# Statistics: mean/expected value of an continuous distribution

1. Sep 22, 2014

### davidhansson

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

2. Sep 22, 2014

I'm not sure. The expected value is this integral:
$$E(X) = \int_0^{10} x f(x) \, dx$$

What do you get when you evaulate it?

3. Sep 22, 2014

### Ray Vickson

We cannot possibly help if you don't show us what you did in detail.

4. Sep 22, 2014

### davidhansson

oops, it's actually 6,67 as the book says,, it was just a computational mistake by me.

thank you anyways guys!

thanks/ David