Question

Suppose f(x) = 3x^2/2 for - 1 < x < 1. Determine the mean and variance of X.

Answer 1

$M(X)=\int_{-\infty}^{+\infty} xf(x)dx={\frac 3 2}\int_{-1}^{1}x^3 dx={\frac {3x^4} 8}|_{-1}^1=0$

$V(X)=M(X^2)-M^2(X)=M(X^2)=\int_{-\infty}^{+\infty} x^2f(x)dx={\frac 3 2}\int_{-1}^{1}x^4 dx={\frac {3x^5} {10}}|_{-1}^1={\frac 3 5}$