S2 June 2008 Q1
1. Jean regularly takes a break from work to go to the post office. The amount of time Jean waits in the queue to be served at the post office has a continuous uniform distribution between 0 and 10 minutes.
Jean visits the post office 5 times.
Jean is in the queue when she receives a message that she must return to work for an urgent meeting. She can only wait in the queue for a further 3 minutes.
Given that Jean has already been queuing for 5 minutes,
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 5\) | B1 |
| \(\mathrm{Var}(X) = \dfrac{1}{12}(10 - 0)^2\) or attempt to use \(\displaystyle\int \dfrac{x^2}{10}\,dx - \mu^2\) | M1 |
| \(= \dfrac{100}{12} = \dfrac{25}{3} = 8\dfrac{1}{3} = 8.\dot{3}\) awrt 8.33 | A1 |
| (3) |
Notes
B1 cao
M1 using the correct formula \(\dfrac{(a - b)^2}{12}\) and subst in 10 or 0
or for an attempt at the integration they must increase the power of \(x\) by 1 and subtract their \(\mathrm{E}(X)\) squared.
A1 cao
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \leqslant 2) = (2 - 0) \times \dfrac{1}{10} = \dfrac{1}{5}\) or \(\dfrac{2}{10}\) or 0.2 | M1 A1 |
| (2) |
Notes
M1 for \(\mathrm{P}(X \leqslant 2)\) or \(\mathrm{P}(X \lt 2)\)
A1 cao
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{1}{5}\right)^5 = 0.00032\) or \(\dfrac{1}{3125}\) or \(3.2 \times 10^{-4}\) o.e. | M1 A1 |
| (2) |
Notes
M1 (their b)5. If the answer is incorrect we must see this. No need to check with your calculator
A1 cao
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \geqslant 8)\) or \(\mathrm{P}(X \gt 8)\) | M1 |
| \(\mathrm{P}(X \geqslant 8 \mid X \geqslant 5) = \dfrac{\mathrm{P}(X \geqslant 8)}{\mathrm{P}(X \geqslant 5)}\) \(= \dfrac{2/10}{5/10}\) | M1 |
| \(= \dfrac{2}{5}\) | A1 |
| (3) | |
| (10 marks) |
Notes
M1 writing \(\mathrm{P}(X \geqslant 8)\) (may use > sign). If they do not write \(\mathrm{P}(X \geqslant 8)\) then it must be clear from their working that they are finding it. 0.2 on its own with no working gets M0
M1 For attempting to use a correct conditional probability.
A1 2/5
Full marks for 2/5 on its own with no incorrect working
alternative
| Scheme | Marks |
|---|---|
| remaining time \(\sim \mathrm{U}[0,5]\) or \(\mathrm{U}[5,10]\) \(\mathrm{P}(X \geqslant 3 \text{ or } 8) = \dfrac{2}{5}\) | M1 M1 A1 |
M1 for \(\mathrm{P}(X \geqslant 3)\) or \(\mathrm{P}(X \geqslant 8)\) may use > sign
M1 using either U[0,5] or U[5,10]
A1 2/5