The probability density function of a random variable X is defined as:

 

\[f(x) = \left\{ \begin{array}{r}ax\cos x,\\0,\end{array} \right.\begin{array}{*{20}{l}}{0 \le x \le {\textstyle{\pi  \over 2}},{\rm{where }}\,a \in \mathbb{R}}\\{{\rm{elsewhere}}}\end{array}\]

 

(a)     Show that \(a = \frac{2}{{\pi  - 2}}\).

(b)     Find \({\text{P}}\left( {X < \frac{\pi }{4}} \right)\).

(c)     Find:

          (i)     the mode of X;

          (ii)     the median of X.

(d)     Find \({\text{P}}\left( {X < \frac{\pi }{8}|X < \frac{\pi }{4}} \right)\).

(a)     \(a\int_0^{\frac{\pi }{2}} {x\cos x{\text{d}}x = 1} \)     (M1)

integrating by parts:

\(u = x\)     \(v' = \cos x\)     M1

\(u' = 1\)     \(v = \sin x\)

\(\int {x\cos x{\text{d}}x = x\sin x + \cos x} \)     A1

\(\left[ {x\sin x + \cos x} \right]_0^{\frac{\pi }{2}} = \frac{\pi }{2} - 1\)     A1

\(a = \frac{1}{{\frac{\pi }{2} - 1}}\)     A1

\( = \frac{2}{{\pi  - 2}}\)     AG

[5 marks]

 

(b)     \({\text{P}}\left( {X < \frac{\pi }{4}} \right) = \frac{2}{{\pi  - 2}}\int_0^{\frac{\pi }{4}} {x\cos x{\text{d}}x = 0.460} \)     (M1)A1

 

Note:     Accept \(\frac{2}{{\pi  - 2}}{\text{ }}\left( { = \frac{{\pi \sqrt 2 }}{8} + \frac{{\sqrt 2 }}{2} - 1} \right)\) or equivalent

 

[2 marks]

 

(c)     (i)     \({\text{mode}} = 0.860\)     A1

          (x-value of a maximum on the graph over the given domain)

          (ii)     \(\frac{2}{{\pi  - 2}}\int_0^m {x\cos x{\text{d}}x = 0.5} \)     (M1)

          \(\int_0^m {x\cos x{\text{d}}x = \frac{{\pi  - 2}}{4}} \)

          \(m\sin m + \cos m - 1 = \frac{{\pi  - 2}}{4}\)     (M1)

          \({\text{median}} = 0.826\)     A1

 

Note:     Do not accept answers containing additional solutions.

 

[4 marks]

 

(d)     \({\text{P}}\left( {X < \frac{\pi }{8}|X < \frac{\pi }{4}} \right) = \frac{{{\text{P}}\left( {X < \frac{\pi }{8}} \right)}}{{{\text{P}}\left( {X < \frac{\pi }{4}} \right)}}\)     M1

\( = \frac{{0.129912}}{{0.459826}}\)

\( = 0.283\)     A1

[2 marks]

 

Total [13 marks]

[N/A]