
nth percentile
I need to find 10th percentile of random variable X with PDF defined with:
$\displaystyle
f(x) = \left\{\begin{array}{cc}c(xx^3)&0< x < 1\\0&x\le 0 \vee x\ge 1 \end{array}\right.
$
where $\displaystyle c\in \mathbb{R}$ is const that should be found too.
By integrating f from 0 to $\displaystyle x$ I got cumulative dist. function $\displaystyle F(x) = c \frac{2x^2  x^4}{4}$, so $\displaystyle c=4$.
First, is this approach OK, and second, how can I calculate required percentile?
Thanks guys

Your method for finding the CDF is correct(but i didn't check you did the integral properly).
the 10 percentile $\displaystyle x_{10}$ satisfies:
$\displaystyle F(x_{10}) = 0.1$

I'm little bit confused with notation $\displaystyle x_{10}$, can you write complete calculation pls?

i have just called the 10th percentile $\displaystyle x_{10}$
The definition of a percentile is:
$\displaystyle P(X<x_{10}) = 0.1$
$\displaystyle F(x_{10}) = 0.1$
So find the value of x that satisfies
$\displaystyle 2x^2  x^4 = 0.1$
Can you finish from there?
Hint: this is a quadratic in $\displaystyle x^2$

Ok, thanks. So we take a root that is in interval $\displaystyle 0<x<1$ (we have 4 solutions, one in that interval)?

yes, since x can not be outside that interval