S2 June 2014 Q6
6. The continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\) given by
\[\mathrm{f}(x) = \begin{cases} \dfrac{2x}{9} & 0 \leqslant x \leqslant 1 \\[1ex] \dfrac{2}{9} & 1 \lt x \lt 4 \\[1ex] \dfrac{2}{3} - \dfrac{x}{9} & 4 \leqslant x \leqslant 6 \\[1ex] 0 & \text{otherwise} \end{cases}\]| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \displaystyle\int_0^1 \frac{2x^2}{9}\,\mathrm{d}x + \int_1^4 \frac{2x}{9}\,\mathrm{d}x + \int_4^6 \frac{2x}{3} - \frac{x^2}{9}\,\mathrm{d}x\) | M1 |
| \(= \left[\dfrac{2x^3}{27}\right]_0^1 + \left[\dfrac{2x^2}{18}\right]_1^4 + \left[\dfrac{x^2}{3} - \dfrac{x^3}{27}\right]_4^6\) | A1 |
| \(= \left[\dfrac{2}{27}\right] + \left[\dfrac{32}{18} - \dfrac{2}{18}\right] + \left[4 - \dfrac{80}{27}\right]\) | M1d |
| \(= 2\frac{7}{9}\) or awrt 2.78 | A1 |
| (4) |
Notes
M1 using \(\displaystyle\int x\mathrm{f}(x)\,\mathrm{d}x\) ignore limits. Must have at least one \(x^n \to x^{n+1}\)
They must add the 3 parts together. Do not allow division by 3.
A1 all integration correct; ignore limits
M1 dependent on previous M being awarded. Subst in correct limits – no need to see zero substituted.
A1 \(2\frac{7}{9}\) oe or awrt 2.78
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\ \dfrac{x^2}{9} & 0 \leqslant x \leqslant 1 \\[1ex] \dfrac{2x}{9} - \dfrac{1}{9} & 1 \lt x \lt 4 \\[1ex] \dfrac{2x}{3} - \dfrac{x^2}{18} - 1 & 4 \leqslant x \leqslant 6 \\ 1 & x \gt 6 \end{cases}\) | B1 M1A1 M1 A1 B1 |
| 1st M1 For \(1 \lt x \lt 4\), \(\mathrm{F}(x) = \displaystyle\int_1^x \frac{2}{9}\,\mathrm{d}x + \frac{1}{9}\) | |
| 2nd M1 For \(4 \leqslant x \leqslant 6\), \(\mathrm{F}(x) = \displaystyle\int_4^x \frac{2}{3} - \frac{x}{9}\,\mathrm{d}x + \frac{7}{9}\) or use +C and F(6) =1 | |
| (6) |
Notes
B1 for 2nd line- allow use of \(\lt\) instead of \(\leqslant\)
M1 For \(1 \lt x \lt 4\), \(\mathrm{F}(x) = \displaystyle\int_1^x \frac{2}{9}\,\mathrm{d}x + \frac{1}{9}\). Limits are needed. or use \(\mathrm{F}(x) = \displaystyle\int_1^x \frac{2}{9}\,\mathrm{d}x + \text{their F}(1)\) need limits or use “their \(\mathrm{F}(1)\)” \(= \displaystyle\int \frac{2}{9}\,\mathrm{d}x + C\) and subst \(x = 1\) into RHS or use “their \(\mathrm{F}(4)\)” \(= \displaystyle\int \frac{2}{9}\,\mathrm{d}x + C\) and subst \(x = 4\) into RHS
A1 for 3rd line allow use of \(\leqslant\) instead of \(\lt\)
M1 For \(4 \leqslant x \leqslant 6\), \(\mathrm{F}(x) = \displaystyle\int_4^x \frac{2}{3} - \frac{x}{9}\,\mathrm{d}x + \frac{7}{9}\). Limits are needed. or use \(\mathrm{F}(x) = \displaystyle\int_4^x \frac{2}{3} - \frac{x}{9}\,\mathrm{d}x + \text{their F}(4)\). Limits are needed. or use “their \(\mathrm{F}(4)\)” \(= \displaystyle\int \frac{2}{3} - \frac{x}{9}\,\mathrm{d}x + C\) and subst \(x = 4\) into RHS or use \(1 = \displaystyle\int \frac{2}{3} - \frac{x}{9}\,\mathrm{d}x + C\) and subst \(x = 6\) into RHS
A1 for 4th line allow use of \(\lt\) instead of \(\leqslant\)
B1 for first and last line - allow use of \(\leqslant\) instead of \(\lt\) and \(\geqslant\) instead of \(\gt\) and “otherwise” for one of \(x \lt 0\) and \(x \gt 6\)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(x) = 0.5\) | M1 |
| \(\dfrac{2m}{9} - \dfrac{1}{9} = 0.5\) | A1ft |
| \(m = 2.75\) | A1 |
| (3) |
Notes
M1 putting any one of their lines = 0.5
A1 their 3rd line = 0.5
A1 2.75
| Scheme | Marks |
|---|---|
| Median < mean therefore positive skew Or Mean \(\approx\) median therefore no skewness | M1A1cao |
| (2) | |
| (15 marks) |
Notes
M1 reason must match their values / a correctly shaped and labelled sketch. Must compare the median and mean, ignore references to mode
A1 no ft Correct answer only from correct values of the mean and median or a correct and fully labelled sketch.