The problem involves finding the median of a probability distribution defined by the function f(x) = C|x-2| for the interval 0 <= x <= 3, with f(x) = 0 otherwise. Participants are exploring how to set up the integral to find the median and are discussing the implications of the absolute value in the function.
The discussion is ongoing, with participants sharing their attempts and questioning the setup of the integral. Some guidance has been offered regarding the integration limits and the behavior of the absolute value function, but no consensus has been reached on the correct approach.
Participants are navigating the challenge of integrating a piecewise function due to the absolute value, and there is uncertainty regarding the placement of m in relation to the value of 2.
If you found C to be 2/5, then the integral should bePhox said: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 said:If you found C to be 2/5, then the integral should be[itex]\displaystyle \int_{M}^{3}\frac{2}{5}|x-2|\,dx \ .[/itex]
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.
Did you graph the function?Phox said: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
Phox said:Yes, and I understand that. What I don't understand is where m lies. Is it in the range of <2 or >2