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I'm not sure about my solution for following question:

A shop is selling item $X_1$ on an average day 20 times for 2\$ with $\sigma_1=5$. Item $X_2$ is sold 40 times for 3\$ with $\sigma_2=10$. You have 100\$ fixed costs. What can you say about the income? (sorry for the bad english)

I started with: (D=daily income)

$D=20 \cdot X_1+40 \cdot X_2 - 100$

$E(D)=E(20 \cdot X_1+40 \cdot X_2 - 100)=E(20 \cdot X_1)+E(40 \cdot X_2)+E(-100)= 20 \cdot E(X_1)+40 \cdot E(X_2) - 100 = 20 \cdot 2 + 40 \cdot 3 - 100 = 60$

60\$ daily income seems logical.

$Var(D)=Var(20 \cdot X_1+40 \cdot X_2 - 100)=Var(20 \cdot X_1)+Var(40 \cdot X_2)+Var(-100)= 400 \cdot Var(X_1) + 1600 \cdot Var(X_2) = 400 \cdot 25 + 1600 \cdot 100 = 170.000$

That seems very unrealistic. Where is my mistake?

Thanks a lot in advance.

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up vote 2 down vote accepted

Let $Y_1$ be the number of items of type $X_1$ sold, and let $Y_2$ be the number of items of type $X_2$ sold.

I think the information supplied is trying to tell us that $Y_1$ has mean $20$ and standard deviation $5$, while $Y_2$ has mean $40$ with standard deviation $10$.

Let $Z$ be the daily net income (after fixed costs). Then $Z=2Y_1+3Y_2-100$.

Then $E(Z)=(2)(20)+(3)(40)-100$.

To compute the variance we need to assume that $Y_1$ and $Y_2$ are independent. Then the variance of $Z$ is $(2^2)\sigma_1^2+(3^2)\sigma_2^2$.

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Ah I see, I misunderstood the information and your idea sounds logical, because the price is always 2 \$, just the amount of sold items can change. – ulead86 Dec 17 '12 at 17:51
Yes, the problem you had was with deciding what the random variables are. – André Nicolas Dec 17 '12 at 17:53
But is $\sigma^2=1000$ of Z not a little bit too high? – ulead86 Dec 17 '12 at 17:56
For standard deviation, you will take the square root, get about $32$. Makes sense, the main contributor is Item $2$. The amount of that sold has standard deviation $10$, and that gets multiplied by $3$ in income calculation, since the price is $3$. – André Nicolas Dec 17 '12 at 18:03
ah of course, thats fine. – ulead86 Dec 17 '12 at 20:28

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