Moderate -0.5 This is a straightforward application of the expectation formula for continuous distributions requiring integration of a constant pdf. While it involves two steps (finding k from normalization, then computing E(R)), both integrations are elementary polynomials with no conceptual difficulty. The question is slightly below average difficulty as it's a standard 'prove the formula' exercise with clear structure, though it does require proper setup and algebraic manipulation.
3 The continuous random variable \(R\) follows a rectangular distribution with probability density function given by
$$f ( r ) = \begin{cases} k & - a \leq r \leq b \\ 0 & \text { otherwise } \end{cases}$$
Prove, using integration, that \(\mathrm { E } ( R ) = \frac { 1 } { 2 } ( b - a )\)
[0pt]
[4 marks]
Completes clear correct rigorous workings to show required result AG. Only award if they have a completely correct solution, which is clear, easy to follow and contains no slips.
Total: 4 marks
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = \frac{1}{b+a}$ | B1 | States value for $k$ |
| $E(R) = \int_{-a}^{b} kr \, dr = \left[k\frac{r^2}{2}\right]_{-a}^{b}$ | M1 | Uses correct integral for $E(R)$ |
| $= \frac{k}{2}(b^2 - a^2) = \frac{1}{2(b+a)}(b-a)(b+a)$ | M1 | Uses limits correctly with 'their' integral |
| $= \frac{(b-a)}{2}$ | R1 | Completes clear correct rigorous workings to show required result **AG**. Only award if they have a completely correct solution, which is clear, easy to follow and contains no slips. |
**Total: 4 marks**
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3 The continuous random variable $R$ follows a rectangular distribution with probability density function given by
$$f ( r ) = \begin{cases} k & - a \leq r \leq b \\ 0 & \text { otherwise } \end{cases}$$
Prove, using integration, that $\mathrm { E } ( R ) = \frac { 1 } { 2 } ( b - a )$\\[0pt]
[4 marks]\\
\hfill \mbox{\textit{AQA Further Paper 3 Statistics Q3 [4]}}