Problem Statement
A continuous random variable $X$ has probability density function given by:
$f(x) = \begin{cases} k x (2 - x), & 0 \le x \le 2 \\ 0, & \text{otherwise} \end{cases}$
(a) Show that $k = \frac{3}{4}$. [3]
(b) Find $\text{Var}(X)$. [4]
Verified Solution & Marking Scheme
(a) Apply total probability condition ∫ f(x) dx = 1
$\int_{0}^{2} k(2x - x^2) \, dx = 1$
$k \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = 1$
$k \left( 4 - \frac{8}{3} \right) = 1 \implies k \left( \frac{4}{3} \right) = 1 \implies k = \frac{3}{4}$
Hence proved.
(b) Find E(X) and E(X²)
By symmetry about $x = 1$, $E(X) = 1$.
Verify:
$E(X) = \frac{3}{4}\int_{0}^{2} x(2x - x^2) \, dx = \frac{3}{4}\int_{0}^{2} (2x^2 - x^3) \, dx = \frac{3}{4}\left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_{0}^{2} = \frac{3}{4}\left(\frac{16}{3} - 4\right) = \frac{3}{4}\left(\frac{4}{3}\right) = 1$
Now find $E(X^2)$:
$E(X^2) = \frac{3}{4}\int_{0}^{2} x^2(2x - x^2) \, dx = \frac{3}{4}\int_{0}^{2} (2x^3 - x^4) \, dx = \frac{3}{4}\left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_{0}^{2}$
$= \frac{3}{4} \left( 8 - \frac{32}{5} \right) = \frac{3}{4} \left( \frac{8}{5} \right) = \frac{6}{5}$
Compute variance Var(X) = E(X²) - [E(X)]²
$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{6}{5} - (1)^2 = \frac{6}{5} - 1 = \frac{1}{5} = 0.2$