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

Find the interquartile range of \(T\).

\(\left| {2 - t} \right|\) correct for \(\left[ {1,{\text{ }}2} \right]\)     A1

\(\left| {2 - t} \right|\) correct for \(\left[ {2,{\text{ }}3} \right]\)     A1

EITHER

let \({q_1}\) be the lower quartile and let \({q_3}\) be the upper quartile

let \(d = 2 - {q_1}{\text{ }}( = {q_3} - 2)\) and so \({\text{IQR}} = 2d\) by symmetry

use of area formulae to obtain \(\frac{1}{2}{d^2} = \frac{1}{4}\)

(or equivalent)     M1A1

\(d = \frac{1}{{\sqrt 2 }}\) or the value of at least one \(q\).     A1

OR

let \({q_1}\) be the lower quartile

consider \(\int_1^{{q_1}} {(2 - t){\text{d}}t = \frac{1}{4}} \)     M1A1

obtain \({q_1} = 2 - \frac{1}{{\sqrt 2 }}\)     A1

THEN

\({\text{IQR}} = \sqrt 2 \)     A1

Note:     Only accept this final answer for the A1.

[4 marks]

Total [6 marks]

The sketched graphs were mostly acceptable, but sometimes scrappy.

Most candidates had some idea about the upper and lower quartiles, but some were rather vague about how to calculate them for this probability density function. Even those who integrated for the lower quartile often made algebraic mistakes in calculating its value.