This Exam P sample reference tests Discrete Random Variables. This is an expected-payment calculation using linearity of expectation. The two benefit components have expectations 32.50 and 70.00, for a total of 102.50 and choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 90.00 replaces the expected filling count by the probability of at least one filling: 50(0.40)+350(0.20)=90. It loses the extra payments from two or three fillings.
CThe value 132.50 is 50(0.65)+500(0.20). It reimburses the second treatment at 100% instead of the stated 70%.
DThe value 250.00 is 3(50)+0.20(500). It treats the maximum of three fillings as certain and also ignores the partial-coverage percentage on the second treatment.
EThe value 400.00 is 50+350, the payment for exactly one of each treatment. It treats that particular outcome as certain instead of averaging over the count distributions.
Original practice · fully worked
Original variant: reimbursement by operating class
A cooperative classifies leased equipment as light-use with probability 0.75 and heavy-use with probability 0.25. The conditional expected numbers of replacement units are 1.2 and 3.6, respectively. A contract reimburses 40 credits per replacement unit and adds a 90-credit calibration allowance for heavy-use equipment. Calculate the expected reimbursement.
A 22.50 credits
B 72.00 credits
C 90.00 credits
D 94.50 credits
E 162.00 credits
Variant answer in brief
Averaging the two conditional replacement counts gives 1.8 units and 72 credits of unit reimbursement. The heavy-use allowance contributes 22.5 more, for 94.5 credits and choice D.
Setup
Setup
Let K be the replacement-unit count and H indicate the heavy-use class. Average the two conditional count means.
E[K]=0.75(1.2)+0.25(3.6)=1.8
E[H]=Pr(H=1)=0.25
Model
Model
Write the contract payment as the unit reimbursement plus the class-specific allowance.
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