| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Standard +0.3 This is a structured multi-part question on unbiased estimators that follows a standard template for S4. Parts (a)-(c) involve routine application of E(aX)=aE(X) and variance formulas. Parts (d)-(f) require understanding of consistency and efficiency but with heavy scaffolding. While it's Further Maths content, the question guides students through each step with minimal novel insight required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(X_1 \sim B(10, p)\) \(\therefore E(X_1) = 10p \Rightarrow E(R_1) = E\!\left(\dfrac{X_1}{10}\right) = \dfrac{10p}{10} = p\) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(X_2 \sim B(n, p)\) \(\therefore E(X_2) = np \Rightarrow E(R_2) = E\!\left(\dfrac{X_2}{n}\right) = \dfrac{np}{n} = p\) | B1 | |
| \(E(Y) = E\!\left(\tfrac{1}{2}[R_1 + R_2]\right) = \tfrac{1}{2}[E(R_1) + E(R_2)] = \tfrac{1}{2}[p + p] = p\) | B1 | (2 marks) |
| Answer | Marks |
|---|---|
| \(\text{Var}(R_2) = \dfrac{1}{n^2}\text{Var}(X_2) = \dfrac{np(1-p)}{n^2} = \dfrac{p(1-p)}{n}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \dfrac{1}{4}\!\left[\dfrac{p(1-p)}{10} + \dfrac{p(1-p)}{n}\right]\) | M1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(\text{Var}(R_2) = \dfrac{p(1-p)}{n} \to 0\) as \(n \to \infty\), \(\therefore R_2\) is consistent | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(Y) = \dfrac{p(1-p)}{4}\!\left[\dfrac{1}{10} + \dfrac{1}{20}\right] = \dfrac{p(1-p)}{80} \times 3 < \text{Var}(R_2)\) | M1 | |
| Since all 3 are unbiased, we select the one with minimum variance, i.e. \(Y\) | A1 | (2 marks) |
| Answer | Marks |
|---|---|
| \(X_1 + X_2 \sim B(n+10, p)\) so consider \(\dfrac{X_1 + X_2}{n+10}\) | B1 |
| \(E\!\left(\dfrac{X_1+X_2}{n+10}\right) = \dfrac{(n+10)p}{(n+10)} = p\) (show unbiased) | M1 |
| \(\text{Var}\!\left(\dfrac{X_1+X_2}{n+10}\right) = \dfrac{p(1-p)}{n+10}\) (find variance) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| and \(\dfrac{p(1-p)}{n+10} < \dfrac{p(1-p)}{n}\) \(\therefore\) always better than \(R_2\) | A1 | both |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Leftrightarrow 0 < (10-n)^2\) Show better than \(Y\); use of \(n=20\) acceptable | M1 | |
| \(\therefore \dfrac{X_1+X_2}{n+10}\) is unbiased and always has smaller variance | A1 cso | (6 marks) (16 marks total) |
# Question 6:
## Part (a)
$X_1 \sim B(10, p)$ $\therefore E(X_1) = 10p \Rightarrow E(R_1) = E\!\left(\dfrac{X_1}{10}\right) = \dfrac{10p}{10} = p$ | B1 | (1 mark)
## Part (b)
$X_2 \sim B(n, p)$ $\therefore E(X_2) = np \Rightarrow E(R_2) = E\!\left(\dfrac{X_2}{n}\right) = \dfrac{np}{n} = p$ | B1 |
$E(Y) = E\!\left(\tfrac{1}{2}[R_1 + R_2]\right) = \tfrac{1}{2}[E(R_1) + E(R_2)] = \tfrac{1}{2}[p + p] = p$ | B1 | (2 marks)
## Part (c)
$\text{Var}(R_2) = \dfrac{1}{n^2}\text{Var}(X_2) = \dfrac{np(1-p)}{n^2} = \dfrac{p(1-p)}{n}$ | B1 |
