Find the median of the probability distribution

[Phox]

Homework Statement

f(x) = C|x-2| for 0 <= x <= 3
f(x) = 0 otherwise

Homework Equations

The Attempt at a Solution

Solved for C, found it to be (2/5).

So.. I'm confused how to set up my integral here. I tried integral(2/3(x-2)dx) from m to 3 = 1/2. That didn't yield the correct result.

 

[SammyS]

Homework Statement

f(x) = C|x-2| for 0 <= x <= 3
f(x) = 0 otherwise

Homework Equations

The Attempt at a Solution

Solved for C, found it to be (2/5).

So.. I'm confused how to set up my integral here. I tried integral(2/3(x-2)dx) from m to 3 = 1/2. That didn't yield the correct result.

If you found C to be 2/5, then the integral should be

\displaystyle \int_{M}^{3}\frac{2}{5}|x-2|\,dx \ .​

Set that equal to 1/2 and solve for M, although it might be easier to integrate from 0 to M.

Graph the integrand to see why.

 

[Phox]

If you found C to be 2/5, then the integral should be

\displaystyle \int_{M}^{3}\frac{2}{5}|x-2|\,dx \ .​

Set that equal to 1/2 and solve for M, although it might be easier to integrate from 0 to m.

Graph the integrand to see why.

Yes, actually that's what I meant to say. I don't understand how I can integrate the function having the absolute value there. I understand how to do it if it were say.. integrating from 0 to 3. But since we don't know what m is I don't understand.

And neither does wolfram alpha lol

 

[SammyS]

Yes, actually that's what I meant to say. I don't understand how I can integrate the function having the absolute value there. I understand how to do it if it were say.. integrating from 0 to 3. But since we don't know what m is I don't understand.

And neither does wolfram alpha lol

Did you graph the function?

If x ≤ 2, then |x - 2| = 2 - x .

If x ≥ 2, then |x - 2| = x - 2 .

 

[Phox]

Yes, and I understand that. What I don't understand is where m lies. Is it in the range of <2 or >2

 

[Ray Vickson]

Yes, and I understand that. What I don't understand is where m lies. Is it in the range of <2 or >2

That is what you need to figure out. If in doubt, try it both ways to see what happens! However, careful examination of your graph should be enough.