Question 444189
The expected value of a discrete random variable x is
{{{E(x) = sum( x[i]p[i], i)}}} where {{{x[i]}}} are the outcomes and {{{p[i]}}} are their probabilities
Using the values x = -2, 4, 10; p = 0.8, 0.15, 0.05, we get
E(x) = -2*0.8 + 4*0.15 + 10*0.05 = -0.5
The variance of a discrete random variable x is
{{{sigma^2 = sum((x[i] - E(x))^2*p[i],i)}}}
So we have 
{{{sigma^2}}} = (-2 - -0.5)^2 + (4 - -0.5)^2 + (10 - -0.5)^2 = 10.35