S2 January 2009 Q7
7. A random variable \(X\) has probability density function given by
\[\mathrm{f}(x) = \begin{cases} -\frac{2}{9}x + \frac{8}{9} & 1 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\](a) Show that the cumulative distribution function \(\mathrm{F}(x)\) can be written in the form \(ax^2 + bx + c\), for \(1 \leqslant x \leqslant 4\) where \(a\), \(b\) and \(c\) are constants. (3)
(b) Define fully the cumulative distribution function \(\mathrm{F}(x)\). (2)
(c) Show that the upper quartile of \(X\) is 2.5 and find the lower quartile. (6)
Given that the median of \(X\) is 1.88
(d) describe the skewness of the distribution. Give a reason for your answer. (2)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x_0) = \displaystyle\int_1^x -\tfrac{2}{9}x + \tfrac{8}{9}\,\mathrm{dx} = \left[-\tfrac{1}{9}x^2 + \tfrac{8}{9}x\right]_1^x\) | M1A1 |
| \(= \left[-\tfrac{1}{9}x^2 + \tfrac{8}{9}x\right] - \left[-\tfrac{1}{9} + \tfrac{8}{9}\right]\) \(= -\tfrac{1}{9}x^2 + \tfrac{8}{9}x - \tfrac{7}{9}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ -\tfrac{1}{9}x^2 + \tfrac{8}{9}x - \tfrac{7}{9} & 1 \leqslant x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\) | B1B1ft |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = 0.75\) ; or \(\mathrm{F}(2.5) = -\tfrac{1}{9} \times 2.5^2 + \tfrac{8}{9} \times 2.5 - \tfrac{7}{9}\) | M1; |
| \(-\tfrac{1}{9}x^2 + \tfrac{8}{9}x - \tfrac{7}{9} = 0.75\) \(4x^2 - 32x + 55 = 0\) \(-x^2 + 8x - 13.75 = 0\) \(x = 2.5\) \(= 0.75\) cso | A1 |
| and \(\mathrm{F}(x) = 0.25\) \(-\tfrac{1}{9}x^2 + \tfrac{8}{9}x - \tfrac{7}{9} = 0.25\) | M1 |
| \(-x^2 + 8x - 7 = 2.25\) \(-x^2 + 8x - 9.25 = 0\) quadratic 3 terms =0 | M1 dep |
| \(x = \dfrac{-8 \pm \sqrt{8^2 - 4 \times -1 \times -9.25}}{2 \times -1}\) | M1 dep |
| \(x = 1.40\) | A1 |
| (6) |
Notes
(corrected from the printed mark scheme: the quadratic is printed as \(4x^2 - 32^x + 55 = 0\))
| Scheme | Marks |
|---|---|
| \(Q_3 - Q_2 \gt Q_2 - Q_1\) Or mode = 1 and mode < median Or mean = 2 and median < mean Sketch of pdf here or be referred to if in a different part of the question Box plot with \(Q_1\), \(Q_2\), \(Q_3\) values marked on | M1 |
| Positive skew | A1 |
| (2) | |
| (13 marks) |
Notes
(corrected from the printed mark scheme: printed “Or mean = 2 and median < mode”; here mode = 1 < median = 1.88 < mean = 2)