| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2019 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Showing estimator is unbiased |
| Difficulty | Challenging +1.2 This is a structured Further Maths statistics question requiring standard techniques: integration for expectation/variance, probability calculation, identifying binomial distribution, and verifying unbiased estimators. While it involves multiple parts and algebraic manipulation, each step follows routine procedures without requiring novel insight. The variance comparison at the end is straightforward once expressions are obtained. Slightly above average difficulty due to the multi-step nature and Further Maths content, but well within standard examination patterns. |
| Spec | 5.02b Expectation and variance: discrete random variables5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \int x\left(1 + \frac{3\lambda x}{2}\right)dx\) | M1 | Attempt to integrate \(xf(x)\) at least one increase in power |
| \(E(X) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x + \frac{3\lambda x^2}{2}\right)dx\) | — | — |
| \(E(X) = \left[\frac{x^2}{2} + \frac{2\lambda x^3}{2}\right]_{-\frac{1}{2}}^{\frac{1}{2}}\) | A1 | Correct integration |
| \(= \frac{\lambda}{8}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x^2 + \frac{3\lambda x^3}{2}\right)dx\) | M1 | Attempt to integrate \(x^2f(x)\) at least one increase in power |
| \(\text{Var}(X) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x^2 + \frac{3\lambda x^3}{2}\right)dx - \left(\frac{\lambda}{8}\right)^2\) | m1 | subtracting 'their \((E(X))^2\)' from 'their \(E(X^2)\)' |
| \(\text{Var}(X) = \left[\frac{x^3}{3} + \frac{3\lambda x^4}{8}\right]_{-\frac{1}{2}}^{\frac{\lambda^2}{64}}\) | — | — |
| \(\text{Var}(X) = \left(\frac{1}{24} + \frac{3\lambda}{128}\right) - \left(\frac{-1}{24} + \frac{3\lambda}{128}\right) - \frac{\lambda^2}{64}\) | — | — |
| \(\text{Var}(X) = \frac{1}{12} - \frac{\lambda^2}{64}\) | A1 | Show substitution of limits and arrive at \(\frac{1}{12} - \frac{\lambda^2}{64}\) convincing |
| \(\text{Var}(X) = \frac{16 - 3\lambda^2}{192}\) | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 0) = \int_0^{\frac{1}{2}}\left(1 + \frac{3\lambda x}{2}\right)dx\) | M1 | Attempt to integrate \(f(x)\) at least one increase in power. With correct limits |
| \(= \left[x + \frac{3\lambda x^2}{4}\right]_0\) | A1 | Convincing |
| \(= \frac{1}{2} + \frac{3\lambda}{l6}\) | — | — |
| \(= \frac{8 + 3\lambda}{16}\) | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| Binomial. \(Y \sim B\left(n, \frac{8+3\lambda}{16}\right)\) | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(T_1) = E\left(\frac{16Y}{3n} - \frac{8}{3}\right)\) | M1 | Use of \(\frac{16E(Y)}{3n} - \frac{8}{3}\) |
| \(= \frac{16E(Y)}{3n} - \frac{8}{3}\) | — | — |
| \(= \frac{16n\left(\frac{8+3\lambda}{16}\right)}{3n} - \frac{8}{3}\) | B1 | Use of \(E(Y) = np\) |
| \(= \frac{8}{3} + \lambda - \frac{8}{3}\) | — | — |
| \(= \lambda\) (therefore unbiased) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(T_1) = \text{Var}\left(\frac{16Y}{3n} - \frac{8}{3}\right)\) | — | — |
| \(= \frac{256n}{9n^2} \cdot \frac{8+3\lambda}{16}\left(1 - \frac{8+3\lambda}{16}\right)\) | M2 | M1 for coeff² M1 for use of npq |
| \(= \frac{256}{9n} \cdot \frac{8+3\lambda}{16} \cdot \frac{8-3\lambda}{16}\) | A1 | — |
| \(= \frac{256}{9n}\left(\frac{64-9\lambda^2}{256}\right)\) | — | — |
| \(= \frac{64-9\lambda^2}{9n}\) | — | convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(T_2) = \text{Var}(8\bar{X})\) | M1 | — |
| \(= 8^2\left(\frac{16-3\lambda^2}{192n}\right)\) | A1 | — |
| \(= \frac{1024-192\lambda^2}{192n} = \frac{16-3\lambda^2}{3n}\) oe | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(T_2) = \frac{48-9\lambda^2}{9n} < \frac{64-9\lambda^2}{9n}\). \(T_2\) is better because it has a smaller variance. | B1 | Must include attempt to compare variances |
## (a)(i)
$E(X) = \int x\left(1 + \frac{3\lambda x}{2}\right)dx$ | M1 | Attempt to integrate $xf(x)$ at least one increase in power
