| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Sample statistics from uniform |
| Difficulty | Standard +0.3 This is a straightforward S3 question requiring standard knowledge of bias in estimators and properties of uniform distributions. Part (a) uses the formula for mean of uniform distribution, part (b) is simple algebra to find k, and part (c) applies the unbiased estimator to find α then uses the upper bound formula. All steps are routine applications of learned techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{E}(\bar{X}) = \mu = \dfrac{2\alpha + 3 + \alpha - 3}{2}\) | M1 | Using the formula \(\left(\dfrac{a+b}{2}\right)\) or getting \(\dfrac{3\alpha}{2}\) |
| \(= \dfrac{3\alpha}{2}\), so \(\bar{X}\) is a biased estimator | A1 | \(\dfrac{3\alpha}{2}\) and concluding. Allow A1 for \(\dfrac{3\alpha}{2} \neq \alpha\). Also allow A1 for bias \(= \pm\dfrac{\alpha}{2} \neq 0\) |
| \(\text{bias} = \dfrac{3\alpha}{2} - \alpha = \pm\dfrac{\alpha}{2}\) | B1 | bias \(= \pm\dfrac{\alpha}{2}\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = \dfrac{2}{3}\) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\alpha = \dfrac{2}{3}\bar{X} = \dfrac{2}{3}(8)\) | M1 | "their \(k\)" \(\times 8\). An attempt to use sample data to find \(\bar{x}\) and multiply by their \(k\). Allow full expression for \(\bar{x}\) or \(\dfrac{\sum x}{n}\). Note: 1st M1 can be implied by correct recovery leading to \(\alpha = \dfrac{16}{3}\) |
| Max value \(= 2\left(\dfrac{16}{3}\right) + 3\) | M1 | \(2 \times\) "their \(\alpha\)" \(+ 3\), where \(\alpha\) is a function of the sample mean found by applying \(\dfrac{\sum x}{n}\) from given data values. Note: \(2(13)+3=39\) is M0M0A0 |
| \(= \dfrac{41}{3}\) | A1 | \(\dfrac{41}{3}\) or \(13\dfrac{2}{3}\) or awrt 13.7 |
| [3] | ||
| Total: 7 |
# Question 2:
$X$ follows a continuous uniform distribution over $[\alpha - 3,\ 2\alpha + 3]$
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{E}(\bar{X}) = \mu = \dfrac{2\alpha + 3 + \alpha - 3}{2}$ | M1 | Using the formula $\left(\dfrac{a+b}{2}\right)$ or getting $\dfrac{3\alpha}{2}$ |
| $= \dfrac{3\alpha}{2}$, so $\bar{X}$ is a biased estimator | A1 | $\dfrac{3\alpha}{2}$ and concluding. Allow A1 for $\dfrac{3\alpha}{2} \neq \alpha$. Also allow A1 for bias $= \pm\dfrac{\alpha}{2} \neq 0$ |
| $\text{bias} = \dfrac{3\alpha}{2} - \alpha = \pm\dfrac{\alpha}{2}$ | B1 | bias $= \pm\dfrac{\alpha}{2}$ |
| | **[3]** | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = \dfrac{2}{3}$ | B1 | |
| | **[1]** | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha = \dfrac{2}{3}\bar{X} = \dfrac{2}{3}(8)$ | M1 | "their $k$" $\times 8$. An attempt to use sample data to find $\bar{x}$ and multiply by their $k$. Allow full expression for $\bar{x}$ or $\dfrac{\sum x}{n}$. Note: 1st M1 can be implied by correct recovery leading to $\alpha = \dfrac{16}{3}$ |
| Max value $= 2\left(\dfrac{16}{3}\right) + 3$ | M1 | $2 \times$ "their $\alpha$" $+ 3$, where $\alpha$ is a function of the sample mean found by applying $\dfrac{\sum x}{n}$ from given data values. Note: $2(13)+3=39$ is M0M0A0 |
| $= \dfrac{41}{3}$ | A1 | $\dfrac{41}{3}$ or $13\dfrac{2}{3}$ or awrt 13.7 |
| | **[3]** | |
| | **Total: 7** | |2. The random variable $X$ follows a continuous uniform distribution over the interval $[ \alpha - 3,2 \alpha + 3 ]$ where $\alpha$ is a constant.\\
The mean of a random sample of size $n$ is denoted by $\bar { X }$
\begin{enumerate}[label=(\alph*)]
\item Show that $\bar { X }$ is a biased estimator of $\alpha$, and state the bias.
Given that $Y = k \bar { X }$ is an unbiased estimator for $\alpha$
\item find the value of $k$.
A random sample of 10 values of $X$ is taken and the results are as follows
$$\begin{array} { l l l l l l l l l l }
3 & 5 & 8 & 12 & 4 & 13 & 10 & 8 & 5 & 12
\end{array}$$
\item Hence estimate the maximum value of $X$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q2 [7]}}