$X$ follows a distribution with distribution function $$F(x)=\begin{cases}1-\left(\frac{2000}{2000+x}\right): &x\ge0\\ 0: &\text{otherwise}\end{cases}$$ Let $Y=max(0,X-500)$. Calculate the $60^{th}$ percentile of positive values of $Y$, ignoring values of $Y$ where $0\le X \le 500$
Here was my solution:
$P(0\le X\le500) = F(500) = 0.2. \\\text{Let } Z \text{ be a random variable with density} f_Z(z) = \begin{cases} 0.8-\left(\frac{2000}{2000+z}\right): &z\ge500\\ 0: &\text{else} \end{cases} \\$
As I'm writing this, I can see that my density function for $Z$ is not a legit density function, and in fact I'm not sure why I made it a density function. Did I mean distribution function? Well, it's not a legit distribution function either. Okay, I give up, I'm ready for some help. If you're curious, my wrong answer was $7500$.
$P(Y=0) = F(500) = 1- \frac{2000}{2500} = 0.2$.
The 60th percentile of the remaining $0.8$ probability is equivalent to $0.2 + 0.6 \cdot 0.8 = 0.68$ quantile of $Y$. So you want to find $y$ such that $P(Y \le y) = 0.68$. But for $y > 500$ we have $P(Y \le y) = P(X \le y+500)$. Can you finish from here?