This Exam P sample reference tests Poisson Distribution. This is a normal approximation to an aggregate of independent Poisson claim amounts. After applying the continuity correction, the standardized premium threshold is 0.964495 and the upper-tail probability is 0.167399, which selects choice C.
Let S be the aggregate payment. Add the means and variances of the 1,000 independent counts, then apply the fixed payment scale.
E[S]=100(1000)(1)=100,000
Var(S)=1002(1000)(1)=10,000,000
SD(S)=1001000=3162.277660…
Model
Model
The loaded premium is 103,000. Payments occur in increments of 100, so exceeding that amount means at least 1,031 claim units; the continuity-corrected boundary is 1,030.5 units, or 103,050 in payment units.
1.03(100,000)=103,000
S>103,000⟺S≥103,100
continuity boundary=103,050
Compute
Compute
Standardize the corrected boundary and evaluate the upper standard-normal tail.
z=3162.277660103,050−100,000=0.9644946864…
Pr(S>103,000)≈1−Φ(0.9644946864)=0.1673989856…
Answer
Answer
The approximated probability rounds to 0.167.
0.167(C)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AA probability near 0.001 is the upper tail beyond about three standard deviations. It results from treating the 3% loading itself as a z-score of 3 rather than converting the monetary margin to standard-deviation units.
BThis is the familiar upper tail 1-Φ(1)=0.1587. It rounds the corrected z-score 0.9645 to 1 before evaluating the tail and loses too much precision.
DThis is approximately 1-Φ(0.03)=0.4880, obtained by using the 3% loading directly as a standardized normal deviation.
EA probability of 0.500 treats the premium threshold as the aggregate mean and ignores the positive loading.
Original practice · fully worked
Original variant: reserve limit for device failures
A service plan covers 400 independent devices, each of which has probability 0.08 of failing during the year. The reserve is exhausted if more than 40 devices fail. Using a normal approximation with continuity correction, calculate the probability that the reserve is exhausted.
A 0.0400
B 0.0586
C 0.0702
D 0.0800
E 0.9414
Variant answer in brief
The failure count has mean 32 and standard deviation √(29.44). More than 40 means at least 41, so the corrected boundary is 40.5 and the normal upper tail is 0.0586, which selects choice B.
Setup
Setup
Model the annual failure count X as binomial and compute its first two moments.
X∼Binomial(400,0.08)
μ=400(0.08)=32
σ=400(0.08)(0.92)=29.44=5.425863987…
Model
Model
Exhaustion means X is at least 41. Replace this discrete event by the normal region above the midpoint 40.5.
Pr(X>40)=Pr(X≥41)≈Pr(Y>40.5)
Compute
Compute
Standardize the corrected boundary and evaluate the upper tail.
z=5.42586398740.5−32=1.566570784…
Pr(X>40)≈1−Φ(1.566570784)=0.0586075256…
Answer
Answer
The continuity-corrected normal approximation is 0.0586.
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