S2 June 2013 Q5
5. The continuous random variable \(X\) has a cumulative distribution function
\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\[1ex] \dfrac{x^3}{10} + \dfrac{3x^2}{10} + ax + b & 1 \leqslant x \leqslant 2 \\[1ex] 1 & x \gt 2 \end{cases}\]where \(a\) and \(b\) are constants.
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(1) = 0, \ \frac{4}{10} + a + b = 0\) | M1 |
| \(a = -\dfrac{3}{5}\) or \(b = \dfrac{1}{5}\) | A1 |
| \(\mathrm{F}(2) = 1, \ 2 + 2a + b = 1\) | M1 |
| Solving gives \(a = -\frac{3}{5}, b = \frac{1}{5}\) | A1 |
| (4) |
Notes
Alt
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(2) - \mathrm{F}(1) = 1, \ 2 + 2a + b - \frac{4}{10} - a - b = 1\) | M1 |
| \(a = -\frac{3}{5}\) | A1 |
| \(\mathrm{F}(2) = 1\) or \(\mathrm{F}(1) = 0\) \(2 - \frac{6}{5} + b = 1\) or \(\frac{4}{10} - \frac{3}{5} + b = 0\) | M1 |
| \(b = \frac{1}{5}\) | A1 |
| (4) |
1st M1 using \(\mathrm{F}(1) = 0\). Clear attempt to form a linear equation for \(a\) and \(b\)
1st A1 either \(a = -0.6\) or \(b = 0.2\) Previous M must be awarded
2nd M1 using \(\mathrm{F}(2) = 1\). Clear attempt to form a second linear equation for \(a\) and \(b\)
2nd A1 if 1st A1 awarded then both \(a\) and \(b\) must be correct otherwise award if either \(a = -0.6\) or \(b = 0.2\)
alt 1st M1 \(\mathrm{F}(2) - \mathrm{F}(1) = 1\). Leading to a value for \(a\): 1st A1 \(a = -0.6\)
2nd M1 using \(\mathrm{F}(2) = 1\) or \(\mathrm{F}(1) = 0\). Leading to a value for \(b\): 2nd A1 \(b = 0.2\)
NB correct values for \(a\) and \(b\) with no working scores no marks.
| Scheme | Marks |
|---|---|
| Differentiating cdf gives \(\mathrm{f}(x) = \frac{3}{10}x^2 + \frac{6}{10}x + a, \quad 1 \leqslant x \leqslant 2\) \(= \dfrac{3}{10}(x^2 + 2x - 2)\) | B1 cso |
| (1) |
Notes
B1 They must differentiate and then factorise. cso
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \displaystyle\int_1^2 \frac{3}{10}(x^3 + 2x^2 - 2x)\,\mathrm{d}x\) | M1 |
| \(= \frac{3}{10}\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - x^2\right]_1^2\) | M1d A1 |
| \(= \dfrac{13}{8}\) | A1 |
| (4) |
Notes
1st M1 for clear attempt to use \(x\mathrm{f}(x)\) with an intention of integrating (Integral sign enough) Ignore limits. Must substitute in \(\mathrm{f}(x)\) or “their \(\mathrm{f}(x)\)”.
2nd M1d dependent on previous M being awarded for some correct integration… at least one correct term with the correct coefficient.
1st A1 for fully correct (possibly unsimplified) integration. Ignore limits
2nd A1 Accept 1.63 and 1.625 or some other exact equivalent
| Scheme | Marks |
|---|---|
| F(1.425) = 0.24355, F(1.435) = 0.25227 | M1A1 |
| 0.25 lies between F(1.425)and F(1.435) hence result. | A1 |
| (3) | |
| (12 marks) |
Notes
M1 expression showing substitution of 1.425 or 1.435 into \(\mathrm{F}(x)\) [or into \(\mathrm{F}(x) - 0.25\)] [or putting their \(\mathrm{F}(x) = 0.25\) and attempting to solve leading to \(x = \ldots\)] May be implied by either pair of the correct answers as given below for the 1st A1
1st A1 awrt 0.244 and awrt 0.252 [or awrt -0.00645 and awrt 0.00227] [or \(x =\) awrt 1.432]
2nd A1 0.25 lies between F(1.425)and F(1.435) [or change in sign therefore root between] [or “1.432” lies between 1.425 and 1.435 therefore root between]. Statement must be true for their method