| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pooled variance estimation |
| Difficulty | Challenging +1.2 This is a multi-part S4 question requiring systematic application of standard estimator properties (bias, variance, consistency). While it involves algebraic manipulation with multiple variables and parameters, each part follows well-established procedures: checking E(estimator)=μ for bias, computing Var() for comparison, and applying consistency definition. The calculations are somewhat lengthy but conceptually straightforward for Further Maths students who have learned these techniques. It's moderately above average difficulty due to the algebraic complexity and multi-step nature, but doesn't require novel insight or proof techniques beyond the syllabus. |
| Spec | 5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(E(A) = \frac{1}{2}\left(3 \cdot \frac{\mu}{3} + 2 \cdot \frac{\mu}{2}\right) = \frac{1}{2}({\mu + \mu}) = \mu\) | M1 | Attempt \(E(A)\) |
| So \(A\) is unbiased | A1 | |
| \(E(B) = \frac{3}{2} \cdot \frac{\mu}{3} + \frac{2}{3} \cdot \frac{\mu}{2} = \frac{\mu}{2} + \frac{\mu}{3} = \frac{5\mu}{6}\) | M1 | Attempt \(E(B)\) |
| So \(B\) is biased, bias \(= \frac{5\mu}{6} - \mu = -\frac{\mu}{6}\) | A1 | |
| \(E(C) = \frac{1}{3}\left(3 \cdot \frac{\mu}{3} + 4 \cdot \frac{\mu}{2}\right) = \frac{1}{3}(\mu + 2\mu) = \mu\) | A1 | So \(C\) is unbiased |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\text{Var}(A) = \frac{1}{4}\left(3 \cdot 3\sigma^2 + 2 \cdot \frac{\sigma^2}{2}\right) = \frac{1}{4}(9\sigma^2 + \sigma^2) = \frac{10\sigma^2}{4} = \frac{5\sigma^2}{2}\) | M1 A1 | Attempt Var of an unbiased estimator |
| \(\text{Var}(C) = \frac{1}{9}\left(9 \cdot 3\sigma^2 + 16 \cdot \frac{\sigma^2}{2}\right) = \frac{1}{9}(27\sigma^2 + 8\sigma^2) = \frac{35\sigma^2}{9}\) | M1 A1 | |
| Since \(\frac{5\sigma^2}{2} < \frac{35\sigma^2}{9}\), \(A\) is the best estimator | A1 | Must compare and conclude |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(E(D) = \frac{1}{k}\left(2n \cdot \frac{\mu}{3} + n \cdot \frac{\mu}{2}\right) = \mu\) | M1 | Set \(E(D) = \mu\) |
| \(\frac{1}{k}\left(\frac{2n\mu}{3} + \frac{n\mu}{2}\right) = \mu\) | M1 | |
| \(k = \frac{2n}{3} + \frac{n}{2} = \frac{7n}{6}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\text{Var}(D) = \frac{1}{k^2}\left(2n \cdot 3\sigma^2 + n \cdot \frac{\sigma^2}{2}\right)\) | M1 | |
| \(= \frac{1}{(7n/6)^2}\left(6n\sigma^2 + \frac{n\sigma^2}{2}\right) = \frac{36}{49n^2} \cdot \frac{13n\sigma^2}{2}\) | M1 A1 | |
| \(= \frac{468\sigma^2}{98n} = \frac{234\sigma^2}{49n} \to 0\) as \(n \to \infty\), and \(D\) is unbiased | A1 | Must state both conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{234\sigma^2}{49n} < \frac{5\sigma^2}{2}\) | M1 | Compare \(\text{Var}(D)\) with \(\text{Var}(A)\) |
| \(n > \frac{468}{245} = 1.91...\) so least value \(n = 2\) | A1 |
# Question 6:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $E(A) = \frac{1}{2}\left(3 \cdot \frac{\mu}{3} + 2 \cdot \frac{\mu}{2}\right) = \frac{1}{2}({\mu + \mu}) = \mu$ | M1 | Attempt $E(A)$ |
| So $A$ is unbiased | A1 | |
| $E(B) = \frac{3}{2} \cdot \frac{\mu}{3} + \frac{2}{3} \cdot \frac{\mu}{2} = \frac{\mu}{2} + \frac{\mu}{3} = \frac{5\mu}{6}$ | M1 | Attempt $E(B)$ |
| So $B$ is biased, bias $= \frac{5\mu}{6} - \mu = -\frac{\mu}{6}$ | A1 | |
