| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Derive or verify variance formula |
| Difficulty | Moderate -0.3 Part (a)(i) is immediate recall, (a)(ii) is a standard bookwork proof using E(X²) - [E(X)]² with straightforward integration, and part (b) is a routine probability calculation with a uniform distribution. While the proof requires integration technique, it's a formula derivation that appears in most S2 textbooks with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks |
|---|---|
| \(E(X) = \frac{1}{2}b\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = \int_0^b \frac{1}{b}x^2 \, dx\) | M1 | |
| \(= \frac{1}{b}\left[\frac{x^3}{3}\right]_0^b\) | A1 | For correct integration |
| \(= \frac{1}{b}\left(\frac{b^3}{3}\right)\) | ||
| \(= \frac{1}{3}b^2\) | A1 | OE |
| \(\text{Var}(X) = \frac{1}{3}b^2 - \left(\frac{b}{2}\right)^2\) | m1 | Depending on using integration to get \(E(X^2)\) |
| \(= \frac{1}{3}b^2 - \frac{1}{4}b^2 = \frac{1}{12}b^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P( | T | > 0.02) = 1 - P(-0.02 < T < 0.02)\) |
| \(= 1 - 0.04 \times 5 = 0.8\) | M1, A1 |
## 5(a)(i)
$E(X) = \frac{1}{2}b$ | B1 |
## 5(a)(ii)
$E(X^2) = \int_0^b \frac{1}{b}x^2 \, dx$ | M1 |
$= \frac{1}{b}\left[\frac{x^3}{3}\right]_0^b$ | A1 | For correct integration
$= \frac{1}{b}\left(\frac{b^3}{3}\right)$ |
$= \frac{1}{3}b^2$ | A1 | OE
$\text{Var}(X) = \frac{1}{3}b^2 - \left(\frac{b}{2}\right)^2$ | m1 | Depending on using integration to get $E(X^2)$
$= \frac{1}{3}b^2 - \frac{1}{4}b^2 = \frac{1}{12}b^2$ | A1 |
## 5(b)
$P(|T| > 0.02) = 1 - P(-0.02 < T < 0.02)$ | M1 |
$= 1 - 0.04 \times 5 = 0.8$ | M1, A1 |
---5
\begin{enumerate}[label=(\alph*)]
\item The continuous random variable $X$ follows a rectangular distribution with probability density function defined by
$$f ( x ) = \begin{cases} \frac { 1 } { b } & 0 \leqslant x \leqslant b \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Write down $\mathrm { E } ( X )$.
\item Prove, using integration, that
$$\operatorname { Var } ( X ) = \frac { 1 } { 12 } b ^ { 2 }$$
\end{enumerate}\item At an athletics meeting, the error, in seconds, made in recording the time taken to complete the 10000 metres race may be modelled by the random variable $T$, having the probability density function
$$f ( t ) = \left\{ \begin{array} { c c }
5 & - 0.1 \leqslant t \leqslant 0.1 \\
0 & \text { otherwise }
\end{array} \right.$$
Calculate $\mathrm { P } ( | T | > 0.02 )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q5 [10]}}