Problem Statement
A continuous random variable $X$ has the probability density function (PDF) given by:
$$f(x) = \begin{cases} kx(2 - x) & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$$
where $k$ is a positive real constant.
**(a)** [2 marks] Show that $k = \frac{3}{4}$.
**(b)** [3 marks] Find the expected value $E(X)$ and the variance $\text{Var}(X)$.
**(c)** [3 marks] Calculate the conditional probability $P\left(X > 1.5 \mid X > 1\right)$.
Verified Solution & Marking Scheme
Part (a) [M1]
Total probability condition: $\int_0^2 f(x)\,dx = 1 \implies k \int_0^2 (2x - x^2)\,dx = 1$
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) [A1]
By symmetry of the parabola $x(2 - x)$ on $[0, 2]$, $E(X) = 1$. (Or $\int_0^2 \frac{3}{4}x^2(2 - x)\,dx = \frac{3}{4}\left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2 = \frac{3}{4}\left(\frac{16}{3} - 4\right) = 1$)
Part (b) [M1]
$E(X^2) = \int_0^2 x^2 f(x)\,dx = \frac{3}{4}\int_0^2 (2x^3 - x^4)\,dx = \frac{3}{4}\left[\frac{x^4}{2} - \frac{x^5}{5}\right]_0^2 = \frac{3}{4}\left(8 - \frac{32}{5}\right) = \frac{3}{4}\left(\frac{8}{5}\right) = \frac{6}{5}$
Part (b) [A1]
$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{6}{5} - 1^2 = \frac{1}{5}$ (or $0.2$)
Part (c) [M1]
Conditional probability formula: $P(X > 1.5 \mid X > 1) = \frac{P(X > 1.5 \cap X > 1)}{P(X > 1)} = \frac{P(X > 1.5)}{P(X > 1)}$
Part (c) [A1]
$P(X > 1) = \frac{1}{2}$ by symmetry. $P(X > 1.5) = \frac{3}{4}\int_{1.5}^2 (2x - x^2)\,dx = \frac{3}{4}\left[x^2 - \frac{x^3}{3}\right]_{3/2}^2 = \frac{3}{4}\left( \frac{4}{3} - \left(\frac{9}{4} - \frac{9}{8}\right) \right) = \frac{3}{4}\left(\frac{4}{3} - \frac{9}{8}\right) = \frac{3}{4}\left(\frac{5}{24}\right) = \frac{5}{32}$
Part (c) [A1]
$P(X > 1.5 \mid X > 1) = \frac{5/32}{1/2} = \frac{5}{16}$