Statistics: mean/expected value of an continuous distribution

davidhansson
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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
 
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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
 
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