Expected Value and integration

[rhyno89]

Homework Statement

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)

Homework Equations

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 can't 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 can't figure out the bounds. Is it the value of 0,1 - the value of .5,1 or something else all together

Thanks in advance

 

[Mark44]

Homework Statement

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)

Homework Equations

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 can't 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 can't figure out the bounds. Is it the value of 0,1 - the value of .5,1 or something else all together

Thanks in advance

Try to be more consistent in your use of variables. You are using x and y interchangeably, and that's bound to lead to confusion.

$$E(x) = \int_{0.5}^1 3x^2 dx$$

You got the right antiderivative, but you should be working with a definite integral. Since you're concerned only with the interval [0.5, 1], you should expect the expected value to be a number somewhere in that interval.