Find the value of $k$.

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

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

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

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

Hence state the interquartile range of $X$.

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

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]

median $= 3$     A1

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

[1 mark]

mode $= 3$     A1

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

[1 mark]

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

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]

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

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

$= \frac{1}{2}$     A1

[2 marks]