Results 1 to 6 of 6

- Jan 3rd 2013, 07:26 AM #1

- Joined
- Jan 2013
- From
- Sverige
- Posts
- 19

- Jan 3rd 2013, 07:38 AM #2

- Jan 3rd 2013, 07:40 AM #3

- Joined
- Apr 2005
- Posts
- 18,448
- Thanks
- 2531

## Re: Var + e

x and y, separately, can be 0 or 1 so x+ y can be 0, 1, or 2.

What must x and y be so that x+ y= 0? What is the probability of that?

What must x and y be so that x+ y= 1? What is the probability of that?

What must x and y be so that x+ y= 2? What is the probability of that?

- Jan 3rd 2013, 07:48 AM #4

- Joined
- Jan 2013
- From
- Sverige
- Posts
- 19

- Jan 3rd 2013, 08:08 AM #5

- Jan 3rd 2013, 09:01 AM #6

- Joined
- Apr 2005
- Posts
- 18,448
- Thanks
- 2531

## Re: Var + e

Your probability distribution for z= x+ y, 0,20 that z= 0, 0,55 that z= 1, and 0,25 that z= 2, is correct. Now, the expected value is 0(0,20)+ 1(0,55)+ 2(0,25)= 1,10+ 0,50= 1,60. The variance, then, is 0,20(0- 1,60)^2+ 0,55(1- 1,60)^2+ 0,25(2- 1,60)^2. You used the values, not the variation from the expected value.