This Exam P sample reference tests Uniform Distribution. For a payment below the cap, the payment CDF equals the uniform loss CDF, x/5. At the cap every possible loss produces a payment no larger than the cap, so the CDF jumps to one; this is choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis form extends the continuous uniform CDF to five and ignores the payment ceiling. The payment cannot exceed four.
CThe slope 0.25 would normalize a uniform variable on [0,4], but the underlying loss is uniform on [0,5] and the remaining probability accumulates at the cap.
DThis combines the wrong slope with the wrong upper endpoint. It neither preserves the loss probabilities below four nor respects the payment ceiling.
EThis shifts the lower support to one even though small positive losses generate equally small positive payments. There is no one-unit deductible.
Original practice · fully worked
Original variant: median of a thresholded rainfall reading
Daily rainfall depth R has cumulative distribution F_R(r)=r²/144 for 0≤r≤12. A drainage display reports Y=max(R-3,0), measured in centimeters. Determine the median of the displayed reading Y.
A 3.000
B 4.485
C 5.485
D 6.000
E 8.485
Variant answer in brief
For positive y, F_Y(y)=F_R(y+3)=(y+3)²/144. Setting this equal to one half gives y=6√2-3=5.485, choice C.
Setup
Setup
The threshold transformation places all rainfall at or below three into a point mass at zero.
Y=max(R−3,0),Pr(Y=0)=FR(3)=161
Model
Model
Because the atom at zero is below one half, the median is positive. For positive y, translate the displayed threshold back to rainfall.
FY(y)=Pr(R≤y+3)=144(y+3)2,0<y<9
Compute
Compute
Set the transformed CDF equal to one half and take the nonnegative root.
144(m+3)2=21
m+3=62,m=62−3=5.485281
Answer
Answer
The median displayed depth is about 5.485 centimeters.
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