7 The random variable \(X\) has probability density function defined by $$f ( x ) = \begin{cases} \frac { 1 } { 2 } & 0 \leqslant x \leqslant 1
\frac { 1 } { 18 } ( x - 4 ) ^ { 2 } & 1 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$

1. State values for the median and the lower quartile of \(X\).
2. Show that, for \(1 \leqslant x \leqslant 4\), the cumulative distribution function, \(\mathrm { F } ( x )\), of \(X\) is given by $$\mathrm { F } ( x ) = 1 + \frac { 1 } { 54 } ( x - 4 ) ^ { 3 }$$ (You may assume that \(\int ( x - 4 ) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 3 } ( x - 4 ) ^ { 3 } + c\).)
3. Determine \(\mathrm { P } ( 2 \leqslant X \leqslant 3 )\).
4. 1. Show that \(q\), the upper quartile of \(X\), satisfies the equation \(( q - 4 ) ^ { 3 } = - 13.5\).
   2. Hence evaluate \(q\) to three decimal places.
      \includegraphics[max width=\textwidth, alt={}]{c31c5c67-834e-42ce-b4af-555890c393d5-19_2484_1709_223_153}