| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2022 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Calculate moments from PDF |
| Difficulty | Challenging +1.2 This is a standard Further Maths statistics question on unbiased estimators and variance minimization. Part (a) requires recognizing that X+Y ~ N(180, 2σ²) and calculating a probability (routine). Parts (b) and (c) involve algebraic verification of unbiasedness and variance calculations, then differentiation to minimize variance—all standard techniques for this topic. While it requires multiple steps and careful algebra, it follows predictable patterns without requiring novel insight or geometric creativity beyond the setup. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((X+Y) \sim N(180, 2\sigma^2)\) | B1 | si |
| \(P(180 - \sigma < X + Y < 180 + \sigma) = P\!\left(\frac{180-\sigma-180}{\sqrt{2\sigma^2}} < Z < \frac{180+\sigma-180}{\sqrt{2\sigma^2}}\right)\) | M1 | |
| \(= P\!\left(\frac{-1}{\sqrt{2}} < Z < \frac{1}{\sqrt{2}}\right)\) | M1 | SC2 for only doing one side leading to 0.7602 or 0.76115 from tables |
| \(= 0.52...\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(T_1) = E\!\left(45 + \frac{1}{4}(3X - Y)\right)\) | ||
| \(E(T_1) = 45 + \frac{3}{4}E(X) - \frac{1}{4}E(180 - X)\) | M1 | M1 for either first line |
| \(E(T_1) = 45 + \frac{3}{4}\alpha - 45 + \frac{1}{4}\alpha\) | M1 | |
| \(E(T_1) = \alpha\) | A1 | Convincing |
| \(T_1\) is an unbiased estimator for \(\alpha\) | ||
| \(\text{Var}(T_1) = \text{Var}\!\left(45 + \frac{1}{4}(3X-Y)\right)\) | ||
| \(\text{Var}(T_1) = \frac{9}{16}\text{Var}(X) + \frac{1}{16}\text{Var}(Y)\) | M1 | |
| \(\text{Var}(T_1) = \frac{5}{8}\sigma^2\) | A1 | |
| \(\text{Var}(T_1) < \sigma^2\) | E1 | FT their \(\text{Var}(T_1) = k\sigma^2\) where \(k < 1\) |
| \(\therefore T_1\) is a better estimator than \(X\) | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(T_2) = E(\lambda X + (1-\lambda)(180° - Y))\) | ||
| \(E(T_2) = \lambda\alpha + (1-\lambda)(180° - \beta)\) | M1 | |
| \(E(T_2) = \lambda\alpha + (1-\lambda)(\alpha)\) | ||
| \(E(T_2) = \lambda\alpha + \alpha - \lambda\alpha = \alpha\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(T_2) = \text{Var}(\lambda X + (1-\lambda)(180° - Y))\) | ||
| \(\text{Var}(T_2) = \lambda^2\text{Var}(X) + (1-\lambda)^2\text{Var}(180° - Y)\) | M1 | |
| \(\text{Var}(T_2) = \lambda^2\sigma^2 + (1-\lambda)^2\sigma^2\) | A1 | oe ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{d\lambda}\text{Var}(T_2) = 2\lambda\sigma^2 - 2(1-\lambda)\sigma^2\) | M1A1 | M1 for attempt to differentiate with at least one decrease in power. A1 FT \(\text{Var}(T_2)\) for equivalent difficulty only. |
| \(\frac{d}{d\lambda}\text{Var}(T_2) = 0\) gives the best estimator | M1 | Use of \(\frac{d}{d\lambda}\text{Var}(T_2) = 0\) |
| \(2\lambda\sigma^2 = 2(1-\lambda)\sigma^2 \Rightarrow 2\lambda = 2 - 2\lambda\) | cao | |
| \(\lambda = \frac{1}{2}\) | A1 | Accept alternative justification |
| \(\frac{d^2}{d\lambda^2}\text{Var}(T_2) = 4\sigma^2 > 0\), therefore minimum | E1 | (9) |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(X+Y) \sim N(180, 2\sigma^2)$ | B1 | si |
| $P(180 - \sigma < X + Y < 180 + \sigma) = P\!\left(\frac{180-\sigma-180}{\sqrt{2\sigma^2}} < Z < \frac{180+\sigma-180}{\sqrt{2\sigma^2}}\right)$ | M1 | |
| $= P\!\left(\frac{-1}{\sqrt{2}} < Z < \frac{1}{\sqrt{2}}\right)$ | M1 | SC2 for only doing one side leading to 0.7602 or 0.76115 from tables |
