| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Standard +0.3 This is a straightforward S4 statistics question requiring standard calculations: finding bias (difference between expectation and parameter), adjusting for unbiasedness, computing variance of a linear transformation, verifying unbiasedness via integration, and comparing estimators by variance. All techniques are routine for Further Maths S4 students with no novel problem-solving required, making it slightly easier than average despite being Further Maths content. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim U[-a, 2a]\), so \(E(X) = \frac{-a+2a}{2} = \frac{a}{2}\) | B1 | Correct \(E(X)\) |
| \(E(\bar{X}) = E\left(\frac{X_1+X_2}{2}\right) = \frac{a}{2}\) | M1 | Using linearity of expectation |
| Since \(E(\bar{X}) = \frac{a}{2} \neq a\), \(\bar{X}\) is biased. Bias \(= \frac{a}{2} - a = -\frac{a}{2}\) | A1 | Correct bias stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k = 2\) | B1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X) = \frac{(2a-(-a))^2}{12} = \frac{9a^2}{12} = \frac{3a^2}{4}\) | M1 A1 | Correct variance of \(X\) |
| \(\text{Var}(Y) = \text{Var}(2\bar{X}) = 4 \cdot \text{Var}(\bar{X}) = 4 \cdot \frac{3a^2/4}{2} = \frac{3a^2}{2}\) | M1 A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(M) = \int_{-a}^{2a} x \cdot \frac{2(x+a)}{9a^2}\,dx\) | M1 | Setting up correct integral |
| \(= \frac{2}{9a^2}\int_{-a}^{2a}(x^2+ax)\,dx\) | M1 | Expanding |
| \(= \frac{2}{9a^2}\left[\frac{x^3}{3}+\frac{ax^2}{2}\right]_{-a}^{2a} = \frac{2}{9a^2}\left[\frac{8a^3}{3}+2a^3-\left(\frac{-a^3}{3}+\frac{a^3}{2}\right)\right]\) | A1 | Correct integration |
| \(= \frac{2}{9a^2} \cdot \frac{9a^3}{2} \cdot \ldots = a\) | A1 | Shown \(E(M)=a\), hence unbiased |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(M) = E(M^2) - [E(M)]^2 = \frac{3}{2}a^2 - a^2 = \frac{a^2}{2}\) | B1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(M\) since \(\text{Var}(M) = \frac{a^2}{2} < \text{Var}(Y) = \frac{3a^2}{2}\), so \(M\) is more efficient. | B1 B1 | State \(M\); give reason comparing variances |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M = \max(5, -1) = 5\), so estimate of \(a = 5\) | B1 | Correct answer |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim U[-a, 2a]$, so $E(X) = \frac{-a+2a}{2} = \frac{a}{2}$ | B1 | Correct $E(X)$ |
| $E(\bar{X}) = E\left(\frac{X_1+X_2}{2}\right) = \frac{a}{2}$ | M1 | Using linearity of expectation |
| Since $E(\bar{X}) = \frac{a}{2} \neq a$, $\bar{X}$ is biased. Bias $= \frac{a}{2} - a = -\frac{a}{2}$ | A1 | Correct bias stated |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k = 2$ | B1 | Correct value |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = \frac{(2a-(-a))^2}{12} = \frac{9a^2}{12} = \frac{3a^2}{4}$ | M1 A1 | Correct variance of $X$ |
| $\text{Var}(Y) = \text{Var}(2\bar{X}) = 4 \cdot \text{Var}(\bar{X}) = 4 \cdot \frac{3a^2/4}{2} = \frac{3a^2}{2}$ | M1 A1 | Correct final answer |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(M) = \int_{-a}^{2a} x \cdot \frac{2(x+a)}{9a^2}\,dx$ | M1 | Setting up correct integral |
| $= \frac{2}{9a^2}\int_{-a}^{2a}(x^2+ax)\,dx$ | M1 | Expanding |
| $= \frac{2}{9a^2}\left[\frac{x^3}{3}+\frac{ax^2}{2}\right]_{-a}^{2a} = \frac{2}{9a^2}\left[\frac{8a^3}{3}+2a^3-\left(\frac{-a^3}{3}+\frac{a^3}{2}\right)\right]$ | A1 | Correct integration |
| $= \frac{2}{9a^2} \cdot \frac{9a^3}{2} \cdot \ldots = a$ | A1 | Shown $E(M)=a$, hence unbiased |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(M) = E(M^2) - [E(M)]^2 = \frac{3}{2}a^2 - a^2 = \frac{a^2}{2}$ | B1 | Correct answer |
## Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $M$ since $\text{Var}(M) = \frac{a^2}{2} < \text{Var}(Y) = \frac{3a^2}{2}$, so $M$ is more efficient. | B1 B1 | State $M$; give reason comparing variances |
## Part (g):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M = \max(5, -1) = 5$, so estimate of $a = 5$ | B1 | Correct answer |\begin{enumerate}
\item A random sample of size $2 , X _ { 1 }$ and $X _ { 2 }$, is taken from the random variable $X$ which has a continuous uniform distribution over the interval $[ - a , 2 a ] , a > 0$\\
(a) Show that $\bar { X } = \frac { X _ { 1 } + X _ { 2 } } { 2 }$ is a biased estimator of $a$ and find the bias.
\end{enumerate}
The random variable $Y = k \bar { X }$ is an unbiased estimator of $a$.\\
(b) Write down the value of the constant $k$.\\
(c) Find $\operatorname { Var } ( Y )$.
The random variable $M$ is the maximum of $X _ { 1 }$ and $X _ { 2 }$\\
The probability density function, $m ( x )$, of $M$ is given by
$$m ( x ) = \left\{ \begin{array} { c l }
\frac { 2 ( x + a ) } { 9 a ^ { 2 } } & - a \leqslant x \leqslant 2 a \\
0 & \text { otherwise }
\end{array} \right.$$
(d) Show that $M$ is an unbiased estimator of $a$.
Given that $\mathrm { E } \left( M ^ { 2 } \right) = \frac { 3 } { 2 } a ^ { 2 }$\\
(e) find $\operatorname { Var } ( M )$.\\
(f) State, giving a reason, whether you would use $Y$ or $M$ as an estimator of $a$.
A random sample of two values of $X$ are 5 and - 1\\
(g) Use your answer to part (f) to estimate $a$.\\
\hfill \mbox{\textit{Edexcel S4 2013 Q4 [16]}}