| $E(C) = \frac{1}{3}\left(3 \cdot \frac{\mu}{3} + 4 \cdot \frac{\mu}{2}\right) = \frac{1}{3}(\mu + 2\mu) = \mu$ | A1 | So $C$ is unbiased |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\text{Var}(A) = \frac{1}{4}\left(3 \cdot 3\sigma^2 + 2 \cdot \frac{\sigma^2}{2}\right) = \frac{1}{4}(9\sigma^2 + \sigma^2) = \frac{10\sigma^2}{4} = \frac{5\sigma^2}{2}$ | M1 A1 | Attempt Var of an unbiased estimator |
| $\text{Var}(C) = \frac{1}{9}\left(9 \cdot 3\sigma^2 + 16 \cdot \frac{\sigma^2}{2}\right) = \frac{1}{9}(27\sigma^2 + 8\sigma^2) = \frac{35\sigma^2}{9}$ | M1 A1 | |
| Since $\frac{5\sigma^2}{2} < \frac{35\sigma^2}{9}$, $A$ is the best estimator | A1 | Must compare and conclude |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $E(D) = \frac{1}{k}\left(2n \cdot \frac{\mu}{3} + n \cdot \frac{\mu}{2}\right) = \mu$ | M1 | Set $E(D) = \mu$ |
| $\frac{1}{k}\left(\frac{2n\mu}{3} + \frac{n\mu}{2}\right) = \mu$ | M1 | |
| $k = \frac{2n}{3} + \frac{n}{2} = \frac{7n}{6}$ | A1 | |
## Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\text{Var}(D) = \frac{1}{k^2}\left(2n \cdot 3\sigma^2 + n \cdot \frac{\sigma^2}{2}\right)$ | M1 | |
| $= \frac{1}{(7n/6)^2}\left(6n\sigma^2 + \frac{n\sigma^2}{2}\right) = \frac{36}{49n^2} \cdot \frac{13n\sigma^2}{2}$ | M1 A1 | |
| $= \frac{468\sigma^2}{98n} = \frac{234\sigma^2}{49n} \to 0$ as $n \to \infty$, and $D$ is unbiased | A1 | Must state both conditions |
## Part (e):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{234\sigma^2}{49n} < \frac{5\sigma^2}{2}$ | M1 | Compare $\text{Var}(D)$ with $\text{Var}(A)$ |
| $n > \frac{468}{245} = 1.91...$ so least value $n = 2$ | A1 | |
These pages (22–24) are blank lined answer/continuation sheets for **Question 6** of a PhysicsAndMathsTutor.com exam paper. They contain **no mark scheme content** — only ruled lines for student responses, the header "Question 6 continued," and the footer noting **(Total 18 marks)** and **TOTAL FOR PAPER: 75 MARKS**.
To extract a mark scheme, you would need to upload the actual mark scheme document rather than the student answer booklet pages.6. A random sample $X _ { 1 } , X _ { 2 } , X _ { 3 } , \ldots , X _ { 2 n }$ is taken from a population with mean $\frac { \mu } { 3 }$ and variance $3 \sigma ^ { 2 }$. A second random sample $Y _ { 1 } , Y _ { 2 } , Y _ { 3 } , \ldots , Y _ { n }$ is taken from a population with mean $\frac { \mu } { 2 }$ and variance $\frac { \sigma ^ { 2 } } { 2 }$, where the $X$ and $Y$ variables are all independent.\\
$A$, $B$ and $C$ are possible estimators of $\mu$, where
$$\begin{aligned}
& A = \frac { X _ { 1 } + X _ { 2 } + X _ { 3 } + Y _ { 1 } + Y _ { 2 } } { 2 } \\
& B = \frac { 3 X _ { 1 } } { 2 } + \frac { 2 Y _ { 1 } } { 3 } \\
& C = \frac { 3 X _ { 1 } + 4 Y _ { 1 } } { 3 }
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Show that two of $A , B$ and $C$ are unbiased estimators of $\mu$ and find the bias of the third estimator of $\mu$.
\item Showing your working clearly, find which of $A$, $B$ and $C$ is the best estimator of $\mu$.
The estimator
$$D = \frac { 1 } { k } \left( \sum _ { i = 1 } ^ { 2 n } X _ { i } + \sum _ { i = 1 } ^ { n } Y _ { i } \right)$$
is an unbiased estimator of $\mu$.
\item Find $k$ in terms of $n$.
\item Show that $D$ is also a consistent estimator of $\mu$.
\item Find the least value of $n$ for which $D$ is a better estimator of $\mu$ than any of $A$, $B$ or $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2015 Q6 [18]}}