Percentiles in a Continous Random Variable distribution

**Pink Lady** · August 4th 2009, 09:46 AM

The probability density function of x is:
f(x) = 4x(1-x^2) when x is a value between 0 and 1. (0 and 1 are included)
= 0 otherwise
What is the 75th percentile of this distribution?

**Flyingdutchman** · August 4th 2009, 01:41 PM

the 75th percentile is the the x that gives the answer f=0.75 I believe

**mr fantastic** · August 4th 2009, 04:11 PM

Originally Posted by Pink Lady

The probability density function of x is:
f(x) = 4x(1-x^2) when x is a value between 0 and 1. (0 and 1 are included)
= 0 otherwise
What is the 75th percentile of this distribution?

By definition:

Find the value of a such that $\int_0^a 4x (1 - x^2) \, dx = 0.75$ .

Originally Posted by Flyingdutchman

the 75th percentile is the the x that gives the answer f=0.75 I believe

Sorry but that's not correct.

**matheagle** · August 4th 2009, 04:59 PM

It's F(a)=.75, not f(a)=.75.

**Plato** · August 4th 2009, 05:17 PM

$P(X \leqslant a) = \left\{ {\begin{array}{*{20}c}<br /> {0,} & {a \leqslant 0} \\<br /> {\int_0^a {4x(1 - x^2 )dx,} } & {0 < a \leqslant 1} \\<br /> {1,} & {1 < a} \\ \end{array} } \right.$

**Flyingdutchman** · August 4th 2009, 10:27 PM

yeah sorry about that, F, F, F, F!!! Bad dutchman!

**matheagle** · August 5th 2009, 10:23 PM

Originally Posted by Flyingdutchman

yeah sorry about that, F, F, F, F!!! Bad dutchman!

don't F me
sry another bad pun

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