1. ## 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.

2. 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:

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

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

Which makes $\displaystyle c=0.75$

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

$\displaystyle F(x_{25})=0.25$

and

$\displaystyle F(x_{75})=0.75$

for roots in $\displaystyle [-1,1]$

CB

3. ## .

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

4. 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

,

,

### questions on calculating quartile from CDF

Click on a term to search for related topics.