S2 June 2014 (R) Q6
6. In an experiment some children were asked to estimate the position of the centre of a circle. The random variable \(D\) represents the distance, in centimetres, between the child’s estimate and the actual position of the centre of the circle. The cumulative distribution function of \(D\) is given by
\[\mathrm{F}(d) = \begin{cases} 0 & d \lt 0 \\ \dfrac{d^2}{2} - \dfrac{d^4}{16} & 0 \leqslant d \leqslant 2 \\ 1 & d \gt 2 \end{cases}\]Justify your answer. (5)
The experiment is conducted on 80 children.
| Scheme | Marks |
|---|---|
| \(\dfrac{d^2}{2} - \dfrac{d^4}{16} = \dfrac{1}{2}\) | M1 |
| \(\left[d^4 - 8d^2 + 8 = 0 \Rightarrow\right] 8 = \left(d^2 - 4\right)^2\) or \(d^2 = \dfrac{8 \pm \sqrt{64 - 32}}{2}\) | M1 |
| \(d^2 = 4 - \sqrt{8}\) | M1d |
| \(d = \sqrt{4 - \sqrt{8}} = 1.08239\ldots\) awrt 1.08 | A1 |
| (4) |
Notes
1st M1 for forming this equation based on \(\mathrm{F}(d) = 0.5\) oe
2nd M1 for attempting to solve (complete the square or use formula) – must be correct for their equation
d3rd M1 for square rooting to get \(d = \ldots\). Do not award for \(d\) = awrt 1.17. Dependent on previous M being awarded.
A1 for awrt 1.08 Must reject any negative answers
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(d) = d - \dfrac{d^3}{4}\) | M1 |
| \(\left[\mathrm{f}^{\prime}(d) = 0 \Rightarrow\right] \quad 1 - \dfrac{3d^2}{4} = 0\) | M1A1 |
| \(\left[d^2 = \dfrac{4}{3} \text{ so}\right] d = 1.154\ldots\) | A1 |
| \(\mathrm{f}^{\prime\prime}(d) = -\dfrac{6d}{4} \lt 0\) so max | B1 |
| (5) |
Notes
1st M1 for attempting to find \(\mathrm{f}(d)\). Some correct differentiation. \(x^n \to x^{n-1}\)
2nd M1 for attempting \(\mathrm{f}^{\prime}(d)\) and setting it = 0 Some correct differentiation \(x^n\) to \(x^{n+1}\)
1st A1 for a correct equation for \(d\)
2nd A1 for awrt 1.15 or 1.155 or \(\sqrt{\dfrac{4}{3}}\) or \(\dfrac{2\sqrt{3}}{3}\) or \(\dfrac{2}{\sqrt{3}}\) oe
B1 for a method confirming that their value gives a max not a min
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(D \lt 1) = \left[\dfrac{1}{2} - \dfrac{1}{16}\right] = \dfrac{7}{16}\) | B1 |
| Number of children \(= 80 \times \dfrac{7}{16},\ = 35\) | M1, A1 |
| (3) | |
| (12 marks) |
Notes
M1 for \(80 \times p\), \(0 \lt p \lt 1\)
A1 for 35 only