| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Construct combined estimator |
| Difficulty | Challenging +1.2 This is a multi-part question on unbiased estimators requiring understanding of expectation and variance properties. Parts (a) and (b) are straightforward applications of E(estimator)=p, while part (c) requires comparing variances and solving an inequality—moderately above average but standard Further Stats material with clear structure. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{E}(Q) = k\!\left(\dfrac{\text{E}(X)}{m} + \dfrac{\text{E}(Y)}{n}\right)\) then \(\text{E}(Q) = k\!\left(\dfrac{mp}{m} + \dfrac{np}{n}\right)\) | M1 | For selecting the correct models for \(X\) and \(Y\) and substituting into \(\text{E}(Q)\). Both \(mp\) and \(np\) must be seen or used as the expected values. Cannot be implied by \(k(p+p)\) |
| \(2kp = p \therefore k = \dfrac{1}{2}\) | A1* | Cao sets their expression in \(k\) and \(p\) equal to \(p\) before achieving the given answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{E}(R) = \dfrac{amp}{m} + \dfrac{bnp}{n}\) | M1 | Using the model to find \(\text{E}(R)\) in terms of \(a\) and \(b\). Both \(mp\) and \(np\) must be seen or used. Must see \(\dfrac{amp}{m} + \dfrac{bnp}{n}\). Cannot be implied by \(ap + bp\) |
| \(\dfrac{amp}{m} + \dfrac{bnp}{n} = p \therefore a + b = 1\) | A1* | Cao sets their expression in \(a\), \(b\) and \(p\) equal to \(p\) before achieving the given answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(Q) = \dfrac{mp(1-p)}{4m^2} + \dfrac{np(1-p)}{4n^2} = \dfrac{1}{4}p(1-p)\!\left(\dfrac{1}{m}+\dfrac{1}{n}\right)\) | M1 | For a correct attempt at \(\text{Var}(Q)\) with at least two of \(m\), \(n\) and 2 being squared on the denominator |
| \(\text{Var}(R) = \dfrac{a^2 mp(1-p)}{m^2} + \dfrac{b^2 np(1-p)}{n^2} = p(1-p)\!\left(\dfrac{a^2}{m}+\dfrac{b^2}{n}\right)\) | M1 | For a correct attempt at \(\text{Var}(R)\) in terms of \(a\) and \(b\) with at least one of \(a\) and \(b\) being squared and at least one of \(m\) and \(n\) being squared |
| \(\left(\dfrac{a^2}{m}+\dfrac{b^2}{n}\right) < \dfrac{1}{4}\!\left(\dfrac{1}{m}+\dfrac{1}{n}\right)\) | M1 | Using their \(\text{Var}(R) <\) their \(\text{Var}(Q)\); condone \(=\) instead of \(<\) (there must be a term in \(a^2\) in their equation or inequality) |
| \(\left(\dfrac{a^2}{100}+\dfrac{(1-a)^2}{200}\right) < \dfrac{1}{4}\!\left(\dfrac{1}{100}+\dfrac{1}{200}\right)\) | M1 | Substituting \(b = 1-a\); may be scored earlier |
| \(12a^2 - 8a + 1 < 0 \Rightarrow a = \ldots\) | M1 | Forming and solving correctly a 3-term quadratic in \(a\); condone \(=\) instead of \(<\) |
| \(a = \dfrac{1}{6}\) or \(\dfrac{1}{2}\) | A1 | Correct values |
| \(\dfrac{1}{6} < a < \dfrac{1}{2}\) | A1ft | Dep. on all previous M marks and for selecting the right range using their values of \(a\) which must be between 0 and 1 |
