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

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

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Answer 1 · 2016-03-16T17:01:23

$Var = sigma^2 = int (x-mu)^2f(x) dx$ 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:
$mu=3int_-1^1 x^3dx = [3/4x^4]_-1^1 = 3/4[1-1]$
$sigma_0^2=3int_-1^1 x^4dx=3/5[x^5]_-1^1 = 6/5$

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

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