This Exam P sample reference tests Uniform Distribution. The retained amount is 40% of the original uniform loss. Its 30th percentile is 0.40 × 4.40, or 1.76, so choice A is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis uses 0.30(0.60)(10)=1.80, multiplying the percentile rank, reimbursement rate, and upper endpoint while ignoring both the lower endpoint and the unreimbursed share.
CThis computes the reimbursed rather than retained amount: 0.60Q_X(0.30)=0.60(4.40)=2.64.
DThis reverses the percentile rank and uses the 70th percentile of U: 0.80+0.70(4.00-0.80)=3.04.
EThis stops at the original-loss percentile Q_X(0.30)=2+0.30(8)=4.40 and never applies the 0.40 retained fraction.
Original practice · fully worked
Original variant: reverse calibration quantile
An optical scanner's raw calibration score S is uniform, but its endpoints are not listed. Engineers report that Q_S(0.20)=18 and Q_S(0.80)=42. The display converts the score to D=50-0.5S. Determine the 25th percentile of D.
A 20
B 25
C 30
D 35
E 40
Variant answer in brief
The two supplied quantiles recover S as uniform on [10,50]. Because the display transformation is decreasing, its 25th percentile uses the 75th percentile of S and equals 30, so choice C is correct.
Setup
Setup
Write S as uniform on [a,a+d]. The gap between the reported percentile ranks determines the support width.
42−18=(0.80−0.20)d
d=40,a=18−0.20(40)=10
Model
Model
The display is a decreasing affine transformation, so lower display quantiles correspond to upper raw-score quantiles.
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