$\text{Var}(R_1) = \dfrac{p(1-p)}{10}$ $\therefore \text{Var}(Y) = \tfrac{1}{4}[\text{Var}(R_1) + \text{Var}(R_2)]$
$= \dfrac{1}{4}\!\left[\dfrac{p(1-p)}{10} + \dfrac{p(1-p)}{n}\right]$ | M1, A1 | (3 marks)
## Part (d)
Since $\text{Var}(R_2) = \dfrac{p(1-p)}{n} \to 0$ as $n \to \infty$, $\therefore R_2$ is consistent | M1, A1 | (2 marks)
## Part (e)
$\text{Var}(R_1) = \dfrac{p(1-p)}{10} > \dfrac{p(1-p)}{20} = \text{Var}(R_2)$
$\text{Var}(Y) = \dfrac{p(1-p)}{4}\!\left[\dfrac{1}{10} + \dfrac{1}{20}\right] = \dfrac{p(1-p)}{80} \times 3 < \text{Var}(R_2)$ | M1 |
Since all 3 are unbiased, we select the one with minimum variance, i.e. $Y$ | A1 | (2 marks)
## Part (f)
$X_1 + X_2 \sim B(n+10, p)$ so consider $\dfrac{X_1 + X_2}{n+10}$ | B1 |
$E\!\left(\dfrac{X_1+X_2}{n+10}\right) = \dfrac{(n+10)p}{(n+10)} = p$ (show unbiased) | M1 |
$\text{Var}\!\left(\dfrac{X_1+X_2}{n+10}\right) = \dfrac{p(1-p)}{n+10}$ (find variance) | M1 |
$\dfrac{p(1-p)}{n+10} < \dfrac{p(1-p)}{10}$ $\therefore$ always better than $R_1$
and $\dfrac{p(1-p)}{n+10} < \dfrac{p(1-p)}{n}$ $\therefore$ always better than $R_2$ | A1 | both
$\dfrac{p(1-p)}{n+10} < \dfrac{p(1-p)}{4}\!\left[\dfrac{n+10}{10n}\right]$
$\Leftrightarrow 40n < 100 + 20n + n^2$
$\Leftrightarrow 0 < (10-n)^2$ Show better than $Y$; use of $n=20$ acceptable | M1 |
$\therefore \dfrac{X_1+X_2}{n+10}$ is unbiased and always has smaller variance | A1 cso | (6 marks) **(16 marks total)**6. A statistics student is trying to estimate the probability, $p$, of rolling a 6 with a particular die. The die is rolled 10 times and the random variable $X _ { 1 }$ represents the number of sixes obtained. The random variable $R _ { 1 } = \frac { X _ { 1 } } { 10 }$ is proposed as an estimator of $p$.
\begin{enumerate}[label=(\alph*)]
\item Show that $R _ { 1 }$ is an unbiased estimator of $p$.
The student decided to roll the die again $n$ times ( $n > 10$ ) and the random variable $X _ { 2 }$ represents the number of sixes in these $n$ rolls. The random variable $R _ { 2 } = \frac { X _ { 2 } } { n }$ and the random variable $Y = \frac { 1 } { 2 } \left( R _ { 1 } + R _ { 2 } \right)$.
\item Show that both $R _ { 2 }$ and $Y$ are unbiased estimators of $p$.
\item Find $\operatorname { Var } \left( R _ { 2 } \right)$ and $\operatorname { Var } ( Y )$.
\item State giving a reason which of the 3 estimators $R _ { 1 } , R _ { 2 }$ and $Y$ are consistent estimators of $p$.
\item For the case $n = 20$ state, giving a reason, which of the 3 estimators $R _ { 1 } , R _ { 2 }$ and $Y$ you would recommend.
The student's teacher pointed out that a better estimator could be found based on the random variable $X _ { 1 } + X _ { 2 }$.
\item Find a suitable estimator and explain why it is better than $R _ { 1 } , R _ { 2 }$ and $Y$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 Q6 [18]}}