Varience

[kuahji]

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

\int (x-1)^2 (1/4) dx with the limits -2 to 2

So it became
1/4\int x^2-2x+1 dx
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 :(.

[HallsofIvy]

Sorry, but it is you. Your "conceptual error" is in the formula for the mean.

Integrating any probability density will give you 1- that's not the "mean", it is the total probability that the result is somewhere in that interval which is, by definition, 1. The mean is the integral of x times the probility density function. Here, that is
\int_{-2}^2 x(1/4) dx.

[kuahji]

Oops. Nevermind, it is my mistake. Forgot to square the x & divide by two. >.<