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Every day, the number of network blackouts has a distribution (probability mass function)
x | 0 | 1 | 2
P(x)| 0.7| 0.2| 0.1
A small internet trading company estimates that each network blackout results in a $500
loss. Compute expectation and variance of this company’s daily loss due to blackouts.

I computed this E(x)= .4 , varX= .44, but I don't know how i can find the answer?

The answer is
E(Y ) = 200 dollars,
Var(Y ) = 110, 000 squared dollars.

Respuesta :

merlynthewhizz
Expected Mean, E(X), is obtained by multiplying each pair of [tex]x[/tex] and its [tex]P(x)[/tex] 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

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