| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Challenging +1.2 This is a structured S4 question on estimator properties requiring standard techniques: showing unbiasedness via E(Y)=a, computing E(M) and Var(M) using the given pdf with straightforward integration, and comparing estimators using bias-variance criteria. While it involves multiple parts and careful algebraic manipulation, each step follows established methods without requiring novel insight—slightly above average difficulty due to the technical computation and S4-level content. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = 2E(\bar{X}) = 2 \times \frac{a}{2} = a\) | M1 A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(M) = \int_0^a \frac{nm^n}{a^n}\, dm\) | M1 | |
| \(= \left[\frac{nm^{n+1}}{a^n(n+1)}\right]_0^a = \frac{na}{n+1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(M) = \int_0^a \frac{nm^{n+1}}{a^n}\, dm - \left(\frac{na}{n+1}\right)^2\) | M1 A1 | |
| \(= \left[\frac{nm^{n+2}}{a^n(n+2)}\right]_0^a - \frac{n^2a^2}{(n+1)^2}\) | M1d | |
| \(= na^2\left(\frac{(n+1)^2 - n(n+2)}{(n+1)^2(n+2)}\right)\) | ||
| \(= \frac{na^2}{(n+2)(n+1)^2}\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(S) = \frac{n+1}{n}E(M) = \frac{n+1}{n}\times\frac{na}{n+1} = a\) | B1 | |
| \(\text{Var}(S) = \left(\frac{n+1}{n}\right)^2 \frac{na^2}{(n+2)(n+1)^2} = \frac{a^2}{n(n+2)}\) | B1 | |
| \(\text{Var}(Y) = 4\text{Var}(\bar{X}) = 4\times\frac{a^2}{12n} = \frac{a^2}{3n}\) | M1 A1 | |
| As \(n>1\), \(n(n+2) > 3n\); therefore \(\text{Var}(S) < \text{Var}(Y)\), \(\therefore S\) is the better estimator | M1 M1 A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2E(\bar{X})\) | M1 | For \(2E(\bar{X})\) |
| \(2 \times \frac{a}{2}\) leading to \(a\) | A1 | For \(2 \times \frac{a}{2}\) leading to \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempting to integrate correct expression | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempting to integrate correct expression for \(E(X^2)\) | M1 | |
| Correct \(E(X^2)\) | A1 | |
| Using correct formula for \(\text{Var}(M)\) | M1d | Dependent on previous M mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{n+1}{n}E(M) = a\) or \(\frac{n+1}{n} \times \frac{na}{n+1} = a\) | B1 | |
| Using \(4\text{Var}(\bar{X})\) | M1 | |
| NB: Failure to show \(S\) is unbiased gains maximum of 5/7 — lose first B1 and final A1 |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = 2E(\bar{X}) = 2 \times \frac{a}{2} = a$ | M1 A1cso | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(M) = \int_0^a \frac{nm^n}{a^n}\, dm$ | M1 | |
| $= \left[\frac{nm^{n+1}}{a^n(n+1)}\right]_0^a = \frac{na}{n+1}$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(M) = \int_0^a \frac{nm^{n+1}}{a^n}\, dm - \left(\frac{na}{n+1}\right)^2$ | M1 A1 | |
| $= \left[\frac{nm^{n+2}}{a^n(n+2)}\right]_0^a - \frac{n^2a^2}{(n+1)^2}$ | M1d | |
| $= na^2\left(\frac{(n+1)^2 - n(n+2)}{(n+1)^2(n+2)}\right)$ | | |
| $= \frac{na^2}{(n+2)(n+1)^2}$ | A1cso | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(S) = \frac{n+1}{n}E(M) = \frac{n+1}{n}\times\frac{na}{n+1} = a$ | B1 | |
| $\text{Var}(S) = \left(\frac{n+1}{n}\right)^2 \frac{na^2}{(n+2)(n+1)^2} = \frac{a^2}{n(n+2)}$ | B1 | |
| $\text{Var}(Y) = 4\text{Var}(\bar{X}) = 4\times\frac{a^2}{12n} = \frac{a^2}{3n}$ | M1 A1 | |
| As $n>1$, $n(n+2) > 3n$; therefore $\text{Var}(S) < \text{Var}(Y)$, $\therefore S$ is the better estimator | M1 M1 A1cso | |
## Question (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2E(\bar{X})$ | M1 | For $2E(\bar{X})$ |
| $2 \times \frac{a}{2}$ leading to $a$ | A1 | For $2 \times \frac{a}{2}$ leading to $a$ |
## Question (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempting to integrate correct expression | M1 | |
## Question (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempting to integrate correct expression for $E(X^2)$ | M1 | |
| Correct $E(X^2)$ | A1 | |
| Using correct formula for $\text{Var}(M)$ | M1d | Dependent on previous M mark |
## Question (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{n+1}{n}E(M) = a$ or $\frac{n+1}{n} \times \frac{na}{n+1} = a$ | B1 | |
| Using $4\text{Var}(\bar{X})$ | M1 | |
| | | NB: Failure to show $S$ is unbiased gains maximum of 5/7 — lose first B1 and final A1 |6. A random sample of size $n$ is taken from the random variable $X$, which has a continuous uniform distribution over the interval [ $0 , a$ ], $a > 0$
The sample mean is denoted by $\bar { X }$
\begin{enumerate}[label=(\alph*)]
\item Show that $Y = 2 \bar { X }$ is an unbiased estimator of $a$
The maximum value, $M$, in the sample has probability density function
$$f ( m ) = \left\{ \begin{array} { c c }
\frac { n m ^ { n - 1 } } { a ^ { n } } & 0 \leqslant m \leqslant a \\
0 & \text { otherwise }
\end{array} \right.$$
\item Find E(M)
\item Show that $\operatorname { Var } ( M ) = \frac { n a ^ { 2 } } { ( n + 2 ) ( n + 1 ) ^ { 2 } }$
The estimator $S$ is defined by $S = \frac { n + 1 } { n } M$\\
Given that $n > 1$
\item state which of $Y$ or $S$ is the better estimator for $a$. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2016 Q6 [15]}}