S2 June 2013 (R) Q2
2. The continuous random variable \(Y\) has cumulative distribution function
\[\mathrm{F}(y) = \begin{cases} 0 & y \lt 0 \\ \dfrac{1}{4}(y^3 - 4y^2 + ky) & 0 \leqslant y \leqslant 2 \\ 1 & y \gt 2 \end{cases}\]where \(k\) is a constant.
(a) Find the value of \(k\). (2)
(b) Find the probability density function of \(Y\), specifying it for all values of \(y\). (3)
(c) Find \(\mathrm{P}(Y \gt 1)\). (2)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(2) = 1\) gives: \(\dfrac{1}{4}\left(2^3 - 4 \times 2^2 + 2k\right) = 1\) | M1 |
| \(\underline{k = 6}\) | A1 |
| (2) |
Notes
M1 for an attempt to use \(\mathrm{F}(2) = 1\). Clear attempt to form a linear equation for \(k\)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(y) = \dfrac{\mathrm{d}}{\mathrm{d}y}(\mathrm{F}(y)) = \dfrac{1}{4}\left(3y^2 - 8y + \text{"6"}\right)\) | M1A1ft |
| \(\mathrm{f}(y) = \begin{cases} \dfrac{1}{4}\left(3y^2 - 8y + 6\right) & 0 \leqslant y \leqslant 2 \\ 0 & \text{otherwise} \end{cases}\) | A1 |
| (3) |
Notes
M1 for some correct differentiation \(y^n \to y^{n-1}\)
1st A1ft for \(3y^2 - 8y + \text{"6"}\), follow through their value of \(k\) or even \(k\) as a letter
2nd A1 for a fully correct solution including the 0 otherwise.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(Y \gt 1) = 1 - \mathrm{F}(1) = 1 - \dfrac{1}{4}\left(1^3 - 4 \times 1^2 + k\right)\) | M1 |
| \(= \dfrac{1}{4}\) (o.e.) | A1 |
| (2) | |
| (7 marks) |
Notes
M1 for clear use of \(1 - \mathrm{F}(y)\) or attempt at integrating \(\mathrm{f}(y)\); at least one correct term with correct coefficient, and using limit of 1 and 2
A1 for \(\frac{1}{4}\) or any exact equivalent