S2 January 2006 Q3
3. The random variable \(X\) is uniformly distributed over the interval \([-1, 5]\).
(a) Sketch the probability density function \(\mathrm{f}(x)\) of \(X\). (3)
Find
(b) \(\mathrm{E}(X)\), (1)
(c) \(\mathrm{Var}(X)\), (2)
(d) \(\mathrm{P}(-0.3 \lt X \lt 3.3)\). (2)
| Scheme | Marks |
|---|---|
![]() | B1 B1 B1 |
| (3) |
Notes
1st B1 horizontal line (shape)
2nd B1 \(-1, 5\)
3rd B1 \(\dfrac{1}{6}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 2\) by symmetry | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(X) = \dfrac{1}{12}(5 + 1)^2 \quad\) or \(\displaystyle\int \frac{x^2}{6}\,\mathrm{d}x - 4 = \left[\frac{x^3}{18}\right]_{-1}^{5} - 4\) | M1 |
| \(= 3\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(-0.3 \lt X \lt 3.3) = \dfrac{3.6}{6} \quad\) or \(\displaystyle\int_{-0.3}^{3.3} \frac{1}{6}\,\mathrm{d}x = \left[\frac{x}{6}\right]_{-0.3}^{3.3}\) | M1 |
| \(= 0.6\) | A1 |
| (2) | |
| (8 marks) |
Notes
M1 full correct method for the correct area
