S2 January 2013 Q7
7. The continuous random variable \(X\) has the following probability density function
\[\mathrm{f}(x) = \begin{cases} a + bx & 0 \leqslant x \leqslant 5 \\ 0 & \text{otherwise} \end{cases}\]where \(a\) and \(b\) are constants.
Given that \(\mathrm{E}(X) = \dfrac{35}{12}\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^5 a + bx\,\mathrm{d}x = 1\) | M1 |
| \(\left[ax + \dfrac{bx^2}{2}\right]_0^5 = 1\) | A1 |
| \(5a + \dfrac{25b}{2} = 1\) | M1dep |
| \(10a + 25b = 2\) | A1cso |
| (4) |
Notes
1st M1 Attempting to integrate with correct limits or for an attempt to find area \(0.5(a + b)h\) or Attempting to integrate and using F(5) = 1
1st A1 Correct integration or correct area
2nd M1 for using =1. This is dependent on the first M1 being awarded.
2nd A1 cso condone missing d\(x\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^5 ax + bx^2\,\mathrm{d}x = \frac{35}{12}\) | M1 |
| \(\left[\dfrac{ax^2}{2} + \dfrac{bx^3}{3}\right]_0^5 = \dfrac{35}{12}\) | A1 |
| \(\dfrac{25a}{2} + \dfrac{125b}{3} = \dfrac{35}{12}\) \(30a + 100b = 7\) | A1 |
| (3) |
Notes
M1 using or writing (limits not needed) \(\displaystyle\int_0^5 ax + bx^2\,\mathrm{d}x = \frac{35}{12}\)
1st A1 correct integration
2nd A1 may be awarded for an unsimplified version \(\dfrac{25a}{2} + \dfrac{125b}{3} = \dfrac{35}{12}\)
| Scheme | Marks |
|---|---|
| \(30a + 100b = 7\) \(10a + 25b = 2\) | M1 |
| \(a = 0.1 \quad b = 0.04\) | A1,A1 |
| (3) |
Notes
M1 attempting to solve “their equations” simultaneously – either using rearranging and substitution or making one of the coefficients the ‘same’ (ignore sign) and either adding or subtracting. May be implied by correct values for \(a\) and \(b\)
1st A1 for 0.1
2nd A1 for 0.04
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^m 0.1 + 0.04x\,\mathrm{d}x = 0.5\) | M1 |
| \(\left[0.1x + \dfrac{0.04x^2}{2}\right]_0^m = 0.5\) | A1ft |
| \(0.1m + 0.02m^2 - 0.5 = 0\) | |
| \(m = \dfrac{-0.1 \pm \sqrt{0.1^2 + 4 \times 0.02 \times 0.5}}{2 \times 0.02}\) | |
| \(m = 3.09, -8.09\) therefore 3.09 | A1 |
| (3) |
Notes
M1 writing or using \(\displaystyle\int_0^m\) “their \(a\)” + “their \(b\)”\(x\,\mathrm{d}x = 0.5\): limits not needed
1st A1 correct integration for their “\(a\)” and “\(b\)”
NB the correct equation simplifies to \(m^2 + m - 25 = 0\)
A1 3.09 only. If they have both roots then they must select 3.09
| Scheme | Marks |
|---|---|
| mean < median (< mode) | B1ft |
| negatively skewed | B1 dep ft |
| (2) | |
| (15 marks) |
Notes
1st B1ft They must compare their values for mean and median correctly. They only need to compare 2 of mean, median and mode. If they compare either the median or mean with the mode only then the value of the mode must be stated. They may draw a sketch that matches their values of ‘\(a\)’ and ‘\(b\)’ for \(0 \leqslant x \leqslant 5\). It must not go below the \(x\)-axis This may be seen in part (a).
2nd B1 dependent f.t. on the previous B being awarded.