S2 June 2008 Q7
7. A random variable \(X\) has probability density function given by
\[\mathrm{f}(x) = \begin{cases} \dfrac{1}{2}x & 0 \leqslant x \lt 1 \\[2mm] kx^3 & 1 \leqslant x \leqslant 2 \\[2mm] 0 & \text{otherwise} \end{cases}\]where \(k\) is a constant.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^1 \dfrac{1}{2}x\ \mathrm{d}x = \left[\dfrac{1}{4}x^2\right]_0^1 = \dfrac{1}{4}\) oe | M1 |
| \(\displaystyle\int_1^2 kx^3\ \mathrm{d}x\ \left[\dfrac{1}{4}kx^4\right]_1^2 = 4k - \dfrac{1}{4}k\) oe | A1 |
| \(\dfrac{1}{4} + 4k - \dfrac{1}{4}k = 1\) | dM1dep on previous M |
| \(\dfrac{15k}{4} = \dfrac{3}{4}\) \(k = \dfrac{1}{5}\) * | A1 |
| (4) |
Notes
M1 attempt to integrate both parts
A1 both answer correct
dM1 adding two answers and putting = 1
(a) M1 attempting to integrate both parts
A1 both answers correct
M1 dependent on the previous M being awarded.. adding the two answers together
A1 cso
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^1 \dfrac{1}{2}x^2\ \mathrm{d}x = \left[\dfrac{1}{6}x^3\right]_0^1 = \dfrac{1}{6}\) | M1 A1 |
| \(\displaystyle\int_1^2 \dfrac{1}{5}x^4\ \mathrm{d}x = \left[\dfrac{1}{25}x^5\right]_1^2 = \dfrac{32}{25} - \dfrac{1}{25}\) \(= \dfrac{31}{25}\) or 1.24 | A1 |
| \(\mathrm{E}(X) = \dfrac{1}{6} + \dfrac{31}{25}\) \(= \dfrac{211}{150} = 1\dfrac{61}{150} = 1.40\dot{6}\) | A1 |
| (4) |
Notes
M1 attempt to integrate \(x\mathrm{f}(x)\) for one part A1 1/6
(b) M1 attempting to use integral of \(x\,\mathrm{f}(x)\) on one part
A1 1/6
A1 31/25
A1 awrt 1.41
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = \displaystyle\int_0^x \dfrac{1}{2}t\ \mathrm{d}t\) (for \(0 \leqslant x \leqslant 1\)) | M1 |
| \(= \dfrac{1}{4}x^2\) | A1 |
| \(\mathrm{F}(x) = \displaystyle\int_1^x \dfrac{1}{5}t^3\ \mathrm{d}t; + \displaystyle\int_0^1 \dfrac{1}{2}t\ \mathrm{d}t\) (for \(1 \lt x \leqslant 2\)) | M1; M1 |
| \(= \dfrac{1}{20}x^4 + \dfrac{1}{5}\) | A1 |
| \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\[1mm] \dfrac{1}{4}x^2 & 0 \leqslant x \leqslant 1 \\[1mm] \dfrac{1}{20}x^4 + \dfrac{1}{5} & 1 \lt x \leqslant 2 \\[1mm] 1 & x \gt 2 \end{cases}\) | B1 ft B1 |
| (7) |
Notes
ignore limits for M; must use limit of 0; need limit of 1 and variable upper limit; need limit 0 and 1; B1 ft middle pair, B1 ends
(c) M1 Att to integrate \(\dfrac{1}{2}t\) (they need to increase the power by 1). Ignore limits for method mark
A1 \(\dfrac{1}{4}x^2\) allow use of t. must have used/implied use of limit of 0. This must be on its own without anything else added
M1 att to integrate \(\displaystyle\int_1^x \dfrac{1}{5}t^3\ \mathrm{d}t\) and correct limits.
M1 \(\displaystyle\int_0^1 \dfrac{1}{2}t\ \mathrm{d}t +\) Att to integrate using limits 0 and 1. no need to see them put 0 in.
they must add this to their \(\displaystyle\int_1^x \dfrac{1}{5}t^3\ \mathrm{d}t\). may be given if they add 1/4
(Alternative method for these last two M marks
M1 for att to \(\displaystyle\int \dfrac{1}{5}t^3\ \mathrm{d}t\) and putting + C
M1 use of F(2) = 1 to find C)
A1 \(\dfrac{1}{20}x^4 + \dfrac{1}{5}\) must be correct
B1 middle pair followed through from their answers. condone them using < or \(\leqslant\) incorrectly they do not need to match up
B1 end pairs. condone them using < or \(\leqslant\). They do not need to match up
NB if they show no working and just write down the distribution. If it is correct they get full marks. If it is incorrect then they cannot get marks for any incorrect part. So if \(0 \lt x \lt 1\) is correct they can get M1 A1 otherwise M0 A0. if \(1 \lt x \lt 2\) is correct they can get M1 A1A1 otherwise M0 A0A0. you cannot award B1ft if they show no working unless the middle parts are correct. (corrected from the printed mark scheme: printed “if \(3 \lt x \lt 4\) is correct”)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(m) = 0.5\) | |
| \(\dfrac{1}{20}m^4 + \dfrac{1}{5} = 0.5\) | M1 A1ft |
| \(m = \sqrt[4]{6}\) or 1.57 or awrt 1.57 | A1 |
| (3) |
Notes
either eq; eq for their \(1 \leqslant x \leqslant 2\)
(d) M1 either of their \(\dfrac{1}{4}x^2\) or \(\dfrac{1}{20}x^4 + \dfrac{1}{5} = 0.5\)
A1 for their \(\mathrm{F}(X)\) \(1 \lt x \lt 2 = 0.5\)
A1 cao
If they add both their parts together and put = 0.5 they get M0
I they work out both parts separately and do not make the answer clear they can get M1 A1 A0
| Scheme | Marks |
|---|---|
| negative skew | B1 |
| This depends on the previous B1 being awarded. One of the following statements which must be compatible with negative skew and their figures. If they use mode then they must have found a value for it Mean < Median Mean < mode Mean < median (< mode) Median < mode Sketch of the pdf. | dB1 |
| (2) | |
| (20 marks) |
Notes
(e) B1 negative skew only
B1 Dependent on getting the previous B1. their reason must follow through from their figures.