$Y = a X + b, L$

Y = 3 X + 4 X \sim N (μ = 2, σ = 5) L = 2 H_{1} - 3 H_{2} + H_{3} H_{i} \sim iid N (μ = 2, σ^{2} = 3)

Finding $E (Y)$ by plugging $μ$ in, $V (Y)$ if iid by directly plugging $σ^{2}$ into the squared coefficients.

$E (Y)$

3 \times 2 + 4 = 10

(2 - 3 + 1) \times 2 = 0

Problem	Formula (the first line)	Information Used (mu)	Calculation	Answer
E(Y)	$E (3 X + 4) = 3 E (X) + 4$	$E (X) = 2$	$3 (2) + 4 = 6 + 4$	10
E(L)	$E (2 H_{1} - 3 H_{2} + H_{3}) = 2 E (H_{1}) - 3 E (H_{2}) + E (H_{3})$	$E (H_{i}) = 2$	$2 (2) - 3 (2) + 1 (2) = 4 - 6 + 2$	0

3^{2} \times 5^{2} = 9 \times 25 = 225

(4 + 9 + 1) \times 3 = 42

Problem	Formula	Information Used (sigma)	Calculation	Answer
V(Y)	$V (3 X + 4) = 3^{2} V (X)$	$V (X) = 5^{2} = 25$ (drop +b term)	$9 \cdot 25$	225
V(L)	$V (2 H_{1} - 3 H_{2} + H_{3}) = 2^{2} V (H_{1}) + (- 3)^{2} V (H_{2}) + 1^{2} V (H_{3})$	$V (H_{i}) = 3$ (and independence $\sim^{iid}$ )	$4 (3) + 9 (3) + 1 (3) = 12 + 27 + 3$	42