# n-th percentile

• Jun 7th 2010, 09:58 AM
losm1
n-th percentile
I need to find 10th percentile of random variable X with PDF defined with:

$\displaystyle f(x) = \left\{\begin{array}{cc}c(x-x^3)&0< x < 1\\0&x\le 0 \vee x\ge 1 \end{array}\right.$

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
• Jun 7th 2010, 10:36 AM
SpringFan25
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:

$\displaystyle F(x_{10}) = 0.1$
• Jun 7th 2010, 10:46 AM
losm1
I'm little bit confused with notation $\displaystyle x_{10}$, can you write complete calculation pls?
• Jun 7th 2010, 11:03 AM
SpringFan25
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 $\displaystyle x^2$
• Jun 7th 2010, 11:16 AM
losm1
Ok, thanks. So we take a root that is in interval $\displaystyle 0<x<1$ (we have 4 solutions, one in that interval)?
• Jun 7th 2010, 11:28 AM
SpringFan25
yes, since x can not be outside that interval