| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 15 |
| 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 routine integration of a piecewise PDF to find the CDF, followed by straightforward probability calculations. Part (b)(i) is a 'show that' requiring basic polynomial integration, while parts (ii)-(iv) involve substitution and solving simple equations. The piecewise nature adds mild complexity but this is typical S2 fare with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| For \(0 \leq x \leq 1\): | ||
| \(F(x) = \int_0^x \frac{1}{5}(2x+1) \, dx\) | M1 | Ignore limits |
| \(= \left[\frac{1}{5}(x^2 + x)\right]_0^x\) | A1 | Ignore limits |
| \(= \frac{1}{5}x(x+1)\) | A1 |
| Answer | Marks |
|---|---|
| \(P(X \leq 1) = F(1) = \frac{2}{5}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geq x) = \frac{17}{20} \Rightarrow F(x) = \frac{3}{20}\) | M1 | |
| \(\frac{1}{5}x(x+1) = \frac{3}{20}\) | m1 | |
| \(x(x+1) = \frac{3}{4}\) | ||
| \(x^2 + x - \frac{3}{4} = 0\) | A1 | |
| \(\left(x - \frac{1}{2}\right)(x + \frac{3}{2}) = 0\) | m1 | Any valid method attempted |
| \(x = \frac{1}{2}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(F(1) = 0.4\), \(q\) lies in \(0 \leq r \leq 1\) | ||
| \(F(q) = \frac{1}{5}(q^2 + q) = 0.25\) | M1 | |
| \(\Rightarrow q^2 + q = 1.25\) | ||
| \(q^2 + q - 1.25 = 0\) | A1 | |
| \(\Rightarrow q = \frac{-1 \pm \sqrt{1 - 4 \times (-1.25)}}{2}\) | m1 | |
| \(q = \frac{1}{2}(\sqrt{6} - 1)\) (\(q > 0\)) | A1 | AWFW (0.724 to 0.725) |
| B2 | for line segment (0,0.2) to (1,0.6); for correctly shaped curve (1,0.6) to (4,0) | 2 |
## 7(b)(i)
For $0 \leq x \leq 1$: |
$F(x) = \int_0^x \frac{1}{5}(2x+1) \, dx$ | M1 | Ignore limits
$= \left[\frac{1}{5}(x^2 + x)\right]_0^x$ | A1 | Ignore limits
$= \frac{1}{5}x(x+1)$ | A1 |
## 7(b)(ii)
$P(X \leq 1) = F(1) = \frac{2}{5}$ | B1 |
## 7(b)(iii)
$P(X \geq x) = \frac{17}{20} \Rightarrow F(x) = \frac{3}{20}$ | M1 |
$\frac{1}{5}x(x+1) = \frac{3}{20}$ | m1 |
$x(x+1) = \frac{3}{4}$ |
$x^2 + x - \frac{3}{4} = 0$ | A1 |
$\left(x - \frac{1}{2}\right)(x + \frac{3}{2}) = 0$ | m1 | Any valid method attempted
$x = \frac{1}{2}$ | A1 | CAO
## 7(b)(iv)
Since $F(1) = 0.4$, $q$ lies in $0 \leq r \leq 1$ |
$F(q) = \frac{1}{5}(q^2 + q) = 0.25$ | M1 |
$\Rightarrow q^2 + q = 1.25$ |
$q^2 + q - 1.25 = 0$ | A1 |
$\Rightarrow q = \frac{-1 \pm \sqrt{1 - 4 \times (-1.25)}}{2}$ | m1 |
$q = \frac{1}{2}(\sqrt{6} - 1)$ ($q > 0$) | A1 | AWFW (0.724 to 0.725)
| B2 | for line segment (0,0.2) to (1,0.6); for correctly shaped curve (1,0.6) to (4,0) | 2 |
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# TOTAL: 75 marks7 The continuous random variable $X$ has probability density function defined by
$$f ( x ) = \begin{cases} \frac { 1 } { 5 } ( 2 x + 1 ) & 0 \leqslant x \leqslant 1 \\ \frac { 1 } { 15 } ( 4 - x ) ^ { 2 } & 1 < x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the cumulative distribution function, $\mathrm { F } ( x )$, for $0 \leqslant x \leqslant 1$ is
$$\mathrm { F } ( x ) = \frac { 1 } { 5 } x ( x + 1 )$$
\item Hence write down the value of $\mathrm { P } ( X \leqslant 1 )$.
\item Find the value of $x$ for which $\mathrm { P } ( X \geqslant x ) = \frac { 17 } { 20 }$.
\item Find the lower quartile of the distribution.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q7 [15]}}