This Exam P sample reference tests Uniform Distribution. The reimbursement has a 25% atom at zero. Its 20th percentile is therefore zero, while its median corresponds to the original loss median 500 and equals 500-250=250. The difference is 250, choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 225 is the difference between the conditional-on-positive median 375 and conditional-on-positive 20th percentile 150. The question includes all losses, including the zero-payment atom.
CThe value 300 is the original loss-quantile difference 500-200. It overlooks that the deductible collapses the entire lowest 25% of losses to a reimbursement of zero.
DThe value 375 is the median reimbursement conditional on a positive reimbursement. Conditioning removes the zero-payment losses and changes the target population.
EThe value 500 is the median loss before applying the deductible. It is not the median reimbursement or the requested difference.
Original practice · fully worked
Original variant: expected reimbursement from a capped layer
A repair cost X is uniformly distributed from 0 to 120 credits. A plan reimburses 50% of the portion of X between 20 and 80 credits, so its maximum reimbursement is 30 credits. Calculate the expected reimbursement.
A 17.500 credits
B 20.833 credits
C 22.500 credits
D 30.000 credits
E 35.000 credits
Variant answer in brief
The plan pays half of a 60-credit layer above 20. The layer's expected amount is the survival integral from 20 to 80, equal to 35, so the expected reimbursement is 17.5 credits and choice A.
Setup
Setup
Express the contract as coinsurance applied to a limited loss layer.
Y=0.50min((X−20)+,60)
Model
Model
Use a survival integral over the attachment-to-limit interval.
E[Y]=0.50∫2080Pr(X>t)dt
Pr(X>t)=1−120t
Compute
Compute
Integrate the uniform survival function and apply the coinsurance factor.
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