This Exam P sample reference tests Uniform Distribution. For a limit m between 10 and 60, the limited payment has mean m-(m-10)²⁄¹⁰⁰. Setting this equal to 31 gives feasible root m=40, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the uncapped mean, (10+60)/2=35. It does not solve for the benefit limit that lowers the expected payment to 31.
BNormalizing the triangular shortfall by the upper endpoint 60 instead of the support width 50 gives a spurious root near 37.1 and can lead to choice 38.
DThe expected reduction is 35-31=4. Setting only the conditional mean excess (60-m)/2 equal to 4 gives m=52 but omits the probability that the limit is reached.
ESetting the maximum possible reduction 60-m equal to the expected reduction 4 gives m=56. A maximum shortfall is not an expected shortfall.
Original practice · fully worked
Original variant: calibrate a sensor saturation point
A sensor's uncapped load reading L is exponential with mean 8 units. The display records D=min(L,c), where c is an unknown saturation point. Engineers measure E[D]=6. Determine c.
A 2.301
B 5.545
C 8.000
D 11.090
E 16.636
Variant answer in brief
The expected limited exponential reading is 8(1-exp(−c/8)). Equating this to 6 gives exp(−c/8)=1/4 and c=8 ln 4=11.090, so choice D is correct.
Setup
Setup
Use the tail-integral formula for a nonnegative reading capped at c.
E[min(L,c)]=∫0cPr(L>t)dt
Pr(L>t)=e−t/8
Model
Model
Integrate the exponential survival function up to the saturation point.
E[D]=∫0ce−t/8dt
E[D]=8(1−e−c/8)
Compute
Compute
Set the limited mean equal to six and invert the exponential.
8(1−e−c/8)=6
e−c/8=41
c=8ln4=11.0903549…
Answer
Answer
The display saturates at approximately 11.090 units.
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