The random variable T has the probability density function

\[f(t) = \frac{\pi }{4}\cos \left( {\frac{{\pi t}}{2}} \right),{\text{ }} - 1 \leqslant t \leqslant 1.\]

Find

(a)     P(T = 0) ;

(b)     the interquartile range.

(a)     Any consideration of \(\int_0^0 {f(x){\text{d}}x} \)     (M1)

0     A1     N2

(b)     METHOD 1

Let the upper and lower quartiles be a and −a

\(\frac{\pi }{4}\int_a^1 {\cos \frac{{\pi t}}{2}{\text{d}}t = 0.25} \)     M1

\( \Rightarrow \left[ {\frac{\pi }{4} \times \frac{2}{\pi }\sin \frac{{\pi t}}{2}} \right]_a^1 = 0.25\)     A1

\( \Rightarrow \left[ {\frac{1}{2}\sin \frac{{\pi t}}{2}} \right]_a^1 = 0.25\)

\( \Rightarrow \left[ {\frac{1}{2} - \frac{1}{2}\sin \frac{{\pi a}}{2}} \right] = 0.25\)     A1

\( \Rightarrow \frac{1}{2}\sin \frac{{\pi a}}{2} = \frac{1}{4}\)

\( \Rightarrow \sin \frac{{\pi a}}{2} = \frac{1}{2}\)

\(\frac{{\pi a}}{2} = \frac{\pi }{6}\)

\(a = \frac{1}{3}\)     A1

Since the function is symmetrical about t = 0 ,

interquartile range is \(\frac{1}{3} - \left( { - \frac{1}{3}} \right) = \frac{2}{3}\)     R1

METHOD 2

\(\frac{\pi }{4}\int_{ - a}^a {\cos \frac{{\pi t}}{2}{\text{d}}t = 0.5 = \frac{\pi }{2}\int_0^a {\cos \frac{{\pi t}}{2}{\text{d}}t} } \)     M1A1

\( \Rightarrow \left[ {\sin \frac{{a\pi }}{2}} \right] = 0.5\)     A1

\( \Rightarrow \frac{{a\pi }}{2} = \frac{\pi }{6}\)

\( \Rightarrow a = \frac{1}{3}\)     A1

The interquartile range is \(\frac{2}{3}\)     R1

[7 marks]

All but the best candidates struggled with part (a). The vast majority either did not attempt it or let t = 1 . There was no indication from any of the scripts that candidates wasted an undue amount of time in trying to solve part (a). Many candidates attempted part (b), but few had a full understanding of the situation and hence were unable to give wholly correct answers.