| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF to PDF derivation |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard CDF operations: finding median by solving F(d)=0.5, finding mode by differentiating to get PDF then setting f'(d)=0, and using F(1) for probability calculation. All techniques are routine for this specification, though the algebra with the quartic term requires care. Slightly easier than average due to being purely procedural with no conceptual challenges. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2}{2} - \frac{d^4}{16} = \frac{1}{2}\) | M1 | For forming equation based on \(F(d) = 0.5\) |
| \(\left[d^4 - 8d^2 + 8 = 0 \Rightarrow\right] 8 = (d^2-4)^2\) or \(d^2 = \frac{8 \pm \sqrt{64-32}}{2}\) | M1 | For attempting to solve (complete the square or use formula) — must be correct for their equation |
| \(d^2 = 4 - \sqrt{8}\) | M1d | For square rooting to get \(d=\ldots\) Do not award for \(d =\) awrt 1.17. Dependent on previous M |
| \(d = \sqrt{4-\sqrt{8}} = 1.08239\ldots\) awrt 1.08 | A1 (4) | Must reject any negative answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(d) = d - \frac{d^3}{4}\) | M1 | For attempting to find \(f(d)\). Some correct differentiation \(x^n \to x^{n-1}\) |
| \(\left[f'(d)=0 \Rightarrow\right] 1 - \frac{3d^2}{4} = 0\) | M1A1 | 2nd M1 for attempting \(f'(d)\) and setting \(=0\). A1 for correct equation for \(d\) |
| \(\left[d^2 = \frac{4}{3}\right]\) so \(d = 1.154\ldots\) | A1 | For awrt 1.15 or 1.155 or \(\sqrt{\frac{4}{3}}\) or \(\frac{2\sqrt{3}}{3}\) or \(\frac{2}{\sqrt{3}}\) |
| \(f''(d) = -\frac{6d}{4} < 0\) so max | B1 (5) | For method confirming maximum not minimum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(D<1) = \left[\frac{1}{2} - \frac{1}{16}\right] = \frac{7}{16}\) | B1 | |
| Number of children \(= 80 \times \frac{7}{16} = 35\) | M1, A1 (3) | M1 for \(80 \times p\), \(0 |
# Question 6:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2}{2} - \frac{d^4}{16} = \frac{1}{2}$ | M1 | For forming equation based on $F(d) = 0.5$ |
| $\left[d^4 - 8d^2 + 8 = 0 \Rightarrow\right] 8 = (d^2-4)^2$ or $d^2 = \frac{8 \pm \sqrt{64-32}}{2}$ | M1 | For attempting to solve (complete the square or use formula) — must be correct for their equation |
| $d^2 = 4 - \sqrt{8}$ | M1d | For square rooting to get $d=\ldots$ Do not award for $d =$ awrt 1.17. Dependent on previous M |
| $d = \sqrt{4-\sqrt{8}} = 1.08239\ldots$ awrt 1.08 | A1 (4) | Must reject any negative answers |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(d) = d - \frac{d^3}{4}$ | M1 | For attempting to find $f(d)$. Some correct differentiation $x^n \to x^{n-1}$ |
| $\left[f'(d)=0 \Rightarrow\right] 1 - \frac{3d^2}{4} = 0$ | M1A1 | 2nd M1 for attempting $f'(d)$ and setting $=0$. A1 for correct equation for $d$ |
| $\left[d^2 = \frac{4}{3}\right]$ so $d = 1.154\ldots$ | A1 | For awrt 1.15 or 1.155 or $\sqrt{\frac{4}{3}}$ or $\frac{2\sqrt{3}}{3}$ or $\frac{2}{\sqrt{3}}$ |
| $f''(d) = -\frac{6d}{4} < 0$ so max | B1 (5) | For method confirming maximum not minimum |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(D<1) = \left[\frac{1}{2} - \frac{1}{16}\right] = \frac{7}{16}$ | B1 | |
| Number of children $= 80 \times \frac{7}{16} = 35$ | M1, A1 (3) | M1 for $80 \times p$, $0<p<1$. A1 for 35 only |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
$$\mathrm { F } ( d ) = \left\{ \begin{array} { c c }
0 & d < 0 \\
\frac { d ^ { 2 } } { 2 } - \frac { d ^ { 4 } } { 16 } & 0 \leqslant d \leqslant 2 \\
1 & d > 2
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find the median of $D$.
\item Find the mode of $D$.
Justify your answer.
The experiment is conducted on 80 children.
\item Find the expected number of children whose estimate is less than 1 cm from the actual centre of the circle.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q6 [12]}}