| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a standard S2 question on continuous probability distributions requiring routine techniques: sketching a linear pdf, calculating E(X) by integration, verifying a given CDF formula, using F(x) for probability calculations, and solving a quadratic for the median. All steps are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Straight line from \((1, 0.2)\) to \((5, 0.3)\) | B2,1 | B2 for st. line from \((1,0.2)\) to \((5,0.3)\) B1 st. line (\(m > 0\)) from \(x = 1\) to \(x = 5\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \frac{1}{40}\int_1^5 x(x+7) dx\) | M1 | Ignore limits |
| \(= \frac{1}{40}\left[\frac{x^3}{3} + \frac{7x^2}{2}\right]_1^5\) | A1 | Ignore limits |
| \(= \frac{1}{40}\left(\frac{125}{3} + \frac{175}{2} - \frac{1}{3} - \frac{7}{2}\right) = 3\frac{2}{15}\) | A1 | cao (accept 3.133 or \(\frac{47}{15}\) oe exact) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x) = \int_1^x \frac{1}{40}(x+7) dx\) | M1 | \(F(x) = \int \left(\frac{x}{40} + \frac{7}{40}\right) dx = \frac{x^2}{80} + \frac{7x}{40} + c\) (M1A1) |
| \(= \frac{1}{40}\left[\frac{x^2}{2} + 7x\right]_1^x\) | A1 | |
| \(= \frac{1}{80}(x^2 + 14x - 1 - 14)\) | F(1) \(= 0 \Rightarrow c = -\frac{1}{80} - \frac{7}{40} = -\frac{15}{80}\) | |
| \(= \frac{1}{80}\left(x^2 + 14x - 15\right)\) | Adep1 | or [use of F(5) = 1] \(\Rightarrow F(x) = \frac{1}{80}(x^2 + 14x - 15)\) |
| \(= \frac{1}{80}(x+15)(x-1)\) | Adep1 | \(F(x) = \frac{1}{80}(x+15)(x-1)\) (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(2.5 \leq X \leq 4.5) = F(4.5) - F(2.5)\) | M1 | |
| \(= \frac{1}{80}(19.5 \times 3.5 - 17.5 \times 1.5)\) | M1 | Trapezium Rule \(\frac{1}{2}\left(\frac{23}{80} - \frac{19}{80}\right) \times 2 = \frac{42}{80} = \frac{21}{40}\) |
| \(= \frac{42}{80} = \frac{21}{40} (0.525)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(m) = \frac{1}{2}\) | B1 | \(\int_1^m \frac{1}{40}(x+7) dx = 0.5\) (B1) |
| \(\Rightarrow \frac{1}{80}(m^2 + 14m - 15) = \frac{1}{2}\) | M1 | Correct equation formed |
| \((\times 80) \Rightarrow m^2 + 14m - 15 = 40\) \(m^2 + 14m - 55 = 0\) | Adep1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = \frac{-14 \pm \sqrt{196 + 220}}{2} = \frac{-14 \pm 20.396}{2}\) | M1 | Correct attempt at solving quadratic (by formula, oe). |
| \(m = \frac{-14 + 20.396}{2}\) (since \(m > 1\)) | A1 | cao |
| \(m = 3.198\) (3dp) |
## (a)
| Straight line from $(1, 0.2)$ to $(5, 0.3)$ | B2,1 | B2 for st. line from $(1,0.2)$ to $(5,0.3)$ B1 st. line ($m > 0$) from $x = 1$ to $x = 5$. |
## (b)
| $E(X) = \frac{1}{40}\int_1^5 x(x+7) dx$ | M1 | Ignore limits |
| $= \frac{1}{40}\left[\frac{x^3}{3} + \frac{7x^2}{2}\right]_1^5$ | A1 | Ignore limits |
| $= \frac{1}{40}\left(\frac{125}{3} + \frac{175}{2} - \frac{1}{3} - \frac{7}{2}\right) = 3\frac{2}{15}$ | A1 | cao (accept 3.133 or $\frac{47}{15}$ oe exact) |
## (c)
| $F(x) = \int_1^x \frac{1}{40}(x+7) dx$ | M1 | $F(x) = \int \left(\frac{x}{40} + \frac{7}{40}\right) dx = \frac{x^2}{80} + \frac{7x}{40} + c$ (M1A1) |
| $= \frac{1}{40}\left[\frac{x^2}{2} + 7x\right]_1^x$ | A1 | |
| $= \frac{1}{80}(x^2 + 14x - 1 - 14)$ | | F(1) $= 0 \Rightarrow c = -\frac{1}{80} - \frac{7}{40} = -\frac{15}{80}$ |
| $= \frac{1}{80}\left(x^2 + 14x - 15\right)$ | Adep1 | or [use of F(5) = 1] $\Rightarrow F(x) = \frac{1}{80}(x^2 + 14x - 15)$ |
| $= \frac{1}{80}(x+15)(x-1)$ | Adep1 | $F(x) = \frac{1}{80}(x+15)(x-1)$ (AG) |
## (d)(i)
| $P(2.5 \leq X \leq 4.5) = F(4.5) - F(2.5)$ | M1 | |
| $= \frac{1}{80}(19.5 \times 3.5 - 17.5 \times 1.5)$ | M1 | Trapezium Rule $\frac{1}{2}\left(\frac{23}{80} - \frac{19}{80}\right) \times 2 = \frac{42}{80} = \frac{21}{40}$ |
| $= \frac{42}{80} = \frac{21}{40} (0.525)$ | A1 | |
## (d)(ii)
| $F(m) = \frac{1}{2}$ | B1 | $\int_1^m \frac{1}{40}(x+7) dx = 0.5$ (B1) |
| $\Rightarrow \frac{1}{80}(m^2 + 14m - 15) = \frac{1}{2}$ | M1 | Correct equation formed |
| $(\times 80) \Rightarrow m^2 + 14m - 15 = 40$ $m^2 + 14m - 55 = 0$ | Adep1 | AG |
## (e)
| $m = \frac{-14 \pm \sqrt{196 + 220}}{2} = \frac{-14 \pm 20.396}{2}$ | M1 | Correct attempt at solving quadratic (by formula, oe). |
| $m = \frac{-14 + 20.396}{2}$ (since $m > 1$) | A1 | cao |
| $m = 3.198$ (3dp) | | |
**Total: 16 marks**
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# **TOTAL: 75 marks**6 The random variable $X$ has probability density function defined by
$$f ( x ) = \begin{cases} \frac { 1 } { 40 } ( x + 7 ) & 1 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item Find the exact value of $\mathrm { E } ( X )$.
\item Prove that the distribution function F , for $1 \leqslant x \leqslant 5$, is defined by
$$\mathrm { F } ( x ) = \frac { 1 } { 80 } ( x + 15 ) ( x - 1 )$$
\item Hence, or otherwise:
\begin{enumerate}[label=(\roman*)]
\item find $\mathrm { P } ( 2.5 \leqslant X \leqslant 4.5 )$;
\item show that the median, $m$, of $X$ satisfies the equation $m ^ { 2 } + 14 m - 55 = 0$.
\end{enumerate}\item Calculate the value of the median of $X$, giving your answer to three decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q6 [16]}}