A random variable has a probability density function given by

$$f(x) = \left\{ {\begin{array}{*{20}{c}}   {kx(2 - x),}&{0 \leqslant x \leqslant 2} \\   {0,}&{{\text{elsewhere}}{\text{.}}} \end{array}} \right.$$

(a)     Show that $k = \frac{3}{4}$ .

(b)     Find ${\text{E}}(X)$ .

(a)     $\int_0^2 {kx(2 - x){\text{d}}x = 1}$     M1A1

Note: Award M1 for LHS and A1 for setting = 1 at any stage.

$\left[ {\frac{{2k}}{2}{x^2} - \frac{k}{3}{x^3}} \right]_0^2 = 1$     A1

$k\left( {4 - \frac{8}{3}} \right) = 1$     A1

$k = \frac{3}{4}$     AG

(b)     ${\text{E}}(X) = \frac{3}{4}\int_0^2 {{x^2}(2 - x){\text{d}}x}$     (M1)

= 1     A1

Note: Accept answers that indicate use of symmetry.

[6 marks]

The integration was particularly well done in this question. A number of students treated the distribution as discrete. On the whole a) was done well once the distribution was recognized although there was a certain amount of fudging to achieve the result. A significant number of students did not initially set the integral equal to 1. Very few noted the symmetry of the distribution in b).