Company sells 2 kind of cars, Lamborghinis and Ferraris, and the sales of the cars are independent. From every sold Lamborghini the company gets 10000 dollar profit and for every Ferrari the company gets 1500 dollar profit. Expected value of sales per month for Lamborghini is 3.2 and expected value of sales per month for Ferrari is 7.5. Standard deviations of sales per month are 1.1 for Lamborghini and 2.9 for Ferrari. What is the expected value of total profit of company in month and what is the standard deviation of that total profit?
The right answers should be: EX = 40050 and D = 10813.53
I started doing it like this: EX = 10000*3.2 + 1500*7.5 = 43250
This obviously is wrong. Apparently I should also use the standard deviations given in the exercise while doing the calculations but I don't know how.
Best Answer
As @Raskolnikov said, your "correct" answers appear to be wrong. In general, when dealing with two random variables, $X$ and $Y$: $$ E(X + Y) = E(X) + E(Y)\\ Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y) $$ Here, $X$ and $Y$, the random variables representing sales-per-month, are independent, their covariance is $0$.
The next thing to remember is that constants are linear in expectation and quadratic in variance. In other words: $$ E(aX) = aE(X)\\ Var(aX) = a^2Var(X) $$ Profit is going to be profit-per-unit (given as $10000$ and $1500$ respectively) times the number of units ($X$ and $Y$).
So, the answer to your question would be: $$ E(aX + bY) = aE(X) + bE(Y) = 10000\cdot3.2 + 1500\cdot7.5 = \mathbf{43,250}\\ Var(aX + bY) = a^2Var(X) +b^2 Var(Y) = 10000^2\cdot1.1^2 + 1500^2\cdot2.9^2 = 139,922,500\\ SD(aX + bY) = \sqrt{Var(aX + bY)} = \sqrt{139,922,500} \approx \mathbf{11828.88} $$