Solved – linear transformations of mean and variance

expected valuevariance

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.

Best Answer

The problem is not as you stated (reflecting the beginning of your solution process). That's where your error is.

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.