For the payment Y=(X-1000) positive part, direct integration gives E[Y]=6533.33 and E[Y^2]=60,977,777.78. Thus SD(Y)=4277.07, which corresponds to choice B.
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How to solve this Transformations of Random Variables question
Answer in brief
For the payment Y=(X-1000) positive part, direct integration gives E[Y]=6533.33 and E[Y^2]=60,977,777.78. Thus SD(Y)=4277.07, which corresponds to choice B.
Setup
Setup
Let X be the modeled loss and Y the insurer payment after applying the ordinary deductible.
X∼Unif(0,15000)
Y=(X−1000)+=max(X−1000,0)
Model
Model
The payment has a point mass at zero, so calculate both raw moments from the original loss density rather than treating Y as an ordinary uniform variable.
E[Yk]=150001∫100015000(x−1000)kdx,k=1,2
Compute
Compute
Evaluate the two integrals and center the second moment.
E[Y]=2(15000)140002=6533.3333…
E[Y2]=3(15000)140003=60977777.7778…
Var(Y)=E[Y2]−E[Y]2=18293333.3333…
SD(Y)=18293333.3333…=4277.0706…
Answer
Answer
Rounding to the precision of the choices gives a payment standard deviation of 4277.
4277(B)
Error diagnosis
Why the other letters tempt
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis nearby underestimate is inconsistent with the exact first and second payment moments; it reflects an excessive centering subtraction or a numerical integration error.
CThe value 15000 divided by sqrt(12), approximately 4330, is the standard deviation of the original uniform loss and ignores the deductible transformation.
DThis exceeds the original loss standard deviation even though the payment map x to (x-1000) positive part is 1-Lipschitz, so it fails a useful variance sanity check.
EThe value 6533 is the expected payment, not its standard deviation; it stops after computing the first raw moment.
Original practice · fully worked
Original variant: workshop repair reimbursement
A makerspace reimburses member equipment repairs only for the portion above 2,000 dollars. For an accepted breakdown report, the total repair cost is modeled as uniform from 0 to 12,000 dollars. Calculate the standard deviation of the reimbursement.
A 2,887 dollars
B 3,227 dollars
C 3,464 dollars
D 4,167 dollars
E 5,270 dollars
Variant answer in brief
The reimbursement has first moment 4166.67 and second moment 27,777,777.78. Its variance is 10,416,666.67, giving a standard deviation of 3227.49 dollars and choice B.
Setup
Setup
Write C for repair cost and R for the reimbursement.
C∼Unif(0,12000)
R=(C−2000)+
Model
Model
Integrate over the cost region in which the reimbursement is positive.
E[Rk]=120001∫200012000(c−2000)kdc,k=1,2
Compute
Compute
Calculate both raw moments before taking the square root of the variance.
E[R]=2(12000)100002=4166.6667…
E[R2]=3(12000)100003=27777777.7778…
Var(R)=10416666.6667…
SD(R)=3227.4861…
Answer
Answer
The reimbursement standard deviation rounds to 3,227 dollars.
3227 dollars(B)
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