## Question 7:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{E}(Q) = k\!\left(\dfrac{\text{E}(X)}{m} + \dfrac{\text{E}(Y)}{n}\right)$ then $\text{E}(Q) = k\!\left(\dfrac{mp}{m} + \dfrac{np}{n}\right)$ | M1 | For selecting the correct models for $X$ and $Y$ and substituting into $\text{E}(Q)$. Both $mp$ and $np$ must be seen or used as the expected values. Cannot be implied by $k(p+p)$ |
| $2kp = p \therefore k = \dfrac{1}{2}$ | A1* | Cao sets their expression in $k$ and $p$ equal to $p$ before achieving the given answer with no errors |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{E}(R) = \dfrac{amp}{m} + \dfrac{bnp}{n}$ | M1 | Using the model to find $\text{E}(R)$ in terms of $a$ and $b$. Both $mp$ and $np$ must be seen or used. Must see $\dfrac{amp}{m} + \dfrac{bnp}{n}$. Cannot be implied by $ap + bp$ |
| $\dfrac{amp}{m} + \dfrac{bnp}{n} = p \therefore a + b = 1$ | A1* | Cao sets their expression in $a$, $b$ and $p$ equal to $p$ before achieving the given answer with no errors |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(Q) = \dfrac{mp(1-p)}{4m^2} + \dfrac{np(1-p)}{4n^2} = \dfrac{1}{4}p(1-p)\!\left(\dfrac{1}{m}+\dfrac{1}{n}\right)$ | M1 | For a correct attempt at $\text{Var}(Q)$ with at least two of $m$, $n$ and 2 being squared on the denominator |
| $\text{Var}(R) = \dfrac{a^2 mp(1-p)}{m^2} + \dfrac{b^2 np(1-p)}{n^2} = p(1-p)\!\left(\dfrac{a^2}{m}+\dfrac{b^2}{n}\right)$ | M1 | For a correct attempt at $\text{Var}(R)$ in terms of $a$ and $b$ with at least one of $a$ and $b$ being squared and at least one of $m$ and $n$ being squared |
| $\left(\dfrac{a^2}{m}+\dfrac{b^2}{n}\right) < \dfrac{1}{4}\!\left(\dfrac{1}{m}+\dfrac{1}{n}\right)$ | M1 | Using their $\text{Var}(R) <$ their $\text{Var}(Q)$; condone $=$ instead of $<$ (there must be a term in $a^2$ in their equation or inequality) |
| $\left(\dfrac{a^2}{100}+\dfrac{(1-a)^2}{200}\right) < \dfrac{1}{4}\!\left(\dfrac{1}{100}+\dfrac{1}{200}\right)$ | M1 | Substituting $b = 1-a$; may be scored earlier |
| $12a^2 - 8a + 1 < 0 \Rightarrow a = \ldots$ | M1 | Forming and solving correctly a 3-term quadratic in $a$; condone $=$ instead of $<$ |
| $a = \dfrac{1}{6}$ or $\dfrac{1}{2}$ | A1 | Correct values |
| $\dfrac{1}{6} < a < \dfrac{1}{2}$ | A1ft | Dep. on all previous M marks and for selecting the right range using their values of $a$ which must be between 0 and 1 |
---\begin{enumerate}
\item Two organisations are each asked to carry out a survey to find out the proportion, $p$, of the population that would vote for a particular political party.
\end{enumerate}
The first organisation finds that out of $m$ people, $X$ would vote for this particular political party.
The second organisation finds that out of $n$ people, $Y$ would vote for this particular political party.
An unbiased estimator, $Q$, of $p$ is proposed where
$$Q = k \left( \frac { X } { m } + \frac { Y } { n } \right)$$
(a) Show that $k = \frac { 1 } { 2 }$
A second unbiased estimator, $R$, of $p$ is proposed where
$$R = \frac { a X } { m } + \frac { b Y } { n }$$
(b) Show that $a + b = 1$
Given that $m = 100$ and $n = 200$ and that $R$ is a better estimator of $p$ than $Q$\\
(c) calculate the range of possible values of $a$ Show your working clearly.
\hfill \mbox{\textit{Edexcel FS2 2024 Q7 [11]}}