$E(X) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x + \frac{3\lambda x^2}{2}\right)dx$ | — | —
$E(X) = \left[\frac{x^2}{2} + \frac{2\lambda x^3}{2}\right]_{-\frac{1}{2}}^{\frac{1}{2}}$ | A1 | Correct integration
$= \frac{\lambda}{8}$ | A1 | cao
## (a)(ii)
$E(X^2) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x^2 + \frac{3\lambda x^3}{2}\right)dx$ | M1 | Attempt to integrate $x^2f(x)$ at least one increase in power
$\text{Var}(X) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x^2 + \frac{3\lambda x^3}{2}\right)dx - \left(\frac{\lambda}{8}\right)^2$ | m1 | subtracting 'their $(E(X))^2$' from 'their $E(X^2)$'
$\text{Var}(X) = \left[\frac{x^3}{3} + \frac{3\lambda x^4}{8}\right]_{-\frac{1}{2}}^{\frac{\lambda^2}{64}}$ | — | —
$\text{Var}(X) = \left(\frac{1}{24} + \frac{3\lambda}{128}\right) - \left(\frac{-1}{24} + \frac{3\lambda}{128}\right) - \frac{\lambda^2}{64}$ | — | —
$\text{Var}(X) = \frac{1}{12} - \frac{\lambda^2}{64}$ | A1 | Show substitution of limits and arrive at $\frac{1}{12} - \frac{\lambda^2}{64}$ convincing
$\text{Var}(X) = \frac{16 - 3\lambda^2}{192}$ | — | —
## (b)
$P(X > 0) = \int_0^{\frac{1}{2}}\left(1 + \frac{3\lambda x}{2}\right)dx$ | M1 | Attempt to integrate $f(x)$ at least one increase in power. With correct limits
$= \left[x + \frac{3\lambda x^2}{4}\right]_0$ | A1 | Convincing
$= \frac{1}{2} + \frac{3\lambda}{l6}$ | — | —
$= \frac{8 + 3\lambda}{16}$ | — | —
**Total: [18]**
## (c)(i)
Binomial. $Y \sim B\left(n, \frac{8+3\lambda}{16}\right)$ | B1 | —
## (c)(ii)
$E(T_1) = E\left(\frac{16Y}{3n} - \frac{8}{3}\right)$ | M1 | Use of $\frac{16E(Y)}{3n} - \frac{8}{3}$
$= \frac{16E(Y)}{3n} - \frac{8}{3}$ | — | —
$= \frac{16n\left(\frac{8+3\lambda}{16}\right)}{3n} - \frac{8}{3}$ | B1 | Use of $E(Y) = np$
$= \frac{8}{3} + \lambda - \frac{8}{3}$ | — | —
$= \lambda$ (therefore unbiased) | A1 | —
## (d)(i)
$\text{Var}(T_1) = \text{Var}\left(\frac{16Y}{3n} - \frac{8}{3}\right)$ | — | —
$= \frac{256n}{9n^2} \cdot \frac{8+3\lambda}{16}\left(1 - \frac{8+3\lambda}{16}\right)$ | M2 | M1 for coeff² M1 for use of npq
$= \frac{256}{9n} \cdot \frac{8+3\lambda}{16} \cdot \frac{8-3\lambda}{16}$ | A1 | —
$= \frac{256}{9n}\left(\frac{64-9\lambda^2}{256}\right)$ | — | —
$= \frac{64-9\lambda^2}{9n}$ | — | convincing
## (d)(ii)
$\text{Var}(T_2) = \text{Var}(8\bar{X})$ | M1 | —
$= 8^2\left(\frac{16-3\lambda^2}{192n}\right)$ | A1 | —
$= \frac{1024-192\lambda^2}{192n} = \frac{16-3\lambda^2}{3n}$ oe | — | —
## (d)(iii)
$\text{Var}(T_2) = \frac{48-9\lambda^2}{9n} < \frac{64-9\lambda^2}{9n}$. $T_2$ is better because it has a smaller variance. | B1 | Must include attempt to compare variances
**Total: [18]**The random variable $X$ has probability density function
$$f(x) = 1 + \frac{3\lambda x}{2} \quad \text{for } -\frac{1}{2} \leqslant x \leqslant \frac{1}{2},$$
$$f(x) = 0 \quad \text{otherwise,}$$
where $\lambda$ is an unknown parameter such that $-1 \leqslant \lambda \leqslant 1$.
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Find E$(X)$ in terms of $\lambda$.
\item Show that $\text{Var}(X) = \frac{16 - 3\lambda^2}{192}$. [6]
\end{enumerate}
\item Show that P$(X > 0) = \frac{8 + 3\lambda}{16}$. [2]
\end{enumerate}
In order to estimate $\lambda$, $n$ independent observations of $X$ are made. The number of positive observations obtained is denoted by $Y$ and the sample mean is denoted by $\overline{X}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item
\begin{enumerate}[label=(\roman*)]
\item Identify the distribution of $Y$.
\item Show that $T_1$ is an unbiased estimator for $\lambda$, where
$$T_1 = \frac{16Y}{3n} - \frac{8}{3}.$$ [4]
\end{enumerate}
\item
\begin{enumerate}[label=(\roman*)]
\item Show that $\text{Var}(T_1) = \frac{64 - 9\lambda^2}{9n}$.
\item Given that $T_2$ is also an unbiased estimator for $\lambda$, where
$$T_2 = 8\overline{X},$$
find an expression for Var$(T_2)$ in terms of $\lambda$ and $n$.
\item Hence, giving a reason, determine which is the better estimator, $T_1$ or $T_2$. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2019 Q8 [18]}}