Suppose that 15 observations are chosen at random for the pdf f(Y) = 3y^2 on the interval 0 to 1. Let X denote the number that lie in the interval (.5,1). Find E(X)
The Attempt at a Solution
Ok so to get the expected value u integrate the pdf times the random variable.
Doing this is the integral 3y^3 and it turns to .75y^4
This is where I get confused. First of all I thought that it would have summed to 1 so that the number of X on that range would be 15, but it cant be it would be 11.25
Aside from that, to find the number over this range i kno is 15x the expected value but i cant figure out the bounds. Is it the value of 0,1 - the value of .5,1 or something else all together
Thanks in advance