S2 January 2006 Q5
5. A continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\) where\[\mathrm{f}(x) = \begin{cases} kx(x - 2), & 2 \leqslant x \leqslant 3, \\ 0, & \text{otherwise,} \end{cases}\]where \(k\) is a positive constant.
(a) Show that \(k = \dfrac{3}{4}\). (4)
Find
(b) \(\mathrm{E}(X)\), (3)
(c) the cumulative distribution function \(\mathrm{F}(x)\). (6)
(d) Show that the median value of \(X\) lies between 2.70 and 2.75. (2)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_2^3 kx(x - 2)\,\mathrm{d}x = 1\) | M1 |
| \(\left[\dfrac{1}{3}kx^3 - kx^2\right]_2^3 = 1\) | M1 A1 |
| \((9k - 9k) - \left(\dfrac{8k}{3} - 4k\right) = 1\) \(k = \dfrac{3}{4} = 0.75\ *\) | A1 |
| (4) |
Notes
1st M1 \(\int f(x) = 1\)
2nd M1 attempt \(\int\); need either \(x^3\) or \(x^2\)
1st A1 correct \(\int\)
2nd A1 cso
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \displaystyle\int_2^3 \frac{3}{4}x^2(x - 2)\,\mathrm{d}x\) | M1 |
| \(= \left[\dfrac{3}{16}x^4 - \dfrac{1}{2}x^3\right]_2^3\) | A1 |
| \(= 2.6875 = 2\dfrac{11}{16} = 2.69\) (3sf) | A1 |
| (3) |
Notes
M1 attempt \(\int xf(x)\)
1st A1 correct \(\int\)
2nd A1 awrt 2.69
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = \displaystyle\int_2^x \frac{3}{4}(t^2 - 2t)\,\mathrm{d}t\) | M1 |
| \(= \left[\dfrac{3}{4}\left(\dfrac{1}{3}t^3 - t^2\right)\right]_2^x\) | A1 |
| A1 | |
| \(= \dfrac{1}{4}(x^3 - 3x^2 + 4)\) | A1 |
| \(\mathrm{F}(x) = \begin{cases} 0 & x \leqslant 2 \\ \dfrac{1}{4}(x^3 - 3x^2 + 4) & 2 \lt x \lt 3 \\ 1 & x \geqslant 3 \end{cases}\) | B1ft, B1 |
| (6) |
Notes
M1 \(\int \mathrm{f}(x)\) with variable limit or \(+C\)
1st A1 correct integral
2nd A1 lower limit of 2 or \(\mathrm{F}(2) = 0\) or \(\mathrm{F}(3) = 1\)
B1ft, B1 middle, ends
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = \dfrac{1}{2}\) \(\dfrac{1}{4}(x^3 - 3x^2 + 4) = \dfrac{1}{2}\) | M1 |
| \(x^3 - 3x^2 + 2 = 0\) \(x = 2.75,\ x^3 - 3x^2 + 2 \gt 0\) \(x = 2.70,\ x^3 - 3x^2 + 2 \lt 0 \Rightarrow\) root between 2.70 and 2.75 | M1 |
| (or \(\mathrm{F}(2.7) = 0.453\), \(\mathrm{F}(2.75) = 0.527 \Rightarrow\) median between 2.70 and 2.75) | |
| (2) | |
| (15 marks) |
Notes
1st M1 their \(\mathrm{F}(x) = 1/2\)