $a$ | M1, A1 |
| [2] |
**(v)** $\text{Var}(T_1) = ^9_{16}\sigma^2$
$\text{Var}(T_2) = ^9_{64}(\sigma^2+\sigma^2) = ^9_{32}\sigma^2$
$\text{Var}(T_2) < \text{Var}(T_1) \Rightarrow T_2$ better | M1 A1, M1 | M1 for $\sigma^2 \not 2$ for either T.
| [4] |7 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 1 } { 4 } ( 1 + a x ) & - 2 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.\\
(i) Show that $| a | \leqslant \frac { 1 } { 2 }$.\\
(ii) Find $\mathrm { E } ( X )$ in terms of $a$.\\
(iii) Construct an unbiased estimator $T _ { 1 }$ of $a$ based on one observation $X _ { 1 }$ of $X$.\\
(iv) A second observation $X _ { 2 }$ is taken. Show that $T _ { 2 }$, where $T _ { 2 } = \frac { 3 } { 8 } \left( X _ { 1 } + X _ { 2 } \right)$, is also an unbiased estimator of a.\\
(v) Given that $\operatorname { Var } ( X ) = \sigma ^ { 2 }$, determine which of $T _ { 1 }$ and $T _ { 2 }$ is the better estimator.
\hfill \mbox{\textit{OCR S4 2012 Q7 [12]}}