This Exam P sample reference tests Normal Distribution. This is a normal-tail inversion after translating a payment threshold back to a loss threshold. The 86th loss percentile is 633.032, so subtracting the 500 payment threshold gives a deductible of 133.032 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AA value near 36 can result from reading 0.14 as an area from the mean, obtaining z near 0.36, and reporting only the standard-deviation displacement rather than the deductible.
BThe value 39 follows from treating the probability 0.14 itself as a z-score: 525+100(0.14)-500=39.
CUsing z near 0.36 inside the otherwise correct threshold equation produces about 61; the required upper-tail quantile is z=1.0803, not 0.36.
DUsing the lower-tail percentile with z=-1.0803 produces a negative deductible near -83; taking its magnitude loses the direction of the tail.
Original practice · fully worked
Original variant: equipment-service reimbursements
A service program models each equipment repair cost as normal with mean 850 dollars and standard deviation 120 dollars. A deductible d applies to each repair. Only 8% of repairs generate a reimbursement greater than 300 dollars. Calculate d.
A 169
B 550
C 681
D 719
E 1019
Variant answer in brief
A reimbursement above 300 requires a repair cost above d+300. The 92nd percentile of the repair-cost distribution is 1018.609, leaving a deductible of 718.609 and choice D.
Setup
Setup
Represent the repair cost by X and the reimbursement by the positive part after the deductible.
X∼N(850,1202),R=(X−d)+
Model
Model
Translate the reimbursement event into a repair-cost event and express the upper tail as a percentile.
Pr(R>300)=Pr(X>d+300)=0.08
Pr(X≤d+300)=0.92
Compute
Compute
Use the 92nd standard-normal percentile and isolate d.
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