| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a standard S2 question requiring integration of a piecewise PDF to find the CDF, then using it to find median and quartiles. The integration is straightforward (polynomial and linear functions), and the problem-solving follows routine procedures. Part (c) involves solving a quadratic equation, which is basic. Slightly easier than average due to the 'show that' structure guiding students through each step. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
6 The continuous random variable $X$ has the probability density function defined by
$$f ( x ) = \begin{cases} \frac { 3 } { 8 } \left( x ^ { 2 } + 1 \right) & 0 \leqslant x \leqslant 1 \\ \frac { 1 } { 4 } ( 5 - 2 x ) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item The cumulative distribution function of $X$ is denoted by $\mathrm { F } ( x )$.
Show that, for $0 \leqslant x \leqslant 1$,
$$\mathrm { F } ( x ) = \frac { 1 } { 8 } x \left( x ^ { 2 } + 3 \right)$$
\item Hence, or otherwise, verify that the median value of $X$ is 1 .
\item Show that the upper quartile, $q$, satisfies the equation $q ^ { 2 } - 5 q + 5 = 0$ and hence that $q = \frac { 1 } { 2 } ( 5 - \sqrt { 5 } )$.
\item Calculate the exact value of $\mathrm { P } ( q < X < 1.5 )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2011 Q6 [14]}}