- #1
kuahji
- 394
- 2
find the variance for f(x)= 1/4 for -2<x<2 & 0 elsewhere
The first thing I did was find the expected value, which was 1 (just integrated the original function from -2 to 2). Then I set up the next part as
[tex]\int (x-1)^2 (1/4) dx[/tex] with the limits -2 to 2
So it became
1/4[tex]\int x^2-2x+1 dx[/tex]
1/4(8/3-4+2)-1/4(-8/3+4-) =1/3
However, the book has the answer 4/3, which is what you get if you don't multiply through by 1/4. Is this a conceptual error on my part, or a book error? Usually it ends up being me who is wrong :(.
The first thing I did was find the expected value, which was 1 (just integrated the original function from -2 to 2). Then I set up the next part as
[tex]\int (x-1)^2 (1/4) dx[/tex] with the limits -2 to 2
So it became
1/4[tex]\int x^2-2x+1 dx[/tex]
1/4(8/3-4+2)-1/4(-8/3+4-) =1/3
However, the book has the answer 4/3, which is what you get if you don't multiply through by 1/4. Is this a conceptual error on my part, or a book error? Usually it ends up being me who is wrong :(.