Problem Statement
A continuous random variable $X$ has probability density function $f(x) = kx(4 - x)$ for $0 \le x \le 4$, and $f(x) = 0$ otherwise.
**(a)** [2 marks] Show that $k = \frac{3}{32}$.
**(b)** [2 marks] State the mode and the mean of $X$ by symmetry.
**(c)** [4 marks] Calculate the probability $P(1 < X < 3 \mid X > 0.5)$.
Verified Solution & Marking Scheme
Part (a) [M1]
$\int_0^4 k(4x - x^2)\,dx = k \left[2x^2 - \frac{x^3}{3}\right]_0^4 = k\left(32 - \frac{64}{3}\right) = k\left(\frac{32}{3}\right) = 1$
Part (a) [A1]
$k = \frac{3}{32}$ (AG)
Part (b) [A1]
Mode: Parabola vertex is at $x = 2$
Part (b) [A1]
Mean: By symmetry about $x = 2$, $E(X) = 2$
Part (c) [M1]
$P(X > 0.5) = \int_{0.5}^4 \frac{3}{32}(4x - x^2)\,dx = 1 - \int_0^{0.5} \frac{3}{32}(4x - x^2)\,dx = 1 - \frac{3}{32}[2(0.25) - 0.125/3] = 1 - 0.043 = 0.957$
Part (c) [M1]
$P(1 < X < 3) = \int_1^3 \frac{3}{32}(4x - x^2)\,dx = \frac{3}{32}\left[2x^2 - \frac{x^3}{3}\right]_1^3 = \frac{3}{32}\left[(18 - 9) - (2 - 1/3)\right] = \frac{3}{32}\left[9 - \frac{5}{3}\right] = \frac{3}{32}\left(\frac{22}{3}\right) = \frac{22}{32} = 0.6875$
Part (c) [A1]
Conditional probability: $\frac{P((1 < X < 3) \cap (X > 0.5))}{P(X > 0.5)} = \frac{P(1 < X < 3)}{P(X > 0.5)} = \frac{0.6875}{0.957}$
Part (c) [A1]
$\approx 0.718$