Question

What is the variance of X if it has the following probability density function?: f(x) = { 3x2 if -1 < x < 1; 0 otherwise}

Answer 1

`Var = sigma^2 = int (x-mu)^2f(x) dx` which cann be written as:
`sigma^2 = intx^2f(x) dx-2mu^2+mu^2=intx^2f(x)dx - mu^2`

`sigma_0^2=3int_-1^1 x^4dx=3/5[x^5]_-1^1 = 6/5`

Explanation:

I am assuming that question meant to say
`f(x)=3x^2 " for " -1 < x < 1; 0 " otherwise"`
Find the variance?
`Var = sigma^2 = int (x-mu)^2f(x) dx`
Expand:
`sigma^2 = intx^2f(x) dx-2mucancel(intxf(x)dx)^mu+mu^2cancel(intf(x) dx)^1`
`sigma^2 = intx^2f(x) dx-2mu^2+mu^2=intx^2f(x)dx - mu^2`

substitute
`sigma^2 = 3int_-1^1 x^2 *x^2dx -mu^2 = sigma_0^2+mu^2`
Where, `sigma_0^2=3int_-1^1 x^4dx` and `mu=3int_-1^1 x^3dx`
So let's calculate `sigma_0^2 " and " mu`
by symmetry `mu=0` let see:
`mu=3int_-1^1 x^3dx = [3/4x^4]_-1^1 = 3/4[1-1]`
`sigma_0^2=3int_-1^1 x^4dx=3/5[x^5]_-1^1 = 6/5`