| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Geometric applications |
| Difficulty | Standard +0.3 This is a straightforward S4 question on continuous uniform distributions requiring standard integration techniques and variance calculations. Parts (a)-(d) involve routine expectation calculations using given formulas, part (e) is similar variance work, and parts (f)-(g) are direct applications. While multi-part with 7 sections, each step follows standard procedures without requiring novel insight or complex problem-solving—slightly easier than average A-level material. |
| Spec | 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 |
| \(E(X^n) = \int_0^t x^n \cdot \frac{1}{t}\, dx = \left[\frac{x^{n+1}}{t(n+1)}\right]_0^t = \left(\frac{t^{n+1}}{t(n+1)} - 0\right) = \frac{t^n}{n+1}\) | M1, A1 | \(\int_0^b x^{n-1}\frac{1}{t}\,dx\), A1 cao (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((E(x) = \frac{t}{2})\), \(E(S) = kE(X)E(Y) = k \cdot \frac{t^2}{4}\) | M1A1 | |
| \(E(S) = t^2 \Rightarrow k = 4\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(XY) = E(X^2)E(Y^2) - [E(XY)]^2\) | M1 | |
| \(= \frac{t^2}{3} \times \frac{t^2}{3} - \left(\frac{t^2}{4}\right)^2 = \left\{\frac{7t^4}{144}\right\}\) | M1 | |
| \(\text{Var}(S) = k^2\,\text{Var}(XY) = 16 \times \frac{7t^4}{144} = \frac{7t^4}{9}\) | A1 cao | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(u) = t^2 \Rightarrow 2E(x^2)q = t^2 \Rightarrow 2\cdot\frac{t^2}{3}\cdot q = t^2 \Rightarrow q = \frac{3}{2}\) | M1, M1, A1 cao | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(u) = q^2[\text{Var}(x^2) + \text{Var}(Y^2)] = 2q^2\,\text{Var}(x^2)\) | M1 | |
| \(\text{Var}(x^2) = E(x^4) - [E(x^2)]^2 = \frac{t^4}{5} - \left(\frac{t^2}{3}\right)^2 = \frac{4t^4}{45}\) | M1, A1 | |
| \(\text{Var}(u) = 2 \times \frac{9}{4} \times \frac{4}{45}\,t^4 = \frac{2}{5}t^4\) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2}{5} < \frac{7}{9}\) \(\therefore U\) is better \(\therefore\) smaller variance | B1ft | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(u\) estimate is: \(\frac{3}{2}(2^2 + 3^2) = \frac{3}{2} \times 13 = \frac{39}{2}\) or \(19.5\) | B1ft | (1) |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^n) = \int_0^t x^n \cdot \frac{1}{t}\, dx = \left[\frac{x^{n+1}}{t(n+1)}\right]_0^t = \left(\frac{t^{n+1}}{t(n+1)} - 0\right) = \frac{t^n}{n+1}$ | M1, A1 | $\int_0^b x^{n-1}\frac{1}{t}\,dx$, A1 cao (3) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(E(x) = \frac{t}{2})$, $E(S) = kE(X)E(Y) = k \cdot \frac{t^2}{4}$ | M1A1 | |
| $E(S) = t^2 \Rightarrow k = 4$ | A1 | (3) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(XY) = E(X^2)E(Y^2) - [E(XY)]^2$ | M1 | |
| $= \frac{t^2}{3} \times \frac{t^2}{3} - \left(\frac{t^2}{4}\right)^2 = \left\{\frac{7t^4}{144}\right\}$ | M1 | |
| $\text{Var}(S) = k^2\,\text{Var}(XY) = 16 \times \frac{7t^4}{144} = \frac{7t^4}{9}$ | A1 cao | (3) |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(u) = t^2 \Rightarrow 2E(x^2)q = t^2 \Rightarrow 2\cdot\frac{t^2}{3}\cdot q = t^2 \Rightarrow q = \frac{3}{2}$ | M1, M1, A1 cao | (3) |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(u) = q^2[\text{Var}(x^2) + \text{Var}(Y^2)] = 2q^2\,\text{Var}(x^2)$ | M1 | |
| $\text{Var}(x^2) = E(x^4) - [E(x^2)]^2 = \frac{t^4}{5} - \left(\frac{t^2}{3}\right)^2 = \frac{4t^4}{45}$ | M1, A1 | |
| $\text{Var}(u) = 2 \times \frac{9}{4} \times \frac{4}{45}\,t^4 = \frac{2}{5}t^4$ | | (3) |
## Part (f)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{5} < \frac{7}{9}$ $\therefore U$ is better $\therefore$ smaller variance | B1ft | (1) |
## Part (g)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $u$ estimate is: $\frac{3}{2}(2^2 + 3^2) = \frac{3}{2} \times 13 = \frac{39}{2}$ or $19.5$ | B1ft | (1) |6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{f7137ba8-5526-4107-bccd-047de235d7d1-5_392_407_281_852}
\end{center}
\end{figure}
Figure 1 shows a square of side $t$ and area $t ^ { 2 }$ which lies in the first quadrant with one vertex at the origin. A point $P$ with coordinates ( $X , Y$ ) is selected at random inside the square and the coordinates are used to estimate $t ^ { 2 }$. It is assumed that $X$ and $Y$ are independent random variables each having a continuous uniform distribution over the interval $[ 0 , t ]$.\\[0pt]
[You may assume that $\mathrm { E } \left( X ^ { n } Y ^ { n } \right) = \mathrm { E } \left( X ^ { n } \right) \mathrm { E } \left( Y ^ { n } \right)$, where $n$ is a positive integer.]
\begin{enumerate}[label=(\alph*)]
\item Use integration to show that $\mathrm { E } \left( X ^ { n } \right) = \frac { t ^ { n } } { n + 1 }$.
The random variable $S = k X Y$, where $k$ is a constant, is an unbiased estimator for $t ^ { 2 }$.
\item Find the value of $k$.
\item Show that $\operatorname { Var } S = \frac { 7 t ^ { 4 } } { 9 }$.
The random variable $U = q \left( X ^ { 2 } + Y ^ { 2 } \right)$, where $q$ is a constant, is also an unbiased estimator for $t ^ { 2 }$.
\item Show that the value of $q = \frac { 3 } { 2 }$.
\item Find Var $U$.
\item State, giving a reason, which of $S$ and $U$ is the better estimator of $t ^ { 2 }$.
The point $( 2,3 )$ is selected from inside the square.
\item Use the estimator chosen in part (f) to find an estimate for the area of the square.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q6 [17]}}