# CDF and Quartiles

• Feb 19th 2010, 07:12 AM
SterlingM
CDF and Quartiles
I am given a probability density function -

f(x) = (1/2)(1-x), for -1 <= x <= 1

For finding the cdf, I distribute the (1/2) and integrate to

F(x) = (1/2)x - (1/4)x^2

My question is how am I able to use this to find the upper and lower quartiles? Every other problem I've done finding the quartiles has just been a linear F(x). Does having a quadratic mean I will have...two upper and low quartiles? That doesn't seem right to me. Any help would be appreciated. Thanks.
• Feb 20th 2010, 01:10 AM
CaptainBlack
Quote:

Originally Posted by SterlingM
I am given a probability density function -

f(x) = (1/2)(1-x), for -1 <= x <= 1

For finding the cdf, I distribute the (1/2) and integrate to

F(x) = (1/2)x - (1/4)x^2

My question is how am I able to use this to find the upper and lower quartiles? Every other problem I've done finding the quartiles has just been a linear F(x). Does having a quadratic mean I will have...two upper and low quartiles? That doesn't seem right to me. Any help would be appreciated. Thanks.

You have the cumulative distribution function wrong.

What you shuold have is:

$F(x)=\frac{2x-x^2}{4}+c$

for $x \in [-1,1]$ and that $F(-1)=0$.

Which makes $c=0.75$

Now to find the first and third quartiles you solve the equations (which are quadratics and solvable using the quadratic formula):

$F(x_{25})=0.25$

and

$F(x_{75})=0.75$

for roots in $[-1,1]$

CB
• Feb 21st 2010, 10:02 AM
SterlingM
.
The integration is definitely right, but how did you get it? Is there a quotient rule for integration?
• Feb 21st 2010, 10:40 AM
CaptainBlack
Quote:

Originally Posted by SterlingM
The integration is definitely right, but how did you get it? Is there a quotient rule for integration?

I got it in exactly the same way you did, I integrated the powers and then tidied up. You left off the constant of integration the value of which is determined by the end point conditions.

CB