| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 3 |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for expectation and variance of linear combinations of independent normal variables. Part (a) requires simple algebra with E(aX̄ + bȲ), part (b) uses Var(aX̄ + bȲ) = a²Var(X̄) + b²Var(Ȳ) with substitution, and part (c) is basic calculus (differentiation to find minimum). While it involves sample means and requires careful algebraic manipulation, it's a routine textbook exercise requiring no novel insight—slightly easier than average for Further Maths statistics. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
**Question 4(a)**
$E(a\bar{X} + b\bar{Y}) = \mu$ [M1]
$E(a\bar{X} + b\bar{Y}) = aE(\bar{X}) + bE(\bar{Y})$ [M1]
$\Rightarrow a\mu + 3b\mu = \mu \Rightarrow a + 3b = 1$ [A1]
**Total: 3 marks**
**Question 4(b)**
$\text{Var}(a\bar{X} + b\bar{Y}) = a^2\text{Var}(\bar{X}) + b^2\text{Var}(\bar{Y}) = a^2\dfrac{\sigma^2}{n} + 4b^2\dfrac{\sigma^2}{n}$ [M1]
$= \dfrac{\sigma^2}{n}(a^2 + 4b^2) = \dfrac{\sigma^2}{n}(1 - 6b + 9b^2 + 4b^2) = \dfrac{\sigma^2}{n}(1 - 6b + 13b^2)$ **AG** [M1A1]
**Total: 3 marks**
**Question 4(c)**
$\dfrac{d}{db}\text{Var}(a\bar{X} + b\bar{Y}) = -6 + 26b = 0 \Rightarrow b = \dfrac{3}{13}$ [M1A1]
$\Rightarrow \text{Var}_{\min}(a\bar{X} + b\bar{Y}) = \dfrac{\sigma^2}{n}\left(1 - 6 \times \dfrac{3}{13} + 13 \times \dfrac{9}{169}\right) = \dfrac{4\sigma^2}{13n}$ [A1]
**Total: 3 marks**4 The independent random variables $X$ and $Y$ have normal distributions where $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ and $Y \sim \mathrm {~N} \left( 3 \mu , 4 \sigma ^ { 2 } \right)$. Two random samples each of size $n$ are taken, one from each of these normal populations.
\begin{enumerate}[label=(\alph*)]
\item Show that $a \bar { X } + b \bar { Y }$ is an unbiased estimator of $\mu$ provided that $a + 3 b = 1$, where $a$ and $b$ are constants and $\bar { X }$ and $\bar { Y }$ are the respective sample means.
In the remainder of the question assume that $a \bar { X } + b \bar { Y }$ is an unbiased estimator of $\mu$.
\item Show that $\operatorname { Var } ( \overline { a X } + b \bar { Y } )$ can be written as $\frac { \sigma ^ { 2 } } { n } \left( 1 - 6 b + 13 b ^ { 2 } \right)$.
\item The value of the constant $b$ can be varied. Find the value of $b$ that gives the minimum of $\operatorname { Var } ( a \bar { X } + b \bar { Y } )$, and hence find the minimum of $\operatorname { Var } ( a \bar { X } + b \bar { Y } )$ in terms of $\sigma$ and $n$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2019 Q4 [3]}}