A2 June 2023 Q6
6. The continuous random variable \(X\) has cumulative distribution function given by
\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\ k\left(x - ax^2\right) & 0 \leqslant x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\]The values of \(a\) and \(k\) are positive constants such that \(\mathrm{P}(X \lt 2) = \dfrac{2}{3}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{F}(2) = \frac{2}{3} \quad\) and \(\quad \mathrm{F}(4) = 1\) | ||
| \(k(2 - 4a) = \frac{2}{3} \qquad k(4 - 16a) = 1\) | M1 | 2.1 |
| \(\dfrac{2}{3(2 - 4a)} = \dfrac{1}{(4 - 16a)} \to a = \dfrac{1}{10},\quad k = \dfrac{5}{12}\) | M1A1 | 1.1b 1.1b |
| \(\frac{5}{12}\left(m - \frac{1}{10}m^2\right) = 0.5 \to m^2 - 10m + 12 = 0\) | M1M1 | 3.1a 1.1b |
| \(m = 5 - \sqrt{13} \qquad (m = 5 + \sqrt{13}\ \text{ reject})\) | A1 | 1.1b |
| (6) |
Notes
M1: using \(\mathrm{F}(2) = \frac{2}{3}\) and \(\mathrm{F}(4) = 1\) to form two correct expressions in \(a\) and \(k\)
M1: solving simultaneously to find a value of \(a\) or \(k\)
A1: both \(a = \dfrac{1}{10}\) and \(k = \dfrac{5}{12}\), may be implied by correct equation or median
M1: setting up \(\mathrm{F}(m) = 0.5\) to form a 3TQ
M1: using an appropriate method to solve quadratic, resulting in a value of \(m\)
A1: selecting the correct value for \(m\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}\big(\mathrm{F}(x)\big)\) | M1 | 1.1b |
| \(\mathrm{f}(x) = \begin{cases} \text{“}\tfrac{5}{12}\text{”}(1 - \text{“}0.2\text{”}x) & 0 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\) | A1ft | 1.1b |
| (2) |
Notes
M1: differentiating \(\mathrm{F}(x)\), at least one term correct for their \(a\) and \(k\)
A1ft: correct ft expression with correct limits
| Scheme | Marks | AO |
|---|---|---|
| Mode is \(X = 0\)… | B1 | 2.2a |
| …since \(\mathrm{f}(x)\) is linear with negative gradient. or …\(\mathrm{f}(x)\) is a decreasing function | dB1 | 2.4 |
| (2) | ||
| (10 marks) |
Notes
B1: deducing that the mode is \(X = 0\)
dB1: correct reasoning
May be given via a diagram of a linear function \(0 \leqslant x \leqslant 4\) with:
- a negative gradient
- that intersects the \(y\)-axis
- is never negative in \(y\)