So, I came across a couple of homework problems on finding percentiles. The first was:
pdf of $X$ is $f(x)=\frac{10}{x^2}$ for $x\gt 10$, and $0$ otherwise. Finding the $75$th percentile here was fine:
$$\int_{10}^{p}\frac{10}{x^2}dx=.75$$
and $p=40$.
The next problem I also had to find the $75$th percentile and I am not certain on how to go about this.
The random variable $X$ has pdf:
$$f(x) =\begin{cases}(1+x), & {-1\lt x \le 0} \\(1-x), & {0\lt x\lt 1} \\\end{cases}$$
A full solution isn't necessary. I'm really just hoping for help on how to set it up. I haven't been asked to find a $75$th percentile on a pdf that has more than one condition before and can't seem to make sense of it. The best I could come up with was setting both up individually and making $p$ the upper limit on each. My idea was that this could potentially show the $75$th percentile for each condition. But, I can't find anything to verify it one way or the other.
Thanks in advanced.
Look at the graph of your function $f(x)$.
In particular, the pdf is symmetric about $ x = 0$, so where does the $50$th percentile lie?
The $75$th percentile of the entire function will be what percentile of the function for $x > 0$?