Problem Statement
A continuous random variable $X$ has probability density function (PDF):
$f(x) = \begin{cases} k x (2 - x) & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$
(a) Show that $k = \frac{3}{4}$. [2 Marks]
(b) Find the median of $X$. [2 Marks]
(c) Calculate $E(X)$ and $\text{Var}(X)$. [3 Marks]
Verified Solution & Marking Scheme
Part (a): Normalization Integral
$\int_0^2 f(x) \, dx = 1 \implies k \int_0^2 (2x - x^2) \, dx = 1$
$k \left[ x^2 - \frac{x^3}{3} \right]_0^2 = k \left( 4 - \frac{8}{3} \right) = k \left( \frac{4}{3} \right) = 1 \implies k = \frac{3}{4} \quad (AG)$
Part (b): Determine Median
Notice that $f(x) = \frac{3}{4}x(2 - x)$ is a symmetric parabola about $x = 1$ on the interval $[0, 2]$.
By symmetry, $\int_0^1 f(x)\,dx = \frac{1}{2}$. Hence the median is $m = 1$.
Part (c): Expectation and Variance
By symmetry, $E(X) = 1$.
To find $\text{Var}(X) = E(X^2) - [E(X)]^2$:
$E(X^2) = \frac{3}{4}\int_0^2 x^2(2x - x^2)\,dx = \frac{3}{4}\int_0^2 (2x^3 - x^4)\,dx$
$= \frac{3}{4}\left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_0^2 = \frac{3}{4}\left( 8 - \frac{32}{5} \right) = \frac{3}{4}\left(\frac{8}{5}\right) = \frac{6}{5} = 1.2$
$\text{Var}(X) = 1.2 - 1^2 = 0.2 = \frac{1}{5}$