S2 June 2017 Q4
4. The continuous random variable \(X\) is uniformly distributed over the interval \([\alpha, \beta]\)
Given that \(\mathrm{E}(X) = 3.5\) and \(\mathrm{P}(X \gt 5) = \dfrac{2}{5}\)
Given that \(\mathrm{P}(X \lt c) = \dfrac{2}{3}\)
A rectangle has a perimeter of 200 cm. The length, \(S\) cm, of one side of this rectangle is uniformly distributed between 30 cm and 80 cm.
| Scheme | Marks |
|---|---|
| \(\left[\mathrm{E}(X) = \dfrac{\alpha + \beta}{2} = 3.5\right], \Rightarrow \alpha + \beta = 7\) | B1 |
| \(\left[\mathrm{P}(X \gt 5) = \dfrac{\beta - 5}{\beta - \alpha} = \dfrac{2}{5}\right],\) \(\Rightarrow 5(\beta - 5) = 2(\beta - \alpha)\) | M1 |
| \(\alpha = -4\) | A1 |
| \(\beta = 11\) | A1 |
| (4) |
Notes
B1 Correct equation. Need not be simplified
M1 a second correct equation, Using simultaneous equations and eliminating \(\alpha\) or \(\beta\) to gain a value of \(\alpha\) and \(\beta\).
1st A1 for –4
2nd A1 for 11
NB Award full marks for \(\alpha = -4, \beta = 11\)
| Scheme | Marks |
|---|---|
| (i) \(\dfrac{c + 4}{15} = \dfrac{2}{3}\) \([c =]\ 6\) | B1 |
| (ii) \(\mathrm{P}(6 \lt X \lt 9) = \dfrac{1}{15}\times(3)\) | M1 |
| \(= 0.2\) | A1cso |
| (3) |
Notes
(i) B1 for 6
(ii) M1 \(\dfrac{1}{\beta - \alpha}\times(9 - c)\) or [F(9) – F(\(c\))] \(= \dfrac{13}{15} - \dfrac{2}{3}\)
SC if 9 > “their \(\beta\)” award for \(1 - \dfrac{2}{3}\)
A1cso 0.2 oe
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(S \lt 45)] = \dfrac{3}{10}\) | B1 |
| \([\mathrm{P}(S \gt 55)] = \dfrac{1}{2}\) | B1 |
| total \(= \dfrac{3}{10} + \dfrac{1}{2} = \dfrac{4}{5}\) | M1A1 |
| (4) | |
| (11 marks) |
Notes
B1 \(\dfrac{3}{10}\) seen – it does not need to be associated with P (\(S\) < 45)]
B1 \(\dfrac{1}{2}\) seen– it does not need to be associated with P (\(S\) > 55)]
M1 for adding their two areas and the total < 1. Do not allow \(2\times\) a single area
A1 \(\dfrac{4}{5}\) oe
NB Award full marks for \(\dfrac{4}{5}\)