Suppose the future revenue of $1$ invested dollar is described by the random variable $X$ where $\mathbb{P}_X$ is given via the probability density function:
$f_X(x) = \begin{cases} \frac{3}{10}x^2 \hspace{1cm} 0 \leq x \leq 1 \\ \frac{10-x}{45} \hspace{1cm} 1 < x \leq 10 \\ 0 \hspace{1.7cm} else\end{cases}$
How can I calculate the value $v \in \mathbb{R}$ that the revenue exceeds with a probability of $0.95$?