Problem Statement
A continuous random variable $X$ has probability density function defined by:
$$f(x) = \begin{cases} k x(2 - x), & 0 \le x \le 2 \\ 0, & \text{otherwise} \end{cases}$$
**(a)** [2 marks] Show that $k = \frac{3}{4}$.
**(b)** [2 marks] Write down the expected value $E(X)$ by symmetry, and verify it using integration.
**(c)** [4 marks] Determine the variance $\text{Var}(X)$.
Verified Solution & Marking Scheme
Part (a) [M1]
Total probability property: $\int_0^2 f(x) \, dx = 1 \implies k \int_0^2 (2x - x^2) \, dx = 1$
Part (a) [A1]
$k \left[ x^2 - \frac{x^3}{3} \right]_0^2 = k \left( 4 - \frac{8}{3} \right) = k \left(\frac{4}{3}\right) = 1 \implies k = \frac{3}{4}$ (AG)
Part (b) [R1]
By parabolic symmetry about $x = 1$ on $[0, 2]$, $E(X) = 1$.
Part (b) [A1]
Verification: $E(X) = \frac{3}{4} \int_0^2 x(2x - x^2) \, dx = \frac{3}{4} \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2 = \frac{3}{4} \left( \frac{16}{3} - 4 \right) = \frac{3}{4} \left(\frac{4}{3}\right) = 1$
Part (c) [M1]
$\text{Var}(X) = E(X^2) - [E(X)]^2$. Evaluate $E(X^2) = \int_0^2 x^2 f(x) \, dx$
Part (c) [M1]
$E(X^2) = \frac{3}{4} \int_0^2 (2x^3 - x^4) \, dx = \frac{3}{4} \left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_0^2 = \frac{3}{4} \left( 8 - \frac{32}{5} \right) = \frac{3}{4} \left( \frac{8}{5} \right) = \frac{6}{5}$
Part (c) [M1]
$\text{Var}(X) = \frac{6}{5} - (1)^2 = \frac{6}{5} - 1$
Part (c) [A1]
$\text{Var}(X) = \frac{1}{5} = 0.2$