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

1 Answer
Mar 16, 2016

#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#