Statistics: mean/expected value of an continuous distribution

davidhansson
Messages
8
Reaction score
0
So, the exercise is to find the expected value of following distribution: f(x) = 0,02x 0<x<10

answer in the book says 6,67

As far as I knowe, the expected value is calculated by the Integral of x * f(x) between 0 and 10, in this case!
It looks like this won't give the result 6,67!

what am I doing wrong?

thanks/ David
 
Physics news on Phys.org
I'm not sure. The expected value is this integral:
<br /> E(X) = \int_0^{10} x f(x) \, dx<br />

What do you get when you evaulate it?
 
davidhansson said:
So, the exercise is to find the expected value of following distribution: f(x) = 0,02x 0<x<10

answer in the book says 6,67

As far as I knowe, the expected value is calculated by the Integral of x * f(x) between 0 and 10, in this case!
It looks like this won't give the result 6,67!

what am I doing wrong?

thanks/ David

We cannot possibly help if you don't show us what you did in detail.
 
oops, it's actually 6,67 as the book says,, it was just a computational mistake by me.

thank you anyways guys!

thanks/ David
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top