# linear transformations of mean and variance

So here is my problem....

At the start of a week, a coal mine has a high-capacity storage bin that is half full. During the week, 20 loads of coal are added to the storage bin. Each load of coal has a volume that is normally distributed with mean 1.50 cubic yards and standard deviation 0.25 cubic yards.

During the same week, coal is removed from the storage bin and loaded into 4 railroad cars. The amount of coal loaded into each railroad car is normally distributed with mean 7.25 cubic yards and standard deviation 0.50 cubic yards.

The amounts added to the storage bin or removed from the storage bin are mutually independent.

Calculate the probability that the storage bin contains more coal at the end of the week than it had at the beginning of the week.

Add=A~$N(1.5,0.0625)$ Remove=R~$N(7.25,0.25)$

I am applying the following linear transformation:

$Q=20A-4R$

My approach:

$E(Q)=20E(A)-4E(R)=20(1.5)-4(7.25)=1$

$Var(Q)=400Var(A)+16Var(R)=400(.0625)+16(.25)=29$

My problem:

Solution shows $Var(Q)= 2.25$

Their solution:

With each load of coal having mean 1.5 and standard deviation 0.25, twenty loads have a mean of 20(1.5) = 30 and a variance of 20(0.0625) = 1.25. The total amount removed is normal with mean 4(7.25) = 29 and standard deviation 4(0.25) = 1. The difference is normal with mean 30 – 29 = 1 and standard deviation sqrt(1.25 + 1) = 1.5. If D is that difference, then $P(D>0)=P(Z>\frac{0-1}{1.5}=-0.67)=0.7486$

My Question

Is the solution wrong? I thought that $Var(Q)=Var(20A-4R)=400Var(A)+16Var(R)$

This is from #235 in SOA handbook for probability problems.

– Sycorax
Commented Aug 5, 2016 at 19:06
• Is that better Abrial? Commented Aug 5, 2016 at 19:09
• Can you either provide an exact statement of the problem, or an accessible link to it? Perhaps there is something lost "in the translation" between what was written and what you have written here. Commented Aug 5, 2016 at 19:13
• I posted the entire problem Commented Aug 5, 2016 at 19:18
• Can someone please explain why are they adding the two variances and not subtracting? I thought we were looking for the difference between A and R? Commented Sep 8, 2017 at 20:25

Per the problem statement, the total amount added to the storage bin for the week is the sum of 20 i.i.d. N(1.5,0.0625) random variables, which is N(30,20 * 0.0625). Your calculation is based on making one draw from a N(1.5,0.0625) random variable, call it $A$, say, and then using $20A$. That is very different than the sum of 20 i.i.d. random variables having the same distribution as A. You made the corresponding error on the removal term.