A continuous random variable X has a probability density function given by

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{{{(x + 1)}^3}}}{{60}},}&{{\text{for }}1 \leqslant x \leqslant 3} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array},} \right.\]

Find

(a)     \({\text{P}}(1.5 \leqslant X \leqslant 2.5)\) ;

(b)     E(X) ;

(c)     the median of X .

(a)     \(\int_{1.5}^{2.5} {\frac{{{{(x + 1)}^3}}}{{60}}{\text{d}}x = 0.4625\,\,\,\,\,{\text{( =  0.463)}}} \)     M1A1

(b)     \({\text{E}}(X) = \int_1^3 {\frac{{x{{(x + 1)}^3}}}{{60}}{\text{d}}x = 2.31} \)     M1A1

(c)     \(\int_1^m {\frac{{{{(x + 1)}^3}}}{{60}}{\text{d}}x = 0.5} \)     M1

\(\left[ {\frac{{{{(x + 1)}^4}}}{{240}}} \right]_1^m = 0.5\)     (A1)

\(m = 2.41\)     A1

[7 marks]

Parts (a) and (b) were reasonably well done in general but (c) caused problems for many candidates where several misconceptions regarding the median were seen. The expectation was that candidates would use their GDCs to solve (a) and (b), and possibly even (c), although in the event most candidates did the integrations by hand. Those candidates using their GDCs made fewer mistakes in general than those doing the integrations analytically.