**8. Correct. The answer is true**

(a) From
Problem 2, we know that the mean of Y is m = E(Y)=1/2.
Then, the variance is:

Var(Y)= E[(Y-m)^{2}] = E[(Y-(1/2))^{2}]

Var(Y)= ∫_{-∞∞}
(y - (1/2))^{2 }f (y) dy

Var(Y)= ∫_{-∞∞}
(y - (1/2))^{2 }(2e^{-2y })
dy = Var(Y) = 1/4^{}

Alternatively, we can calculate the variance by using a theorem which establishes that:

Var[Y] = E[Y^{2}] - (E[Y])^{2}

Var(Y) = E[(Y-m)^{2}]

Var(Y) = E(Y^{2}) - [E(y)]^{2}

Var(Y) = (1/2) - (1/2)^{2} =1/4

(b) s = [Var(Y)]^{0.5} = 1/2 (the
standard deviation is equal to 500 barrels).