17M.1.AHL.TZ1.H_10

pestleMathematicsAIHLPaper 117M· sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

The continuous random variable X has a probability density function given by

$f(x) = \left\{ {\begin{array}{*{20}{l}} {k\sin \left( {\frac{{\pi x}}{6}} \right),}&{0 \leqslant x \leqslant \,6} \\ {0,}&{{\text{otherwise}}} \end{array}} \right.$ .

Find the value of $k$ .

[4]

By considering the graph of f write down the mean of $X$ ;

[1]

b.i.

By considering the graph of f write down the median of $X$ ;

[1]

b.ii.

By considering the graph of f write down the mode of $X$ .

[1]

b.iii.

Show that $P(0 \leqslant X \leqslant 2) = \frac{1}{4}$ .

[4]

c.i.

Hence state the interquartile range of $X$ .

[2]

c.ii.

Calculate $P(X \leqslant 4|X \geqslant 3)$ .

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to equate integral to 1 (may appear later) M1

$k\int\limits_0^6 {\sin \left( {\frac{{\pi x}}{6}} \right){\text{d}}x = 1}$

correct integral A1

$k\left[ { - \frac{6}{\pi }\cos \left( {\frac{{\pi x}}{6}} \right)} \right]_0^6 = 1$

substituting limits M1

$- \frac{6}{\pi }( - 1 - 1) = \frac{1}{k}$

$k = \frac{\pi }{{12}}$ A1

[4 marks]

mean $= 3$ A1

Note: Award A1A0A0 for three equal answers in $(0,{\text{ }}6)$ .

[1 mark]

b.i.

median $= 3$ A1

Note: Award A1A0A0 for three equal answers in $(0,{\text{ }}6)$ .

[1 mark]

b.ii.

mode $= 3$ A1

Note: Award A1A0A0 for three equal answers in $(0,{\text{ }}6)$ .

[1 mark]

b.iii.

$\frac{\pi }{{12}}\int\limits_0^2 {\sin } \left( {\frac{{\pi x}}{6}} \right){\text{d}}x$ M1

$= \frac{\pi }{{12}}\left[ { - \frac{6}{\pi }\cos \left( {\frac{{\pi x}}{6}} \right)} \right]_0^2$ A1

Note: Accept without the $\frac{\pi }{{12}}$ at this stage if it is added later.

$\frac{\pi }{{12}}\left[ { - \frac{6}{\pi }\left( {\cos \frac{\pi }{3} - 1} \right)} \right]$ M1

$= \frac{1}{4}$ AG

[4 marks]

c.i.

from (c)(i) ${Q_1} = 2$ (A1)

as the graph is symmetrical about the middle value $x = 3 \Rightarrow {Q_3} = 4$ (A1)

so interquartile range is

$4 - 2$

$= 2$ A1

[3 marks]

c.ii.

$P(X \leqslant 4|X \geqslant 3) = \frac{{P(3 \leqslant X \leqslant 4)}}{{P(X \geqslant 3)}}$

$= \frac{{\frac{1}{4}}}{{\frac{1}{2}}}$ (M1)

$= \frac{1}{2}$ A1

[2 marks]

Examiners’ report

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b.i.

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b.ii.

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b.iii.

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c.i.

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c.ii.

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