A2 June 2025 Q5
5. The random variable \(X \sim \mathrm{U}[1, 4]\)
A random sample of 10 observations of \(X\) is taken.
The random variable \(M\) represents the maximum value of these 10 observations.
The cumulative distribution function of \(M\), \(\mathrm{F}(y)\), is given by
\[\mathrm{F}(y) = \begin{cases} 0 & y \lt 1 \\ \left(\dfrac{y-1}{3}\right)^{10} & 1 \leqslant y \leqslant 4 \\ 1 & y \gt 4 \end{cases}\]| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{P}(1.8 \lt X \lt 3.2) = \left[\dfrac{3.2 - 1.8}{4 - 1}\right] = \dfrac{7}{15}\) oe | B1 | 3.4 |
| (ii) \(\mathrm{P}(X \gt 3.2 \mid X \gt 1.8) = \dfrac{\mathrm{P}(X \gt 3.2)}{\mathrm{P}(X \gt 1.8)}\) or \(\dfrac{\frac{4}{15}}{\frac{11}{15}}\) or \(\dfrac{0.8}{2.2}\) oe | M1 | 2.1 |
| \(= \dfrac{4}{11}\) | A1 | 1.1b |
| (3) |
Notes
(i) B1: for \(\frac{7}{15}\) or exact equivalent isw
(ii) M1: for a correct ratio of prob expressions (must be \(\mathrm{P}(X \gt 3.2)\) on num) or values or awrt 0.364
A1: for \(\frac{4}{11}\) or exact equivalent isw
| Scheme | Marks | AO |
|---|---|---|
| \([1 - \mathrm{F}(3.75)] = 1 - \left(\dfrac{3.75 - 1}{3}\right)^{10} = 0.581096\) awrt 0.581 | B1 | 3.4 |
| (1) |
Notes
B1: for awrt 0.581
| Scheme | Marks | AO |
|---|---|---|
(i) \(\mathrm{f}(y) = \dfrac{\mathrm{dF}(y)}{\mathrm{d}y} = \dfrac{10}{3} \times \left(\dfrac{y-1}{3}\right)^9\) or sketch of correct shape | M1 | 1.1b |
| Correct sketch showing \(y = 1\) and 4 and \(\mathrm{f}(1) = 0\) | A1 | 1.1b |
| (ii) [From sketch mode of \(M\) is] 4 | B1 | 2.2a |
| (3) |
Notes
(i) M1: for correct expression or a sketch of correct shape with positive increasing gradient
A1: for a fully correct sketch with 1 and 4 correctly indicated (dashed line not needed).
Condone curves which appear almost linear provided this was not the intention.
Ignore any labelling of the axes or values indicated on the vertical axis.
(ii) B1: for 4
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(M) = k\displaystyle\int y \times 10(y-1)^9\,\mathrm{d}y = k\int y\,\mathrm{d}(y-1)^{10}\) or \(K\displaystyle\int (3u+1)u^9\,\mathrm{d}u\) | M1 | 2.1 |
| \(= k\Big[y(y-1)^{10}\Big]_1^4 - k\displaystyle\int_1^4 (y-1)^{10}\,\mathrm{d}y\) or \(K\left\{\displaystyle\int_0^1 (3u^{10} + u^9)\,\mathrm{d}u = \left[\dfrac{3u^{11}}{11} + \ldots\right]\right\}\) | M1 | 1.1b |
| \(= \left[y\left(\dfrac{y-1}{3}\right)^{10}\right]_1^4 - \dfrac{3}{11}\left[\left(\dfrac{y-1}{3}\right)^{11}\right]_1^4\) oe or \(10\left[\dfrac{3u^{11}}{11} + \dfrac{u^{10}}{10}\right]_0^1\) oe | M1 | 1.1b |
| \(= \left[4 - \dfrac{3}{11}\right] = \dfrac{41}{11}\) | A1 | 1.1b |
| (4) | ||
| (11 marks) |
Notes
1st M1: for a correct expression for \(\mathrm{E}(M)\) and an attempt to start to integrate. May be implied by further work. May use substitution e.g. \(3u = y - 1\) so forms the integral expression in \(u\).
Allow any constant \(k\) (or \(K\))
2nd M1: for a correct first step of integration. Allow any \(k\) (or \(K\)) and still ignore limits.
3rd M1: for a correctly integrated expression including limits (need not be substituted in)
A1: dep on all previous method marks for \(\frac{41}{11}\) or exact equivalent