S2 January 2010 Q4
4. The continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\) given by
\[\mathrm{f}(x) = \begin{cases} k(x^2 - 2x + 2) & 0 \lt x \leqslant 3 \\ 3k & 3 \lt x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\]where \(k\) is a constant.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^3 k(x^2 - 2x + 2)\,\mathrm{d}x + \int_3^4 3k\,\mathrm{d}x = 1\) | M1 |
| \(k\left[\dfrac{1}{3}x^3 - x^2 + 2x\right]_0^3 + \left[3kx\right]_3^4 \ (=1)\) or \(k\left[\dfrac{1}{3}x^3 - x^2 + 2x\right]_0^3 + 3k \ (=1)\) | A1 M1 dep |
| \(9k = 1\) \(k = \dfrac{1}{9}\) **given** cso | A1 |
| (4) |
Notes
1st M1 attempting to integrate at least one part (at least one \(x^n \to x^{n+1}\)) (ignore limits)
1st A1 Correct integration. Limits not needed.
2nd M1 dependent on the previous M being awarded. Adding the two answers together, putting equal to 1 and have the correct limits.
2nd A1 cso
| Scheme | Marks |
|---|---|
| For \(0 \lt x \leqslant 3,\ \mathrm{F}(x) = \displaystyle\int_0^x \frac{1}{9}(t^2 - 2t + 2)\,\mathrm{d}t\) | M1 |
| \(= \dfrac{1}{9}\left(\dfrac{1}{3}x^3 - x^2 + 2x\right)\) | A1 |
| For \(3 \lt x \leqslant 4,\ \mathrm{F}(x) = \displaystyle\int_3^x 3k\,\mathrm{d}t + \frac{2}{3}\) | M1 |
| \(= \dfrac{x}{3} - \dfrac{1}{3}\) | A1 |
| \(\mathrm{F}(x) = \begin{cases} 0 & x \leqslant 0 \\ \dfrac{1}{27}(x^3 - 3x^2 + 6x) & 0 \lt x \leqslant 3 \\ \dfrac{x}{3} - \dfrac{1}{3} & 3 \lt x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\) | B1 ft B1 |
| (6) |
Notes
1st M1 Att to integrate \(\dfrac{1}{9}\left(t^2 - 2t + 2\right)\) (at least one \(x^n \to x^{n+1}\)). Ignore limits for method mark
1st A1 \(\dfrac{1}{9}\left(\dfrac{x^3}{3} - x^2 + 2x\right)\) allow use of \(t\). Must have used/implied use of limit of 0. This must be on its own without anything else added
2nd M1 attempting to find \(\displaystyle\int_3^x 3k + \ldots\) (must get \(3kt\) or \(3kx\)) and they must use the correct limits and add \(\displaystyle\int_0^3 \frac{1}{9}\left(t^2 - 2t + 2\right)\) or \(\dfrac{2}{3}\) or use \(+\,\mathrm{C}\) and use \(\mathrm{F}(4) = 1\)
2nd A1 \(\dfrac{x}{3} - \dfrac{1}{3}\) must be correct
1st B1 middle pair followed through from their answers. condone them using < or \(\leqslant\) incorrectly they do not need to match up
2nd 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 \(0 \lt x \leqslant 3\) is correct they can get M1 A1 otherwise M0 A0. If \(3 \lt x \leqslant 4\) is correct they can get M1 A1 otherwise M0 A0. you cannot award B1ft if they show no working unless the middle parts are correct.
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \displaystyle\int_0^3 \frac{x}{9}(x^2 - 2x + 2)\,\mathrm{d}x + \int_3^4 \frac{x}{3}\,\mathrm{d}x\) | M1 |
| \(= \dfrac{1}{9}\left[\dfrac{1}{4}x^4 - \dfrac{2}{3}x^3 + x^2\right]_0^3 + \left[\dfrac{1}{6}x^2\right]_3^4\) | A1 |
| \(= \dfrac{29}{12}\) or 2.416 or awrt 2.42 | A1 |
| (3) |
Notes
1st M1 attempting to use integral of \(x\,\mathrm{f}(x)\) on one part
1st A1 Correct Integration for both parts added together. Ignore limits.
2nd A1 cao or awrt 2.42
(corrected from the printed mark scheme: the first integral is printed with “d\(t\)” in place of d\(x\))
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(m) = 0.5\) | M1 |
| \(\mathrm{F}(2.6) = \dfrac{1}{27}(2.6^3 - 3 \times 2.6^2 + 6 \times 2.6) = \text{awrt } 0.48\) | M1 |
| \(\mathrm{F}(2.7) = \dfrac{1}{27}(2.7^3 - 3 \times 2.7^2 + 6 \times 2.7) = \text{awrt } 0.52\) | A1 |
| Hence median lies between 2.6 and 2.7 | A1 dA |
| (4) | |
| (17 marks) |
Notes
1st M1 for using \(\mathrm{F}(X) = 0.5\). This may be implied by subst into \(\mathrm{F}(X)\) and comparing answers with 0.5.
2nd M1 for substituting both 2.6 and 2.7 into “their \(\mathrm{F}(X)\)” – 0.5 or “their \(\mathrm{F}(X)\)”
1st A1 awrt 0.48 and 0.52 if using “their \(\mathrm{F}(X)\)” and awrt \(-0.02\) and 0.02 or if using “their \(\mathrm{F}(X)\)” \(-\) 0.5
Other values possible. You may need to check their values for their correct equation
NB these last two marks are B1 B1 on ePEN but mark as M1 A1
2nd A1 for conclusion but only award if it follows from their numbers. Dependent on previous A mark being awarded
SC using calculators
M1 for sign of a suitable equation
M1 A1 for awrt 2.66 provided equation is correct
A1 correct comment