S2 June 2018 Q6
6. The continuous random variable \(X\) has the following cumulative distribution function
\[\mathrm{F}(x) = \begin{cases} 0 & x \leqslant 1 \\ \dfrac{4}{15}(x - 1) & 1 \lt x \leqslant 2 \\ k\left(\dfrac{ax^3}{3} - \dfrac{x^4}{4}\right) + b & 2 \lt x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\]where \(k\), \(a\) and \(b\) are constants.
Given that the mode of \(X\) is \(\dfrac{8}{3}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}^{\prime}(x) = k\left(ax^2 - x^3\right)\) oe | M1 |
| \(\mathrm{F}^{\prime\prime}(x) = k\left(2xa - 3x^2\right)\) oe | M1 |
| \(2xka - 3kx^2 = 0\) \(kx(2a - 3x) = 0\) | |
| \(a = \dfrac{3}{2}\times\dfrac{8}{3}\) or \(2\times4 - 3\times\dfrac{8}{3} = 0\) | M1d |
| \(a = 4*\) | A1cso* |
| (4) |
Notes
M1 attempting to find \(\mathrm{F}^{\prime}(x)\), \(x^n \to x^{n-1}\) condone missing \(k\). Implied by correct \(\mathrm{F}^{\prime\prime}(x)\)
M1 attempting to find \(\mathrm{F}^{\prime\prime}(x)\), \(x^n \to x^{n-1}\) condone missing \(k\)
M1d dependent on the 2nd M being awarded. Putting "their \(2a - 3x\)" = 0 and substituting \(x = 8/3\)
A1* cso fully correct solution with no errors. Must differentiatial including the \(k\). Make sure there is no incorrect notation
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(2) = \dfrac{4}{15} \Rightarrow k\left(\dfrac{32}{3} - 4\right) + b = \dfrac{4}{15}\) or \(\dfrac{20}{3}k + b = \dfrac{4}{15}\) oe | M1 |
| \(\mathrm{F}(4) = 1 \Rightarrow k\left(\dfrac{256}{3} - 64\right) + b = 1\) or \(\dfrac{64}{3}k + b = 1\) oe | M1 |
| \(\dfrac{44}{3}k = \dfrac{11}{15}\) | M1dd |
| \(k = \dfrac{1}{20}\) or \(b = -\dfrac{1}{15}\) | A1 |
| \(\mathrm{F}(2.5) = \text{"their } k\text{"}\left(\dfrac{4}{3}\times2.5^3 - \dfrac{2.5^4}{4}\right) + (\text{"their } b\text{"})\) | M1 |
| \(= \dfrac{623}{1280}\) or \(0.4867\ldots\) awrt 0.487 | A1cso |
| (6) | |
| (10 marks) |
Notes
M1 Form the correct equation in terms of the two unknowns \(k\) and \(b\) using F(2) = 4/15
M1 Form the correct equation in terms of the two unknowns \(k\) and \(b\) using F(4) = 1
M1dd dependent on first two method marks being awarded. Solving the two equations simultaneously by eliminating either \(k\) or \(b\)
A1 one of \(k\) or \(b\) correct.
Alternative to find \(k\)
| Scheme | Marks |
|---|---|
| \(f(x) = \dfrac{4}{15} \quad 1 \lt x \leqslant 2\) \(f(x) = k(4x^2 - x^3) \quad 2 \lt x \leqslant 4\) | (M1) |
| \(\displaystyle\int_1^2 \frac{4}{15}\,\mathrm{d}x + \int_2^4 \text{"}k(4x^2 - x^3)\text{"}\,\mathrm{d}x = 1\) | (M1) |
| \(\left[\dfrac{4}{15}x\right]_1^2 + k\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_2^4 = 1\) or \(k\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_2^4 = \dfrac{11}{15}\) | (M1dd) |
| \(\left[\dfrac{8}{15} - \dfrac{4}{15}\right] + k\left[\dfrac{4\times4^3}{3} - \dfrac{4^4}{4}\right] - k\left[\dfrac{4\times2^3}{3} - \dfrac{2^4}{4}\right] = 1\) | |
| \(k = \dfrac{1}{20}\) | (A1) |
M1 for 4/15 and attempt at differentiating third line \(x^n \to x^{n-1}\) and must have \(k\).
M1 for using their pdf equations with correct limits, adding and setting equal to 1
NB these first two marks can be implied by \(\left[\dfrac{4}{15}x\right]_1^2 + k\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_2^4 = 1\) or \(k\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_2^4 = \dfrac{11}{15}\)
dd M1 dependent of previous method marks being awarded. Correct integration and attempt to substitute limits
A1 \(k\) correct
M1 correct method for finding F(2.5) using their values for \(k\) and \(b\) or allow with the letters \(a\)(or 4), \(k\) and \(b\). May be implied by a correct answer otherwise working must be shown.
A1cso all previous method marks must be awarded \(\dfrac{623}{1280}\) or awrt 0.487
Alternative to find F(2.5)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_1^2 \frac{4}{15}\,\mathrm{d}x + \int_2^{2.5} \text{"}\frac{1}{20}(4x^2 - x^3)\text{"}\,\mathrm{d}x = \left[\frac{4}{15}x\right]_1^2 + k\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_2^{2.5}\) | (M1) |
| \(= \dfrac{623}{1280}\) or \(0.4867\ldots\) | (A1 cso) |
M1 correct method for finding F(2.5) using their value for \(k\) or allow with the letters \(a\)(or 4) and \(k\). May be implied by a correct answer otherwise working must be shown.
A1cso all previous method marks must be awarded \(\dfrac{623}{1280}\) or awrt 0.487