S2 June 2018 Q3
3. The length of time, \(T\), minutes, spent completing a particular task has probability density function
\[\mathrm{f}(t) = \begin{cases} \dfrac{1}{2}(t - 1) & 1 \lt t \leqslant 2 \\ \dfrac{1}{16}(14t - 3t^2 - 8) & 2 \lt t \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\]Given that \(\mathrm{E}(T^2) = \dfrac{267}{40}\)
Given that a person has already spent 1.5 minutes on the task,
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(T) = \displaystyle\int_1^2 \frac{1}{2}t(t - 1)\,\mathrm{d}t + \int_2^4 \frac{1}{16}t(14t - 3t^2 - 8)\,\mathrm{d}t\) | M1 |
| \(= \left[\dfrac{t^3}{6} - \dfrac{t^2}{4}\right]_1^2 + \left[\dfrac{14t^3}{48} - \dfrac{3t^4}{64} - \dfrac{8t^2}{32}\right]_2^4\) | A1 |
| \(= \dfrac{5}{12} + \dfrac{25}{12}\) | M1dep |
| \(= 2.5\) or \(\dfrac{5}{2}\) oe | A1 |
| (4) |
Notes
M1: Using \(\displaystyle\int t\mathrm{f}(t)\) for both parts, attempt to multiply out and an attempt at integration. \(x^n \to x^{n+1}\) Ignore limits.
A1: correct integration for both parts
M1dep : dep on previous method being awarded. For adding the 2 parts together and substituting the correct limits in to each part.
A1: 2.5 do not ISW. You will need to check that they have used Algebriac integration
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(T) = 6.675 - (2.5)^2\) | M1 |
| \(= \dfrac{17}{40}\) or 0.425 | A1 |
| (2) |
Notes
M1: \(\dfrac{267}{40}\) – [“their E(\(T\))”]2, NB must see \(-1^2\) if their E(\(T\)) = 1
A1: 0.425
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(t) = \begin{cases} 0 & t \leqslant 1 \\ \dfrac{1}{4}t^2 - \dfrac{1}{2}t + \dfrac{1}{4} \quad \text{or} \quad \dfrac{(t - 1)^2}{4} & 1 \lt t \leqslant 2 \\ \dfrac{1}{16}(7t^2 - t^3 - 8t) & 2 \lt t \leqslant 4 \\ 1 & t \gt 4 \end{cases}\) | M1 A1 M1 A1 B1 |
| (5) |
Notes
M1: \(\displaystyle\int_1^t \frac{1}{2}(x - 1)\,\mathrm{d}x\) with correct limits or \(\displaystyle\int \frac{1}{2}(x - 1)\,\mathrm{d}x\) and F(1) = 0
There must be an attempt to integrate for either method; \(x^n \to x^{n+1}\)
A1: 2nd line oe allow in terms of \(x\). Must be in the cdf
M1: \(\displaystyle\int_2^t \frac{1}{16}(14x - 3x^2 - 8)\,\mathrm{d}x +\) using "their F(2)" or \(\displaystyle\int \frac{1}{16}(14x - 3x^2 - 8)\) and using F(4) = 1
There must be an attempt to integrate for either method; \(x^n \to x^{n+1}\)
A1: 3rd line oe allow in terms of \(x\). Correct Method must be shown to award the A1 Must be in the cdf
B1: fully correct all in terms of \(t\) (allow < instead of \(\leqslant\) and vice versa ditto > and \(\geqslant\))
NB fully correct answer with no working can gain M1A1M0A0B1
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{4}t^2 - \dfrac{1}{2}t + \dfrac{1}{4} = 0.2\) | M1 |
| \(t^2 - 2t + 1 = 0.8\) \(t^2 - 2t + 0.2 = 0\) \(t = \dfrac{2 \pm \sqrt{2^2 - 4\times1\times0.2}}{2}\) o.e | M1 |
| \(t = 1.894\ldots\) awrt 1.89 | A1 |
| (3) |
Notes
M1: their cdf for \(1 \lt t \leqslant 2\) = 0.2 or \(\displaystyle\int_1^t \frac{1}{2}(t - 1)\,\mathrm{d}t = 0.2\) and attempt at integration \(x^n \to x^{n+1}\)
M1: Correct method for solving their 3 term quadratic equation ie correct use of formula or correct completion of the square
A1: awrt 1.89 allow \(\dfrac{5 + 2\sqrt{5}}{5}\) or \(1 + \dfrac{2}{\sqrt{5}}\) oe must be only one answer given.
| Scheme | Marks |
|---|---|
| \(1 - \mathrm{F}(1.5) = 1 - \left(\dfrac{1}{4}\times1.5^2 - \dfrac{1}{2}\times1.5 + \dfrac{1}{4}\right)\) | M1 |
| \(= \dfrac{15}{16}\) or 0.9375 awrt 0.938 | A1 |
| (2) |
Notes
M1: attempt at 1 – F(1.5) must subst 1.5 into their line for \(1 \lt t \leqslant 2\) or \(\displaystyle\int_1^{1.5} \frac{1}{2}(t - 1)\,\mathrm{d}t\) and attempt at integration \(x^n \to x^{n+1}\) oe
A1: 0.9375
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(T \gt 3) = 0.25\) | |
| \(\mathrm{P}(T \gt 3 \mid T \gt 1.5) = \dfrac{\text{"}0.25\text{"}}{\text{"}0.9375\text{"}}\) | M1 |
| \(= \dfrac{4}{15}\) or awrt 0.267 | A1 |
| (2) | |
| (18 marks) |
Notes
M1: 0.25/ “their (e)” or [1– “their F(3)”]/ “their (e)”
NB if they have written a value for P(\(T\) > 3) allow this as the numerator if 0 < P(\(T\) > 3) < 1
A1: 4/15 or awrt 0.267