S2 January 2012 Q1
1. The time in minutes that Elaine takes to checkout at her local supermarket follows a continuous uniform distribution defined over the interval [3, 9].
Find
Given that Elaine has already spent 4 minutes at the checkout,
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \dfrac{9 + 3}{2} = 6\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(X) = \dfrac{(9 - 3)^2}{12} = 3\) | M1A1 |
| (2) |
Notes
M1 \(\dfrac{(9 - 3)^2}{12}\) or \(\dfrac{(9 + 3)^2}{12}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \gt 7) = (9 - 7) \times \dfrac{1}{6} = \dfrac{1}{3}\) | M1A1 |
| (2) |
Notes
M1 \(\dfrac{(9 - 7)}{6}\) or \(1 - \dfrac{(7 - 3)}{6}\) or \(\displaystyle\int_7^9 \frac{1}{6}\,\mathrm{d}x\) or \(1 - \displaystyle\int_3^7 \frac{1}{6}\,\mathrm{d}x\)
A1 Also acceptable \(0.\dot{3}\), \(0.3\dot{3}\) and awrt 0.333
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \lt 6 \mid X \gt 4) = \dfrac{\mathrm{P}(4 \lt X \lt 6)}{\mathrm{P}(X \gt 4)}\) | M1A1 |
| \(= \dfrac{\frac{2}{6}}{\frac{5}{6}} = \dfrac{2}{5}\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1 \(\dfrac{\mathrm{P}(4 \lt X \lt 6)}{P(X \gt 4)}\) or \(\dfrac{\mathrm{P}(X \lt 6)}{P(X \gt 4)}\) or \(\dfrac{2/6}{5/6}\) or \(\dfrac{3/6}{5/6}\) or \(1 - \dfrac{\mathrm{P}(X \gt 6)}{P(X \gt 4)}\) or \(\dfrac{6 - 4}{9 - 4}\) or \(\dfrac{3}{5}\)
A1 \(\dfrac{\mathrm{P}(4 \lt X \lt 6)}{P(X \gt 4)}\) or \(\dfrac{2/6}{5/6}\) or \(1 - \dfrac{\mathrm{P}(X \gt 6)}{P(X \gt 4)}\) or \(\dfrac{6 - 4}{9 - 4}\)
An answer of \(\dfrac{2}{5}\) gains all 3 marks.
NB \(\leqslant\) and \(\geqslant\) are accepted in the above formulae