Question

Every day, the number of network blackouts has a distribution (probability mass function)

Answer 1

Expected Mean, E(X), is obtained by multiplying each pair of $x$ and its $P(x)$ and add up the answers

E(X) = (0×0.7) + (1×0.2) + (2×0.1) = 0.4

The formula to calculate the variance, Var(X), is given by E(X)² - (E(X))²

E(X²) = (0²×0.7) + (1²×0.2) + (2²×0.1) = 0+0.2+0.4 = 0.6
(E(X))² = (0.4)² = 0.16

Var(X) = 0.6 - 0.16 = 0.44

Translating these answers into the context we have

E(Y) = 0.4×500 = $200
Var(Y) = $110