CAIE S2 2017 November — Question 5 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2017
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeFind median or percentiles
DifficultyModerate -0.3 This is a straightforward S2 question requiring standard techniques: integration to find E(X) and solving F(x)=0.5 for the median. The pdf is a simple linear function making both integrations routine. While it requires careful algebraic manipulation, it involves no conceptual difficulty beyond applying learned formulas, making it slightly easier than average.
Spec5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles

5 A continuous random variable, \(X\), has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 4 } ( x + 1 ) & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
  1. Find \(\mathrm { E } ( X )\).
    ................................................................................................................................. .
  2. Find the median of \(X\).

Question 5(i):
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{1}{4}\int_0^2 (x^2 + x)\,dx\) \(\left(= \frac{1}{4}\left[\frac{x^3}{3} + \frac{x^2}{2}\right]_0^2\right)\)M1 Attempt integ \(xf(x)\), ignore limits
\(= \frac{1}{4}\left(\frac{8}{3} + 2\right)\ (-0)\)A1 Subst correct limits in correct integration
\(= \frac{7}{6}\) OE or \(1.17\) (3 sf)A1
Total: 3
Question 5(ii):
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{1}{4}\int_0^m (x+1)\,dx = 0.5\) \(\left(=\frac{1}{4}\left[\frac{x^2}{2}+x\right]_0^m = 0.5\right)\)M1 attempt integ \(f(x)\), limits \(0\) to unknown (or unknown to \(2\)) and \(= 0.5\)
\(\frac{1}{4}\left(\frac{m^2}{2}+m\right) = 0.5\), \(m^2+2m-4=0\), \(m = \frac{-2 \pm \sqrt{4+16}}{2}\) OEA1 a correct equation in \(m\) (any form); or \(\sqrt{5}-1\)
\(m = 1.24\)A1 must reject the negative value if there
Total: 3
## Question 5(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{4}\int_0^2 (x^2 + x)\,dx$ $\left(= \frac{1}{4}\left[\frac{x^3}{3} + \frac{x^2}{2}\right]_0^2\right)$ | M1 | Attempt integ $xf(x)$, ignore limits |
| $= \frac{1}{4}\left(\frac{8}{3} + 2\right)\ (-0)$ | A1 | Subst correct limits in correct integration |
| $= \frac{7}{6}$ OE or $1.17$ (3 sf) | A1 | |
| **Total: 3** | | |

## Question 5(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{4}\int_0^m (x+1)\,dx = 0.5$ $\left(=\frac{1}{4}\left[\frac{x^2}{2}+x\right]_0^m = 0.5\right)$ | M1 | attempt integ $f(x)$, limits $0$ to unknown (or unknown to $2$) and $= 0.5$ |
| $\frac{1}{4}\left(\frac{m^2}{2}+m\right) = 0.5$, $m^2+2m-4=0$, $m = \frac{-2 \pm \sqrt{4+16}}{2}$ OE | A1 | a correct equation in $m$ (any form); or $\sqrt{5}-1$ |
| $m = 1.24$ | A1 | must reject the negative value if there |
| **Total: 3** | | |

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5 A continuous random variable, $X$, has probability density function given by

$$f ( x ) = \begin{cases} \frac { 1 } { 4 } ( x + 1 ) & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$

(i) Find $\mathrm { E } ( X )$.\\
................................................................................................................................. .\\

(ii) Find the median of $X$.\\

\hfill \mbox{\textit{CAIE S2 2017 Q5 [6]}}