| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Finding unbiased estimator constraints |
| Difficulty | Standard +0.8 This S4 question requires understanding of unbiased estimators and variance minimization through calculus. Part (a) is straightforward verification, but parts (b)-(c) involve algebraic manipulation with the unbiasedness constraint (a+b=2) and optimization via differentiation. While systematic, it demands careful handling of multiple concepts (expectation, variance properties, constrained optimization) making it moderately challenging for Further Maths students. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(E\left(\frac{2}{3}X_1 + \frac{1}{2}X_2 + \frac{5}{6}X_3\right) = \frac{2}{3}\times\frac{k}{2} + \frac{1}{2}\times\frac{k}{2} + \frac{5}{6}\times\frac{k}{2} = k\) | M1 A1 | |
| \(E(X_1 + X_2 + X_3) = k \Rightarrow\) unbiased | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(E(aX_1 + bX_2) = a\frac{k}{2} + b\frac{k}{2} = k\) | M1 | |
| \(a + b = 2\) | A1 | |
| \(\text{Var}(aX_1 + bX_2) = a^2\frac{k^2}{12} + b^2\frac{k^2}{12}\) | M1A1 | |
| \(= a^2\frac{k^2}{12} + (2-a)^2\frac{k^2}{12}\) | M1 | |
| \(= (2a^2 - 4a + 4)\frac{k^2}{12}\) | ||
| \(= (a^2 - 2a + 2)\frac{k^2}{6}\) | A1 cso | (*) since answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Min value when \((2a-2)\frac{k^2}{6} = 0\) | M1A1 | \(\frac{d}{da}(\text{Var})=0\), all correct, condone missing \(\frac{k^2}{6}\) |
| \(\Rightarrow 2a - 2 = 0\), \(a = 1, b = 1\) | A1A1 | |
| \(\frac{d^2(\text{Var})}{da^2} = \frac{2k^2}{6} > 0\) since \(k^2 > 0\) therefore it is a minimum | M1 | |
| min variance \(= (1 - 2 + 2)\frac{k^2}{6} = \frac{k^2}{6}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{k^2}{6}(a-1)^2 - \frac{k^2}{6} + \frac{2k^2}{6}\) | M1 A1 | |
| \(\frac{k^2}{6}(a-1)^2 + \frac{k^2}{6}\) | M1 | |
| Min when \(\frac{k^2}{6}(a-1)^2 = 0\) | A1A1 | |
| \(a = 1,\ b = 1\) | B1 | |
| min var \(= k^2/6\) |
## Question 6:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $E\left(\frac{2}{3}X_1 + \frac{1}{2}X_2 + \frac{5}{6}X_3\right) = \frac{2}{3}\times\frac{k}{2} + \frac{1}{2}\times\frac{k}{2} + \frac{5}{6}\times\frac{k}{2} = k$ | M1 A1 | |
| $E(X_1 + X_2 + X_3) = k \Rightarrow$ unbiased | B1 | |
**(3 marks total)**
---
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $E(aX_1 + bX_2) = a\frac{k}{2} + b\frac{k}{2} = k$ | M1 | |
| $a + b = 2$ | A1 | |
| $\text{Var}(aX_1 + bX_2) = a^2\frac{k^2}{12} + b^2\frac{k^2}{12}$ | M1A1 | |
| $= a^2\frac{k^2}{12} + (2-a)^2\frac{k^2}{12}$ | M1 | |
| $= (2a^2 - 4a + 4)\frac{k^2}{12}$ | | |
| $= (a^2 - 2a + 2)\frac{k^2}{6}$ | A1 cso | (*) since answer given |
**(6 marks total)**
---
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Min value when $(2a-2)\frac{k^2}{6} = 0$ | M1A1 | $\frac{d}{da}(\text{Var})=0$, all correct, condone missing $\frac{k^2}{6}$ |
| $\Rightarrow 2a - 2 = 0$, $a = 1, b = 1$ | A1A1 | |
| $\frac{d^2(\text{Var})}{da^2} = \frac{2k^2}{6} > 0$ since $k^2 > 0$ therefore it is a minimum | M1 | |
| min variance $= (1 - 2 + 2)\frac{k^2}{6} = \frac{k^2}{6}$ | B1 | |
**(6 marks total)**
**Alternative method:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{k^2}{6}(a-1)^2 - \frac{k^2}{6} + \frac{2k^2}{6}$ | M1 A1 | |
| $\frac{k^2}{6}(a-1)^2 + \frac{k^2}{6}$ | M1 | |
| Min when $\frac{k^2}{6}(a-1)^2 = 0$ | A1A1 | |
| $a = 1,\ b = 1$ | B1 | |
| min var $= k^2/6$ | | |6. A continuous uniform distribution on the interval $[ 0 , k ]$ has mean $\frac { k } { 2 }$ and variance $\frac { k ^ { 2 } } { 12 }$. A random sample of three independent variables $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$ is taken from this distribution.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { 2 } { 3 } X _ { 1 } + \frac { 1 } { 2 } X _ { 2 } + \frac { 5 } { 6 } X _ { 3 }$ is an unbiased estimator for $k$.
An unbiased estimator for $k$ is given by $\hat { k } = a X _ { 1 } + b X _ { 2 }$ where $a$ and $b$ are constants.
\item Show that $\operatorname { Var } ( \hat { k } ) = \left( a ^ { 2 } - 2 a + 2 \right) \frac { k ^ { 2 } } { 6 }$
\item Hence determine the value of $a$ and the value of $b$ for which $\hat { k }$ has minimum variance, and calculate this minimum variance.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2009 Q6 [15]}}