S2 June 2018 Q4
4. David aims to catch the train to work each morning. The scheduled departure time of the train is 08 30
The number of minutes after 08 30 that the train departs may be modelled by the random variable \(X\). Given that \(X\) has a continuous uniform distribution over \([\alpha, \beta]\) and that \(\mathrm{E}(X) = 4\) and \(\mathrm{Var}(X) = 12\)
Each morning, the probability that David oversleeps is 0.05
If David oversleeps he will be late for work.
If he does not oversleep he will be in time to catch the train, but will be late for work if the train departs after 08 35
Given that David is late for work,
| Scheme | Marks |
|---|---|
| \(\dfrac{\beta + \alpha}{2} = 4, \qquad \dfrac{(\beta - \alpha)^2}{12} = 12\) | B1 |
| \(\beta + \alpha = 8\) and \((\beta - \alpha) = 12\) or \(\alpha^2 - 8\alpha - 20 = 0\) or \(\beta^2 - 8\beta - 20 = 0\) | B1 |
| \(2\beta = 20\) | M1d |
| \(\beta = 10\) | A1 |
| \(\alpha = -2\) | A1 |
| (5) |
Notes
B1 \(\dfrac{\alpha + \beta}{2} = 4\) and \(\dfrac{(\beta - \alpha)^2}{12} = 12\) oe
B1 A pair of correct linear equations or a correct single equation in \(\alpha\) or \(\beta\)
M1d dep on 1st B mark being awarded. Correct method to solve their simultaneous equations by eliminating \(\alpha\) or \(\beta\) or a correct method to solve their quadratic equation.
A1 cao must state it is \(\beta = 10\) not just write 10 or written as […, 10]
A1 cao must state it is \(\alpha = -2\) not just write -2 or written as [-2, …]
| Scheme | Marks |
|---|---|
| P(David late) \(= 0.05 + 0.95\times\left(\dfrac{\text{"}10\text{"} - 5}{\text{"}12\text{"}}\right)\) | M1, B1ft |
| \(= \dfrac{107}{240}\) or \(0.4458333\ldots\) awrt 0.446 | A1 |
| (3) |
Notes
M1 \(0.05 + 0.95\times(p)\) \(0 \lt p \lt 1\)
B1ft \(\left(\dfrac{10 - 5}{12}\right)\) or \(\dfrac{5}{12}\) or awrt 0.417 or \(\dfrac{\text{"their}\beta\text{"} - 5}{\text{"their}\beta\text{"} - \text{"their}\alpha\text{"}}\)
A1 awrt 0.446 or \(\dfrac{107}{240}\)
NB only award these marks in part(b)
| Scheme | Marks |
|---|---|
| P(missed train | late) \(= \dfrac{0.05}{0.446}\) | M1 |
| \(= \dfrac{12}{107}\) or \(0.1121\ldots\) awrt 0.112 | A1 |
| (2) | |
| (10 marks) |
Notes
M1 \(\dfrac{0.05}{\text{their (b)}}\)
A1 awrt 0.112 or \(\dfrac{12}{107}\)