S2 January 2008 Q4
4. The continuous random variable \(Y\) has cumulative distribution function \(\mathrm{F}(y)\) given by
\[\mathrm{F}(y) = \begin{cases} 0 & y \lt 1 \\ k(y^4 + y^2 - 2) & 1 \leqslant y \leqslant 2 \\ 1 & y \gt 2 \end{cases}\](a) Show that \(k = \dfrac{1}{18}\). (2)
(b) Find \(\mathrm{P}(Y \gt 1.5)\). (2)
(c) Specify fully the probability density function \(\mathrm{f}(y)\). (3)
| Scheme | Marks |
|---|---|
| \(K(2^4 + 2^2 - 2) = 1\) | M1 |
| \(K = 1/18\) | A1 |
| (2) |
Notes
M1 putting \(\mathrm{F}(2) = 1\) or \(\mathrm{F}(2) - \mathrm{F}(1) = 1\)
A1 cso. Must show substituting \(y = 2\) and the 1/18
| Scheme | Marks |
|---|---|
| \(1 - \mathrm{F}(1.5) = 1 - \dfrac{1}{18}(1.5^4 + 1.5^2 - 2)\) | M1 |
| \(= 0.705\) or \(\dfrac{203}{288}\) | A1 |
| (2) |
Notes
M1 either attempting to find \(1 - \mathrm{F}(1.5)\) may write and use \(\mathrm{F}(2) - \mathrm{F}(1.5)\)
A1 awrt 0.705
| Scheme | Marks |
|---|---|
| \(f(y) = \begin{cases} \dfrac{1}{9}(2y^3 + y) & 1 \leqslant y \leqslant 2 \\[2mm] 0 & \textit{otherwise} \end{cases}\) | M1 A1 B1 |
| (3) | |
| (7 marks) |
Notes
M1 attempting to differentiate. Must see either a \(y^n \to y^{n-1}\) at least once
A1 for getting \(\dfrac{1}{9}(2y^3 + y)\) o.e and \(1 \leqslant y \leqslant 2\) allow \(1 \lt y \lt 2\)
B1 for the 0 otherwise. Allow 0 for \(y \lt 1\) and 0 for \(y \gt 2\)
Allow them to use any letter