A2 June 2022 Q8
8. The continuous random variable \(X\) has cumulative distribution function given by
\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ 1.5x - 0.25x^2 - 1.25 & 1 \leqslant x \leqslant 3 \\ 1 & x \gt 3 \end{cases}\]The random variable \(Y = \dfrac{1}{X}\)
| Scheme | Marks | AO |
|---|---|---|
| \(1.5m - 0.25m^2 - 1.25 = 0.5 \quad (\rightarrow 0.25m^2 - 1.5m + 1.75 = 0)\) | M1 | 1.1b |
| \(m = 3 - \sqrt{2}\) (reject \(m = 3 + \sqrt{2}\)) | A1 | 2.2a |
| (2) |
Notes
M1: Correct equation for \(m\)
A1: \(m = 3 - \sqrt{2}\) only (isw if exact answer is given then rounded)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(X \lt 1.6 \mid X \gt 1.2) = \dfrac{\mathrm{P}(1.2 \lt X \lt 1.6)}{\mathrm{P}(X \gt 1.2)}\) | M1 | 3.1a |
| \(\dfrac{\mathrm{F}(1.6) - \mathrm{F}(1.2)}{1 - \mathrm{F}(1.2)}\) | M1 | 1.1b |
| \(= \dfrac{32}{81}\) | A1 | 1.1b |
| (3) |
Notes
M1: Determining the two probabilities required to find the probability
M1: Correct ratio of probabilities
A1: allow awrt 0.395
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(Y \leqslant y) = \mathrm{P}\left(\dfrac{1}{X} \leqslant y\right)\) | ||
| \(= \mathrm{P}\left(X \geqslant \tfrac{1}{y}\right) = 1 - \mathrm{F}\left(\tfrac{1}{y}\right)\) | M1 | 3.1a |
| \(= 1 - \left(\tfrac{1.5}{y} - 0.25\left(\tfrac{1}{y}\right)^2 - 1.25\right)\) | M1 | 1.1b |
| \(\mathrm{F}(y) = \begin{cases} 0 & y \lt \frac{1}{3} \\ 2.25 - \frac{1.5}{y} + 0.25\left(\frac{1}{y}\right)^2 & \frac{1}{3} \leqslant y \leqslant 1 \\ 1 & y \gt 1 \end{cases}\) | A1 A1 | 2.1 1.1b |
| (4) |
Notes
M1: Realising that \(\mathrm{P}\left(X \geqslant \tfrac{1}{y}\right)\) is necessary
M1: Correct use of \(\mathrm{F}(x)\)
A1: Determining correct limits (allow \(\leqslant\) for \(\lt\) etc.)
A1: All lines of cdf correct (ignoring limits) must be in terms of \(y\)
SC: If 0 scored, then [\(\mathrm{F}(y)\) =] \(\tfrac{1.5}{y} - 0.25\left(\tfrac{1}{y}\right)^2 - 1.25 \quad \tfrac{1}{3} \leqslant y \leqslant 1\) scores M0M0A1A0
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(y) = \dfrac{\mathrm{d}}{\mathrm{d}y}(\mathrm{F}(y)) = 1.5y^{-2} - 0.5y^{-3} \rightarrow \mathrm{f}'(y) = -3y^{-3} + 1.5y^{-4}\) | M1 | 3.1a |
| \(\mathrm{f}'(y) = -3y^{-3} + 1.5y^{-4} = 0\) | depM1 | 1.1b |
| \(1.5y^{-4} = 3y^{-3} \rightarrow \tfrac{1.5}{y^4} = \tfrac{3}{y^3}\) | ||
| [Mode of \(Y\) =] 0.5 (since \(\mathrm{f}''(0.5) = -48 \lt 0\)) | A1 | 1.1b |
| (3) | ||
| (12 marks) |
Notes
M1: Realising that the cdf must be differentiated twice
depM1: (dep on previous M1) Equating their \(\mathrm{f}'(y) = 0\) with attempt to solve
A1: 0.5 cao