## anonymous 4 years ago Can somebody give me advice on how to find variance and the expected value; when one considers the function f(Y) = 1/9 is when 0 <= y <= 9 and the probability is 0 everywhere else when y is not in that interval(0-9). please note that this is a uniform distribution function and Y is continous. Im struggling with the integrals. How can you find an expected value for Y, when all values in the interval [0,9] have the same probability? thanks for your help!

1. dumbcow

$E(X) = \int\limits_{0}^{9}p(y)*y = \int\limits_{0}^{9}\frac{y}{9} dy$

2. dumbcow

Variance is E(X^2) - E(X)^2

3. anonymous

why isn't this oh, what dumbcow said

4. anonymous

But when I put in 9 for Y: 1/9*9 = 1. But I think the answer is 4.5. Note that I am having trouble with integrals, even basic ones. Is the expected value equal to the average value in this case?

5. dumbcow

yes it is same as average value since it is uniformly distributed. E(X) = 4.5

6. anonymous

ah I see, thank you!

7. dumbcow

no problem :)