Expectation is linear, meaning
E(a X + b Y) = E(a X) + E(b Y)
= a E(X) + b E(Y)
If X = 1 and Y = 0, we see that the expectation of a constant, E(a), is equal to the constant, a.
Use this property to compute the expectations:
E(X + 10) = E(X) + E(10) = $110
E(5Y) = 5 E(Y) = $450
E(X + Y) = E(X) + E(Y) = $190
Variance has a similar property:
V(a X + b Y) = V(a X) + V(b Y) + Cov(X, Y)
= a^2 V(X) + b^2 V(Y) + Cov(X, Y)
where "Cov" denotes covariance, defined by
E[(X - E(X))(Y - E(Y))] = E(X Y) - E(X) E(Y)
Without knowing the expectation of X Y, we can't determine the covariance and thus variance of the expression a X + b Y.
However, if X and Y are independent, then E(X Y) = E(X) E(Y), which makes the covariance vanish, so that
V(a X + b Y) = a^2 V(X) + b^2 V(Y)
and this is the assumption we have to make to find the standard deviations (which is the square root of the variance).
Also, variance is defined as
V(X) = E[(X - E(X))^2] = E(X^2) - E(X)^2
and it follows from this that, if X is a constant, say a, then
V(a) = E(a^2) - E(a)^2 = a^2 - a^2 = 0
Use this property, and the assumption of independence, to compute the variances, and hence the standard deviations:
V(X + 10) = V(X) ==> SD(X + 10) = SD(X) = $90
V(5Y) = 5^2 V(Y) = 25 V(Y) ==> SD(5Y) = 5 SD(Y) = $40
V(X + Y) = V(X) + V(Y) ==> SD(X + Y) = √[SD(X)^2 + SD(Y)^2] = √8164 ≈ $90.35