Problem

The continuous random variable $X$ has probability density function

$f(x) = \begin{cases} \dfrac{1}{4} - \dfrac{1}{64}x^3 & 0 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that the median of $X$ is 2.165 to 3 decimal places.

The continuous random variable $Y$ has probability density function

$f(y) = \begin{cases} \dfrac{4}{255}y^3 & 1 \leqslant y \leqslant 4 \\ 0 & \text{otherwise} \end{cases}$

(b) Show that $\text{Var}\left(\dfrac{1}{6}Y\right) = \dfrac{66}{7225}$

(Solutions relying entirely on calculator technology are not acceptable.)

(c) Hence find $\text{Var}\left(\dfrac{5}{4} - Y\right)$

(Total for Question 7 is 12 marks)