| $= 0.52...$ | A1 | (4) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(T_1) = E\!\left(45 + \frac{1}{4}(3X - Y)\right)$ | | |
| $E(T_1) = 45 + \frac{3}{4}E(X) - \frac{1}{4}E(180 - X)$ | M1 | M1 for either first line |
| $E(T_1) = 45 + \frac{3}{4}\alpha - 45 + \frac{1}{4}\alpha$ | M1 | |
| $E(T_1) = \alpha$ | A1 | Convincing |
| $T_1$ is an unbiased estimator for $\alpha$ | | |
| $\text{Var}(T_1) = \text{Var}\!\left(45 + \frac{1}{4}(3X-Y)\right)$ | | |
| $\text{Var}(T_1) = \frac{9}{16}\text{Var}(X) + \frac{1}{16}\text{Var}(Y)$ | M1 | |
| $\text{Var}(T_1) = \frac{5}{8}\sigma^2$ | A1 | |
| $\text{Var}(T_1) < \sigma^2$ | E1 | FT their $\text{Var}(T_1) = k\sigma^2$ where $k < 1$ |
| $\therefore T_1$ is a better estimator than $X$ | | (6) |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(T_2) = E(\lambda X + (1-\lambda)(180° - Y))$ | | |
| $E(T_2) = \lambda\alpha + (1-\lambda)(180° - \beta)$ | M1 | |
| $E(T_2) = \lambda\alpha + (1-\lambda)(\alpha)$ | | |
| $E(T_2) = \lambda\alpha + \alpha - \lambda\alpha = \alpha$ | A1 | Convincing |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(T_2) = \text{Var}(\lambda X + (1-\lambda)(180° - Y))$ | | |
| $\text{Var}(T_2) = \lambda^2\text{Var}(X) + (1-\lambda)^2\text{Var}(180° - Y)$ | M1 | |
| $\text{Var}(T_2) = \lambda^2\sigma^2 + (1-\lambda)^2\sigma^2$ | A1 | oe ISW |
### Part (c)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{d\lambda}\text{Var}(T_2) = 2\lambda\sigma^2 - 2(1-\lambda)\sigma^2$ | M1A1 | M1 for attempt to differentiate with at least one decrease in power. A1 FT $\text{Var}(T_2)$ for equivalent difficulty only. |
| $\frac{d}{d\lambda}\text{Var}(T_2) = 0$ gives the best estimator | M1 | Use of $\frac{d}{d\lambda}\text{Var}(T_2) = 0$ |
| $2\lambda\sigma^2 = 2(1-\lambda)\sigma^2 \Rightarrow 2\lambda = 2 - 2\lambda$ | | cao |
| $\lambda = \frac{1}{2}$ | A1 | Accept alternative justification |
| $\frac{d^2}{d\lambda^2}\text{Var}(T_2) = 4\sigma^2 > 0$, therefore minimum | E1 | (9) |7.\\
\includegraphics[max width=\textwidth, alt={}, center]{65369843-222f-48b2-b8cd-a1c304eac3d9-6_707_718_347_660}
The diagram above shows a cyclic quadrilateral $A B C D$, where $\widehat { B A D } = \alpha , \widehat { B C D } = \beta$ and $\alpha + \beta = 180 ^ { \circ }$. These angles are measured.\\
The random variables $X$ and $Y$ denote the measured values, in degrees, of $\widehat { B A D }$ and $\widehat { B C D }$ respectively. You are given that $X$ and $Y$ are independently normally distributed with standard deviation $\sigma$ and means $\alpha$ and $\beta$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Calculate, correct to two decimal places, the probability that $X + Y$ will differ from $180 ^ { \circ }$ by less than $\sigma$.
\item Show that $T _ { 1 } = 45 ^ { \circ } + \frac { 1 } { 4 } ( 3 X - Y )$ is an unbiased estimator for $\alpha$ and verify that it is a better estimator than $X$ for $\alpha$.
\item Now consider $T _ { 2 } = \lambda X + ( 1 - \lambda ) \left( 180 ^ { \circ } - Y \right)$.
\begin{enumerate}[label=(\roman*)]
\item Show that $T _ { 2 }$ is an unbiased estimator for $\alpha$ for all values of $\lambda$.
\item Find $\operatorname { Var } \left( T _ { 2 } \right)$ in terms of $\lambda$ and $\sigma$.
\item Hence determine the value of $\lambda$ which gives the best unbiased estimator for $\alpha$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2022 Q7 [19]}}