This Exam P sample reference tests Insurance Payment Variables. This is an expected excess-loss calculation for a uniform loss. A deductible d leaves expected payment (100-d)²⁄²⁰⁰; setting that expression equal to 32 gives a remaining width of 80 and therefore d=20, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 9 is approximately √(80), obtained by taking an unnecessary second square root after finding the correct excess width 100-d=80.
BThe value 18 solves 50-d=32, treating payment as the unconditional mean loss minus the deductible and ignoring losses that do not exceed d.
DThe value 36 solves (100-d)/2=32, using the mean payment conditional on exceeding the deductible but omitting the probability of exceeding it.
EThe value 52 adds the correct deductible 20 to the expected payment 32. The question asks for the deductible itself, not their sum.
Original practice · fully worked
Original variant: expected payment under a mixed deductible
A repair loss is uniformly distributed from 0 to 10. Independently of the loss, a service contract uses a deductible of 2 with probability 0.40 and a deductible of 6 with probability 0.60. The contract pays the loss above the selected deductible. Calculate the expected payment.
A 0.800
B 1.280
C 1.760
D 2.000
E 3.200
Variant answer in brief
The expected payments under deductibles 2 and 6 are 3.2 and 0.8. Weighting them by probabilities 0.40 and 0.60 gives 1.76, choice C.
Setup
Setup
Use the uniform stop-loss formula at each possible deductible.
E[(X−d)+]=20(10−d)2,0≤d≤10
Model
Model
Evaluate the two conditional expected payments.
E[Y∣D=2]=2082=3.2
E[Y∣D=6]=2042=0.8
Compute
Compute
Average the conditional payments over the independent deductible selection.
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