Problem Statement
The continuous random variable $X$ has probability density function given by:
$$f(x) = \begin{cases} k x(2 - x) & \text{for } 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$$
**(a)** [3 marks] Show that $k = \frac{3}{4}$.
**(b)** [4 marks] Find the expected value $E(X)$ and the variance $\text{Var}(X)$.
Verified Solution & Marking Scheme
Part (a) [M1]
Set total probability $\int_0^2 f(x) \, dx = 1 \implies k \int_0^2 (2x - x^2) \, dx = 1$
Part (a) [A1]
Evaluate: $k \left[ x^2 - \frac{x^3}{3} \right]_0^2 = k\left(4 - \frac{8}{3}\right) = k\left(\frac{4}{3}\right) = 1$
Part (a) [A1]
$k = \frac{3}{4}$ (AG)
Part (b) [M1, A1]
By symmetry about $x = 1$ (or by integration $\frac{3}{4}\int_0^2 (2x^2 - x^3)dx = \frac{3}{4}[\frac{2(8)}{3} - \frac{16}{4}] = \frac{3}{4}(\frac{16}{3} - 4) = 1$), $E(X) = 1$.
Part (b) [M1]
$E(X^2) = \frac{3}{4}\int_0^2 x^2(2x - x^2)dx = \frac{3}{4}\int_0^2 (2x^3 - x^4)dx = \frac{3}{4}\left[\frac{x^4}{2} - \frac{x^5}{5}\right]_0^2 = \frac{3}{4}\left(8 - \frac{32}{5}\right) = \frac{3}{4}\left(\frac{8}{5}\right) = \frac{6}{5}$
Part (b) [A1]
$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{6}{5} - 1^2 = \frac{1}{5} = 0.2$