This Exam P sample reference tests Limited Loss Variables. Split the capped payment at the limit. The two expectation pieces total 1.9, which agrees with official choice D.
Let B be the payment obtained by capping the loss Y at 10. The density is supported above 1, so losses below the cap and losses above the cap must be handled separately.
B=min(Y,10)
fY(y)=2y−3,y>1
Model
Model
For losses between 1 and 10 the payment equals the loss, whereas above 10 the payment is exactly 10. Splitting the expectation at the cap accounts for both regions.
E[B]=∫110y2y−3dy+10∫10∞2y−3dy
Compute
Compute
The below-cap first-moment integral contributes 1.8. The probability above the cap is 0.01, so capped losses contribute a further 10 × 0.01 = 0.1.
∫1102y−2dy=1.8
10Pr(Y>10)=10(10−2)=0.1
E[B]=1.8+0.1=1.9
Answer
Answer
Adding the two disjoint contributions gives an expected capped payment of 1.9, which is choice D.
1.9(D)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 1.0 retains only the certain first unit in the survival-integral representation and omits the positive tail contribution from 1 to 10.
CThe value 1.8 is the below-cap integral alone. It omits the cap payment 10P(Y>10)=0.1 made when the loss exceeds 10.
EThe value 2.0 is the uncapped mean E[Y]. It ignores the reduction produced by limiting every payment to 10.
Original practice · fully worked
Original variant: capped emergency-power demand
A remote clinic’s emergency energy demand D has survival function P(D>x)=(3/x)² for x at least 3 kilowatt-hours. A battery can supply at most 12 kilowatt-hours. Find the expected energy actually supplied.
C 3.750 kWh
B 4.500 kWh
A 5.250 kWh
D 5.625 kWh
E 6.000 kWh
Variant answer in brief
Integrating the survival probability up to the battery ceiling gives 3 plus 9 times the integral of x⁻² from 3 to 12, or 5.25 kWh.
Setup
Setup
Let the supplied energy be the smaller of D and 12. The survival probability is one below the minimum demand 3 and equals 9 divided by x squared thereafter.
SD(x)=1(0≤x<3),SD(x)=9x−2(x≥3)
Model
Model
For a nonnegative demand, the expected amount delivered up to capacity 12 is the integral of its survival function from zero to 12.
E[min(D,12)]=∫012SD(x)dx
Compute
Compute
The interval from 0 to 3 contributes 3. Integrating the inverse-square survival function from 3 to 12 contributes 2.25, for a total of 5.25.
E[min(D,12)]=3+9∫312x−2dx=3+9(31−121)=5.25
Answer
Answer
Therefore the battery supplies 5.250 kWh on average, corresponding to choice A.
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