S2 June 2017 Q6
6. The continuous random variable \(X\) has a probability density function
\[\mathrm{f}(x) = \begin{cases} k(x - 2) & 2 \leqslant x \leqslant 3 \\ k & 3 \lt x \lt 5 \\ k(6 - x) & 5 \leqslant x \leqslant 6 \\ 0 & \text{otherwise} \end{cases}\]where \(k\) is a positive constant.
| Scheme | Marks |
|---|---|
 | B1 B1 |
| (2) |
Notes
B1 correct shape with the end points on the \(x\)-axis
B1 correct shape with \(k\), 2,3,5,6 marked on in the correct places. Allow \(^1/_3\) for \(k\)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}\times k + 2\times k + \dfrac{1}{2}\times k = 1\) | M1 |
| \(3k = 1\) \(k = \dfrac{1}{3}*\) | A1 cso |
| (2) |
Notes
M1 An attempt to find area using any correct method and putting equal to 1
A1 cso. AG Method must be shown and there must be no incorrect working. Need to have these 3 lines as a minimum.
Alternative
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_2^3 k(x - 2)\,\mathrm{d}x + \int_3^5 k\,\mathrm{d}x + \int_5^6 k(6 - x)\,\mathrm{d}x = 1\) \(\left[\dfrac{kx^2}{2} - 2kx\right]_2^3 + [kx]_3^5 + k\left[6x - \dfrac{x^2}{2}\right]_5^6 = 1\) | M1 |
| \(\left(-\dfrac{3}{2}k + 2k\right) + (5k - 3k) + \left(18k - \dfrac{35}{2}k\right) = 1\) | |
| \(3k = 1\) \(k = \dfrac{1}{3}\) | A1 cso |
M1 Correct integration to find the whole area, put = 1 and an attempt to integrate, ignore limits for attempt \(x^n \to x^{n+1}\)
A1 cso Method must be shown – at least one step between integration and \(k\) = 1/3 and there must be no incorrect working.
SC For using verification they could get M1 A0 if there are no errors
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 2 \\ \dfrac{x^2}{6} - \dfrac{2x}{3} + \dfrac{2}{3} & 2 \leqslant x \leqslant 3 \\ \dfrac{x}{3} - \dfrac{5}{6} & 3 \lt x \lt 5 \\ 2x - \dfrac{x^2}{6} - 5 & 5 \leqslant x \leqslant 6 \\ 1 & x \gt 6 \end{cases}\) Alternative \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 2 \\ \dfrac{1}{6}(x - 2)^2 & 2 \leqslant x \leqslant 3 \\ \dfrac{x}{3} - \dfrac{5}{6} & 3 \lt x \lt 5 \\ 1 - \dfrac{1}{6}(6 - x)^2 & 5 \leqslant x \leqslant 6 \\ 1 & x \gt 6 \end{cases}\) | M1A1 M1A1 M1A1 B1 |
| (7) |
Notes
1st M1 For \(2 \leqslant x \leqslant 3\), \(\displaystyle\int_2^x \frac{1}{3}(t - 2)\,\mathrm{d}t = \left[\frac{t^2}{6} - \frac{2t}{3}\right]_2^x\) and attempt to subst 2 and \(x\)
Or \(\mathrm{F}(x) = \dfrac{x^2}{6} - \dfrac{2x}{3} + C\) and using F(2) = 0
1st A1 for the second row in the above F(\(x\)) oe. Condone < instead of \(\leqslant\) and vice versa
2nd M1 For \(3 \lt x \lt 5\), \(\displaystyle\int_3^x \frac{1}{3}\,\mathrm{d}t + \text{"}\tfrac{1}{6}\text{"} = \left[\frac{t}{3}\right]_3^x + \text{"}\tfrac{1}{6}\text{"}\) and attempt to subst 3 and \(x\). Allow F(3) instead of “\(\frac{1}{6}\)”
or \(\mathrm{F}(x) = \dfrac{x}{3} + C\) and using \(\mathrm{F}(3) = \dfrac{1}{6}\) or \(\mathrm{F}(5) = \dfrac{5}{6}\)
2nd A1 for the third row in the above F(\(x\)) oe. Condone \(\leqslant\) instead of < and vice versa
3rd M1 For \(5 \leqslant x \leqslant 6\), \(\displaystyle\int_5^x 2 - \frac{t}{3}\,\mathrm{d}t + \text{"}\tfrac{5}{6}\text{"} = \left[2t - \frac{t^2}{6}\right]_5^x + \text{"}\tfrac{5}{6}\text{"}\) and subst 5 and \(x\). Allow F(5) instead of “\(\frac{5}{6}\)”
or \(\mathrm{F}(x) = 2x - \dfrac{x^2}{6} + C\) and using F(6) = 1
3rd A1 for the fourth row in the above F(\(x\)) oe. Condone < instead of \(\leqslant\) and vice versa
B1 For both Top line of F(\(x\)) ie 0 \(x \lt 2\) and Bottom line of F(\(x\)) ie 1 \(x \gt 6\)
Condone \(\leqslant\) instead of < and vice versa. Allow one of the lines to have otherwise as its range
| Scheme | Marks |
|---|---|
| \(2x - \dfrac{x^2}{6} - 5 = 0.9\) | M1 |
| \(\dfrac{x^2}{6} - 2x + 5.9 = 0\) \(x = \dfrac{2 \pm \sqrt{4 - 4\times\frac{1}{6}\times5.9}}{\frac{1}{3}}\) | M1 |
| \(x\) = awrt 5.23 | A1 |
| (3) |
Notes
1st M1 using their cdf for \(5 \leqslant x \leqslant 6 = 0.9\)
2nd M1 using either the quadratic formula or completing the square or factorising or any correct method to solve their 3 term quadratic which must have been correctly rearranged. If they write the formula down then allow a slip. If no formula written down then it must be correct for their equation. May be implied by awrt 5.23 or 6.77
A1 awrt 5.23 – (allow \(\frac{30 - \sqrt{15}}{5}\)). If they have 6.77… this must be eliminated
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 4\) \(\mathrm{F}(5.5) - \mathrm{F}(4) = \dfrac{11}{24}\) | M1 A1 |
| (2) | |
| (16 marks) |
Notes
M1 for writing or attempting to find F(5.5) – F(4) or \(\mathrm{P}(X \leqslant 5.5) - \mathrm{P}(x \leqslant 4)\) or \(\mathrm{P}(X \lt 5.5) - \mathrm{P}(x \lt 4)\) or F(5.5) – 0.5 or \(\displaystyle\int_4^5 k\,\mathrm{d}x + \int_5^{5.5} k(6 - x)\,\mathrm{d}x\) with correct limits and \(x^n \to x^{n+1}\). May be implied by a correct answer.
A1 \(\dfrac{11}{24}\) oe or awrt 0.458