| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Given PGF manipulation and properties |
| Difficulty | Challenging +1.3 This is a Further Maths statistics question requiring multiple standard techniques (expectation, variance, unbiased estimators, comparing estimators via variance) across several parts. While it involves extended working and careful algebra, each component follows well-established methods without requiring novel insight. The constraint on θ and the algebraic manipulation add moderate complexity, placing it above average difficulty but not exceptionally challenging for Further Maths students. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05b Unbiased estimates: of population mean and variance |
| Value of \(X\) | 1 | 3 | 5 |
| Probability | \(\theta\) | \(1 - 3\theta\) | \(2\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \theta + 3(1 - 3\theta) + 5 \times 2\theta = 2\theta + 3\) | M1 | AO1 |
| A1 | AO1 | |
| \(\text{Var}(X) = \theta + 9(1 - 3\theta) + 25 \times 2\theta - (2\theta + 3)^2\) | M1 | AO2 |
| \(= \theta + 9 - 27\theta + 50\theta - 4\theta^2 - 12\theta - 9 = 4\theta(3 - \theta)\) | A1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| Consider \(E(V) = \frac{E(\bar{X}) - 3}{2} = \frac{2\theta + 3 - 3}{2} = \theta\) | M1 | AO2 |
| A1 | AO2 | |
| (Therefore V is unbiased) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(V) = \frac{\text{Var}(\bar{X})}{4} = \frac{\theta(3 - \theta)}{n}\) | M1 | AO3 |
| A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Y is \(B(n, \theta)\) So \(E(Y) = n\theta\) | M1 | AO3 |
| A1 | AO2 | |
| \(E(W) = E\left(\frac{Y}{n}\right) = \theta\) | A1 | AO2 |
| (Therefore W is unbiased) | ||
| \(\text{Var}(W) = \frac{\text{Var}(Y)}{n^2} = \frac{\theta(1 - \theta)}{n}\) | M1 | AO2 |
| A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\text{Var}(V)}{\text{Var}(W)} = \frac{\theta(3 - \theta)}{n} \div \frac{\theta(1 - \theta)}{n} = \frac{3 - \theta}{1 - \theta}\) | M1 | AO3 |
| A1 | AO1 | |
| It follows that W is the better estimator since it has the smaller variance | B1 | AO2 |
| B1 | AO2 | |
| [17] |
## 7(a)
$E(X) = \theta + 3(1 - 3\theta) + 5 \times 2\theta = 2\theta + 3$ | M1 | AO1 |
| A1 | AO1 |
$\text{Var}(X) = \theta + 9(1 - 3\theta) + 25 \times 2\theta - (2\theta + 3)^2$ | M1 | AO2 |
$= \theta + 9 - 27\theta + 50\theta - 4\theta^2 - 12\theta - 9 = 4\theta(3 - \theta)$ | A1 | AO2 |
## 7(b)(i)
Consider $E(V) = \frac{E(\bar{X}) - 3}{2} = \frac{2\theta + 3 - 3}{2} = \theta$ | M1 | AO2 |
| A1 | AO2 |
(Therefore V is unbiased) | | |
## 7(b)(ii)
$\text{Var}(V) = \frac{\text{Var}(\bar{X})}{4} = \frac{\theta(3 - \theta)}{n}$ | M1 | AO3 |
| A1 | AO1 |
## 7(c)
Y is $B(n, \theta)$ So $E(Y) = n\theta$ | M1 | AO3 |
| A1 | AO2 |
$E(W) = E\left(\frac{Y}{n}\right) = \theta$ | A1 | AO2 |
(Therefore W is unbiased) | | |
$\text{Var}(W) = \frac{\text{Var}(Y)}{n^2} = \frac{\theta(1 - \theta)}{n}$ | M1 | AO2 |
| A1 | AO1 |
## 7(d)
$\frac{\text{Var}(V)}{\text{Var}(W)} = \frac{\theta(3 - \theta)}{n} \div \frac{\theta(1 - \theta)}{n} = \frac{3 - \theta}{1 - \theta}$ | M1 | AO3 |
| A1 | AO1 |
It follows that W is the better estimator since it has the smaller variance | B1 | AO2 |
| B1 | AO2 |
| [17] | |The discrete random variable $X$ has the following probability distribution, where $\theta$ is an unknown parameter belonging to the interval $\left(0, \frac{1}{3}\right)$.
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
Value of $X$ & 1 & 3 & 5 \\
\hline
Probability & $\theta$ & $1 - 3\theta$ & $2\theta$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Obtain an expression for $E(X)$ in terms of $\theta$ and show that
$$\text{Var}(X) = 4\theta(3 - \theta).$$ [4]
In order to estimate the value of $\theta$, a random sample of $n$ observations on $X$ was obtained and $\bar{X}$ denotes the sample mean.
\item
\begin{enumerate}[label=(\roman*)]
\item Show that
$$V = \frac{\bar{X} - 3}{2}$$
is an unbiased estimator for $\theta$.
\item Find an expression for the variance of $V$. [4]
\end{enumerate}
\item Let $Y$ denote the number of observations in the random sample that are equal to 1.
Show that
$$W = \frac{Y}{n}$$
is an unbiased estimator for $\theta$ and find an expression for $\text{Var}(W)$. [5]
\item Determine which of $V$ and $W$ is the better estimator, explaining your method clearly. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 Q7 [17]}}