| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Moment generating function problems |
| Difficulty | Challenging +1.2 This is a standard Further Statistics 2 question on properties of estimators (bias and variance). Parts (a)-(c) involve routine application of E(aX+bY) and Var(aX+bY) formulas with independent binomial variables. Part (d) requires systematic algebraic manipulation to find optimal coefficients, but follows a predictable template. While it's Further Maths content, it's a textbook-style multi-part question with clear structure and no novel insight required—moderately above average difficulty due to the algebraic work and Further Maths context. |
| Spec | 5.02b Expectation and variance: discrete random variables5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim B(n,p)\), \(E(Y) = np\) | M1 | 3.3 - use of mean \(np\) |
| \(E(\hat{p}_1) = \frac{np + 3np - 2np}{2n} = p\) | M1 | 3.4 - use of binomial model to find expected value |
| \(E(\hat{p}_2) = \frac{2np + 3np + np}{6n} = p\), therefore both are unbiased | A1cso | 1.1b - both expectations evaluated to \(p\) with correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(\hat{p}_1) = \frac{np(1-p) + 9np(1-p) + 4np(1-p)}{4n^2}\) | M1 | 2.1 - use of sum of variances, all terms positive, sight of \(3^2\) and \(2^2\) |
| \(\text{Var}(\hat{p}_1) = \frac{7p(1-p)}{2n}\) | A1 | 1.1b - correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| [Both unbiased so...] \(\text{Var}(\hat{p}_2) < \text{Var}(\hat{p}_1)\) | M1 | 2.4 - correct explanation ft their \(\text{Var}(\hat{p}_1)\) |
| \(\hat{p}_2\) is the better estimator | A1ft | 2.2a - correct deduction consistent with their \(\text{Var}(\hat{p}_1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\hat{p}_3\) is unbiased so \(b = a + 4\) | M1 | 1.1b - use of \(E(\hat{p}_3) = p\) to set up equation in \(a\) and \(b\) only |
| Require \(\text{Var}(\hat{p}_3) < \text{Var}(\hat{p}_2)\) so \(\frac{a^2+10}{b^2} < \frac{7}{18}\) | M1 | 2.1 - setting up inequality using correct expression for \(\text{Var}(\hat{p}_3)\) |
| \(18(a^2+10) < 7(a+4)^2 \rightarrow 11a^2 - 56a + 68 < 0\) | M1 | 1.1b - eliminating one variable and setting up 3TQ inequality |
| \(2 < a < \frac{34}{11}\) | A1ft | 1.1b - correct solution, must choose "inside" region |
| \(a = 3\) and \(b = 7\) | A1 | 2.2a - correctly deducing integer values of \(a\) and \(b\) |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim B(n,p)$, $E(Y) = np$ | M1 | 3.3 - use of mean $np$ |
| $E(\hat{p}_1) = \frac{np + 3np - 2np}{2n} = p$ | M1 | 3.4 - use of binomial model to find expected value |
| $E(\hat{p}_2) = \frac{2np + 3np + np}{6n} = p$, therefore both are unbiased | A1cso | 1.1b - both expectations evaluated to $p$ with correct conclusion |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(\hat{p}_1) = \frac{np(1-p) + 9np(1-p) + 4np(1-p)}{4n^2}$ | M1 | 2.1 - use of sum of variances, all terms positive, sight of $3^2$ and $2^2$ |
| $\text{Var}(\hat{p}_1) = \frac{7p(1-p)}{2n}$ | A1 | 1.1b - correct expression |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| [Both unbiased so...] $\text{Var}(\hat{p}_2) < \text{Var}(\hat{p}_1)$ | M1 | 2.4 - correct explanation ft their $\text{Var}(\hat{p}_1)$ |
| $\hat{p}_2$ is the better estimator | A1ft | 2.2a - correct deduction consistent with their $\text{Var}(\hat{p}_1)$ |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\hat{p}_3$ is unbiased so $b = a + 4$ | M1 | 1.1b - use of $E(\hat{p}_3) = p$ to set up equation in $a$ and $b$ only |
| Require $\text{Var}(\hat{p}_3) < \text{Var}(\hat{p}_2)$ so $\frac{a^2+10}{b^2} < \frac{7}{18}$ | M1 | 2.1 - setting up inequality using correct expression for $\text{Var}(\hat{p}_3)$ |
| $18(a^2+10) < 7(a+4)^2 \rightarrow 11a^2 - 56a + 68 < 0$ | M1 | 1.1b - eliminating one variable and setting up 3TQ inequality |
| $2 < a < \frac{34}{11}$ | A1ft | 1.1b - correct solution, must choose "inside" region |
| $a = 3$ and $b = 7$ | A1 | 2.2a - correctly deducing integer values of $a$ and $b$ |\begin{enumerate}
\item A bag contains a large number of marbles of which an unknown proportion, $p$, is yellow.
\end{enumerate}
Three random samples of size $n$ are taken, and the number of yellow marbles in each sample, $Y _ { 1 } , Y _ { 2 }$ and $Y _ { 3 }$, is recorded.
Two estimators $\hat { \mathrm { p } } _ { 1 }$ and $\hat { \mathrm { p } } _ { 2 }$ are proposed to estimate the value of $p$
$$\begin{aligned}
& \hat { p } _ { 1 } = \frac { Y _ { 1 } + 3 Y _ { 2 } - 2 Y _ { 3 } } { 2 n } \\
& \hat { p } _ { 2 } = \frac { 2 Y _ { 1 } + 3 Y _ { 2 } + Y _ { 3 } } { 6 n }
\end{aligned}$$
(a) Show that $\hat { \mathrm { p } } _ { 1 }$ and $\hat { \mathrm { p } } _ { 2 }$ are both unbiased estimators of $p$\\
(b) Find the variance of $\hat { p } _ { 1 }$
The variance of $\hat { \mathrm { p } } _ { 2 }$ is $\frac { 7 p ( 1 - p ) } { 18 n }$\\
(c) State, giving a reason, which is the better estimator.
The estimator $\hat { p } _ { 3 } = \frac { Y _ { 1 } + a Y _ { 2 } + 3 Y _ { 3 } } { b n }$ where $a$ and $b$ are positive integers.\\
(d) Find the pair of values of $a$ and $b$ such that $\hat { \mathrm { p } } _ { 3 }$ is a better unbiased estimator of $p$ than both $\hat { \mathrm { p } } _ { 1 }$ and $\hat { \mathrm { p } } _ { 2 }$\\
You must show all stages of your working.
\hfill \mbox{\textit{Edexcel FS2 2023 Q8 [12]}}