Quote (furbyjs @ Oct 8 2013 09:48pm)
W is just a variable, in this case it represents WEIGHT, which is what's normally distributed (weight of pizzas in the problem).
So would E(W^2 + 2W) just be the same as E(W), since it's originally Normally Distributed?
So we know expectation is a linear operator right?
So first split this into
E(W^2) + E(2W)
also by linearity of expectation
=E(W^2) + 2E(W)
now we know that variance is E(W^2) - E(W)^2 right?
so this is the same as
= variance(W) + E(W)^2 + 2E(W)