2. Incorrect. The answer is true, not false
Applying the definition of expectation for a continuous random variable:
E(Y) = ∫0∞ yf(y)dy
E(Y) = ∫0∞ y(2e-2y)
dy
E(Y) = 2 [ ye-2y/(-2)-2
e-2y/4] 0∞
E(Y) = 1/2(equivalent to 500 barrels of oil)
Regarding E(Y2), we have that:
E(Y2) = ∫0∞
y2f(y)dy
E(Y2) = 2∫0∞
y2e-2ydy
E(Y2) = 2 [y2e-2y/(-2)-2y
e-2y/(-8)]0∞
E(Y2) = 1/2