Question

Find the variance of the expected value of the probability distribution, X={1, 2, 3, 4, 5} P(X)={.10, .35, .15, .23, .17}

Answer 1

$EX=1⋅0.1+2⋅0.35+3⋅0.15+4⋅0.23+5⋅0.17=3.02\\ EX^2=1^2\cdot 0.1+2^2\cdot 0.35+3^2\cdot 0.15+4^2\cdot 0.23+5^2\cdot 0.17=10.78\\DX=EX^2-\left( EX \right) ^2=10.78-3.02^2=1.6596$