| (a) | $\frac{1}{2} \times 8 \times 4a = 1 \Rightarrow a = \frac{1}{16}$ * | B1* |  |
| | or $\int_{[0]}^{[4]} ax \, dx = 0.5 \Rightarrow \left[\frac{ax^2}{2}\right]_0 = 0.5 \Rightarrow a = \frac{1}{16}$ * | | (1) |
| (b)(i) | (By symmetry) $b = -\frac{1}{16}$ | B1 | (1) |
| (ii) | At $(8, 0)$ $0 = -\frac{1}{16} \times 8 + c \Rightarrow c = \frac{1}{2}$ or at $(4, 0.25)$ $0.25 = -\frac{1}{16} \times 4 + c \Rightarrow c = \frac{1}{2}$ | M1 A1 | (3) |
| (c) | $E(X) = 4$ | B1 |  |
| | $E(X^2) = \int_0^4 x^2 \left(\frac{1}{16}x\right) dx + \int_4^8 x^2 \left(-\frac{1}{16}x + \frac{1}{2}\right) dx$ | M1 |  |
| | $= \frac{1}{64}\left[x^4\right]_0^4 + \left[\begin{matrix} \frac{1}{64}x^4 + \frac{1}{6}x^3 \end{matrix}\right]_4$ | A1ft |  |
| | $= 4 + \left[\left(-64 + \frac{256}{3}\right) - \left(-4 + \frac{32}{3}\right)\right] - \frac{56}{3}$ | dM1A1 |  |
| | $\text{Var}(X) = \frac{56}{3} - 4^2 = \frac{8}{3}$ * | A1* | (6) |
| (d) | $\frac{1}{2} \times Q_1 \times \frac{1}{16} \times Q_1 = \frac{1}{4}$ or $\int_0^{Q_1} \frac{1}{16}x \, dx = 0.25 \rightarrow \frac{Q_1^2}{32} = 0.25$ | M1 |  |
| | $Q_1 = \sqrt{8} = 2.828...$ or $Q_3 = 8 - \sqrt{8} = 5.171...$ awrt 2.83 or awrt 5.17 | A1 |  |
| | $Q_1 = \sqrt{8} = 2.828...$ and $Q_3 = 8 - \sqrt{8} = 5.171...$ awrt 2.83 and awrt 5.17 | A1 | (3) |
| (e) | 50% lies between $Q_1$ and $Q_3$ | M1 |  |
| | Statistician's claim: $p\left('4'- \sqrt{\frac{8}{3}} < X < '4' + \sqrt{\frac{8}{3}}\right) = P(2.37 < X < 5.63)$ | M1 |  |
| | as this is outside $Q_1$ and $Q_3$, $> 0.5/$ statistician's claim is correct* | A1* | (2) |
| | or $P(2.37 < X < 5.63) = 0.6498... > 0.5 /$ statistician's claim is correct* | | |
| | **Total** | **15** |  |

**Notes for Question 7:**

- **(a) B1*:** Allow any correct equivalent method. E.g. $\frac{1}{2} \times 4 \times 4a = \frac{1}{2} \Rightarrow a = \frac{1}{16}$, integration, use of gradients, etc.
  - Answer is given so a complete correct method with no incorrect working must be seen
- **(b) B1:** Cao
- **(b) M1:** Use of equation of line to find $c$ e.g. $y - 0 = -\frac{1}{16}(x-8)$ or use of integration or any valid method
- **(b) A1:** Cao correct answer scores M1A1
- **(c) B1:** For $E(X) = 4$ This may be seen at any point in the solution
- **(c) M1:** For use of $\int x^2 f(x) dx$ or $x^6 \rightarrow x^{6+1}$ for both parts of pdf (ignore limits) ft their values of $b$ and $c$
- **(c) A1ft:** For correct integration of **either** of the 2 parts, ft their values of $b$ and $c$
  - For use of correct limits in **either** part (dep on previous M1)
  - may be implied by sight of 4 or $\frac{44}{3}$ but not implied by $\frac{56}{3}$
  - allow ft on their values of $b$ and $c$ which you may need to check
- **(c) dM1:** For complete correct substitution $4 + \left[\left(-64 + \frac{256}{3}\right) - \left(-4 + \frac{32}{3}\right)\right]$ or $\frac{56}{3}$ allow = $4 + \frac{44}{3}$
- **(c) A1:** For correct method for **either** $Q_1$ or $Q_3$
  - e.g. $\frac{1}{2} \times Q_1 \times \frac{1}{16} \times Q_1 = \frac{1}{4}$ or $\int_0^{Q_1} \frac{1}{16}