Sketch the graph of $y = f(t)$.

Use your sketch to find the mode of $T$.

Find the mean of $T$.

Find the variance of $T$.

Find the probability that $T$ lies between the mean and the mode.

(i)     Find $\int_0^\pi  {f(t){\text{d}}t}$ where $0 \leqslant T \leqslant \frac{\pi }{2}$.

(ii)     Hence verify that the lower quartile of $T$ is $\frac{\pi }{2}$.

two enclosed regions ($0 \leqslant t \leqslant \frac{\pi }{2}$ and $\frac{\pi }{2} \leqslant t \leqslant \pi$) bounded by the curve and the $t$-axis     A1

correct non-symmetrical shape for $0 \leqslant t \leqslant \frac{\pi }{2}$ and

$\frac{\pi }{2} < {\text{mode of }}T < \pi {\text{ clearly apparent}}$    A1

[2 marks]

${\text{mode}} = 2.46$    A1

[1 mark]

${\text{E}}(T) = \frac{1}{\pi }\int_0^\pi  {{t^2}\left| {\sin 2t} \right|{\text{d}}t}$    (M1)

$= 2.04$    A1

[2 marks]

EITHER

${\text{Var}}(T) = \int_0^\pi  {(t - } 2.03788 \ldots {)^2}\left( {\frac{{t\left| {\sin 2t} \right|}}{\pi }} \right){\text{d}}t$     (M1)(A1)

OR

${\text{Var}}(T) = \int_0^\pi  {{t^2}} \left( {\frac{{t\left| {\sin 2t} \right|}}{\pi }} \right){\text{d}}t - {(2.03788 \ldots )^2}$     

(M1)(A1)

THEN

${\text{Var}}(T) = 0.516$    A1

[3 marks]

$\frac{1}{\pi }\int_{{\text{2.03788}} \ldots }^{{\text{2.456590}} \ldots } {t\left| {\sin 2t} \right|{\text{d}}t = {\text{0.285}}}$    (M1)A1

[2 marks]

(i)     attempting integration by parts     (M1)

$(u = t,{\text{ d}}u = {\text{d}}t,{\text{ d}}v = \sin 2t{\text{ d}}t$ and $v =  - \frac{1}{2}\cos 2t)$

$\frac{1}{\pi }\left[ {t\left( { - \frac{1}{2}\cos 2t} \right)} \right]_0^r - \frac{1}{\pi }\int_0^r {\left( { - \frac{1}{2}\cos 2t} \right){\text{d}}t}$    A1

Note:     Award A1 if the limits are not included.

$= \frac{{\sin 2T}}{{4\pi }} - \frac{{T\cos 2T}}{{2\pi }}$    A1

(ii)     $\frac{{\sin \pi }}{{4\pi }} - \frac{{\frac{\pi }{2}\cos \pi }}{{2\pi }} = \frac{1}{4}$     A1

as ${\text{P}}\left( {0 \leqslant T \leqslant \frac{\pi }{2}} \right) = \frac{1}{4}$ (or equivalent), then the lower quartile of $T$ is $\frac{\pi }{2}$     R1AG

[5 marks]

This question was generally accessible to the large majority of candidates. A substantial number of candidates were able to neatly and accurately sketch a non-symmetric bimodal continuous probability density function and to calculate its mean, mode and variance. Quite a few candidates unfortunately attempted this question with their GDC set in degrees.

This question was generally accessible to the large majority of candidates. A substantial number of candidates were able to neatly and accurately sketch a non-symmetric bimodal continuous probability density function and to calculate its mean, mode and variance. Quite a few candidates unfortunately attempted this question with their GDC set in degrees.

This question was generally accessible to the large majority of candidates. A substantial number of candidates were able to neatly and accurately sketch a non-symmetric bimodal continuous probability density function and to calculate its mean, mode and variance. Quite a few candidates unfortunately attempted this question with their GDC set in degrees.

This question was generally accessible to the large majority of candidates. A substantial number of candidates were able to neatly and accurately sketch a non-symmetric bimodal continuous probability density function and to calculate its mean, mode and variance. Quite a few candidates unfortunately attempted this question with their GDC set in degrees.

This question was generally accessible to the large majority of candidates. A substantial number of candidates were able to neatly and accurately sketch a non-symmetric bimodal continuous probability density function and to calculate its mean, mode and variance. Quite a few candidates unfortunately attempted this question with their GDC set in degrees.

This question was generally accessible to the large majority of candidates. A substantial number of candidates were able to neatly and accurately sketch a non-symmetric bimodal continuous probability density function and to calculate its mean, mode and variance. Quite a few candidates unfortunately attempted this question with their GDC set in degrees.