AS June 2023 Q2
2. A continuous random variable \(X\) has probability density function
\[\mathrm{f}(x) = \begin{cases} \dfrac{x}{16}(9 - x^2) & 1 \leqslant x \leqslant 3 \\ 0 & \text{otherwise} \end{cases}\]| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int_{[1]}^{[t]} \frac{9x}{16} - \frac{x^3}{16}\,\mathrm{d}x = \left[\frac{9x^2}{32} - \frac{x^4}{64}\right]_{[1]}^{[t]}\) or \(\dfrac{9x^2}{32} - \dfrac{x^4}{64}\,[+c]\) | M1 | 1.1b |
| \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ \dfrac{9x^2}{32} - \dfrac{x^4}{64} - \dfrac{17}{64} & 1 \leqslant x \leqslant 3 \\ 1 & x \gt 3 \end{cases}\) | B1 A1 | 1.1b 1.1b |
| (3) |
Notes
M1: for correct method to integrate \(\mathrm{f}(x)\). e.g. \(\dfrac{(9 - x^2)^2}{-64}\) or \(1 - \dfrac{(9 - x^2)^2}{64}\) Allow one sign error.
B1: 1st and 3rd line correct need \(x \lt 1\) and \(x \gt 3\)
A1: for correct middle line including \(1 \leqslant x \leqslant 3\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(X \gt 1.8) = 1 - \mathrm{F}(1.8)\) oe | M1 | 1.1b |
| \(= 0.5184\) | A1 | 1.1b |
| (2) |
Notes
M1: for a correct method to find prob. Must have 1 – … and 1.8 substituted into their \(\mathrm{F}(x)\)
A1: awrt 0.518 or \(\frac{324}{625}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(X^{-1}) = \displaystyle\int_1^3 x^{-1}\left(\frac{9}{16}x - \frac{1}{16}x^3\right)\mathrm{d}x\) or \(\displaystyle\int_1^3 \frac{9}{16} - \frac{1}{16}x^2\,\mathrm{d}x\) | M1 | 1.1b |
| \(\mathrm{E}(3X^{-1} + 2) = 3 \times \left[\dfrac{9}{16}x - \dfrac{1}{48}x^3\right]_1^3 + 2\) or \(3 \times \dfrac{7}{12} + 2\) | dM1 | 1.1b |
| \(= 3.75\) | A1 | 1.1b |
| (3) |
Notes
M1: for a correct expression to find \(\mathrm{E}(X^{-1})\) or \(\mathrm{E}(3X^{-1} + 2)\)
dM1 for an attempt to integrate their correct expression (at least 1 term correct)
A1: 3.75 oe [NB \(\mathrm{E}(X) = 1.85\) leading to \(\mathrm{E}(3X^{-1} + 2) = \frac{134}{37} = 3.62\ldots\) scores M0M0A0]
Alternative
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(3X^{-1} + 2) = \displaystyle\int_1^3 \left(\frac{9}{16}x - \frac{1}{16}x^3\right)\left(3x^{-1} + 2\right)\mathrm{d}x\) | M1 |
| \(= \left[\dfrac{27}{16}x - \dfrac{1}{16}x^3 + \dfrac{9}{16}x^2 - \dfrac{1}{32}x^4\right]_1^3\) | M1 |
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}\mathrm{f}(x)}{\mathrm{d}x} = \dfrac{9}{16} - \dfrac{3}{16}x^2\) | M1 | 1.1b |
| \(\dfrac{9}{16} - \dfrac{3}{16}x^2 = 0\) | dM1 | 1.1b |
| \(\dfrac{3}{16}x^2 = \dfrac{9}{16} \Rightarrow x = \sqrt{3}\)* and either a sketch or full statement | A1cso* | 2.1 |
| (3) | ||
| (11 marks) |
Notes
M1: for differentiating \(\mathrm{f}(x)\) with at least one term correct
dM1 for setting up an equation to find the mode. Allow subst of \(\sqrt{3}\)
A1 cso* \(x = \sqrt{3}\) and either a sketch through (0, 0), (3, 0) and showing max between them or statement to show mode not on the boundary \(\mathrm{f}(\sqrt{3}) = 0.6495\ldots \gt 0.5\) \(\mathrm{f}(1) = 0.5\) and \(\mathrm{f}(3) = 0\)