This Exam P sample reference tests Deductibles. A single deductible payout has mean 5,625 and variance 24,609,375. The 200-loss aggregate is approximately normal, and the two bounds give interval probability 0.8201, choice D.
For a Uniform(0,20000) loss, apply the 5000 deductible to form the payment Y and calculate its first two moments.
X∼Unif(0,20000)
Y=(X−5000)+
Model
Model
The single-payment mean is 5625 and variance is 24,609,375. Independence determines the mean and standard deviation of the 200-payment aggregate.
E[Y]=5625
E[Y2]=56250000
Var(Y)=24609375
Compute
Compute
The aggregate bounds standardize to -1.7817 and 1.0690. The normal cumulative-probability difference between them is 0.8201.
E[S]=1125000,SD(S)=200(24609375)=70156.1
zL=−1.7817,zU=1.0690
Φ(1.0690)−Φ(−1.7817)=0.8201
Answer
Answer
The approximate interval probability is 0.8201, corresponding to choice D.
0.8201(D)
Test the method on unseen problems
Probability Univariate Distributions Practice Workbook
30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
CThe value 0.1799 is the complement of the desired interval probability: 1-0.8201.
EThe value 0.8575 is approximately the normal cumulative probability at the upper z-bound 1.0690 alone. It fails to subtract the cumulative probability below the lower bound.
Original practice · fully worked
Original variant: aggregate payout from a discrete protection plan
A protection plan pays nothing for half of devices, 100 dollars for three tenths, and 300 dollars for the rest. For 100 independent devices, approximate the probability total payout is between 11,000 and 15,000 dollars.
B 0.0392
A 0.1056
C 0.2578
D 0.7422
E 0.9608
Answer and calculation
One payout has mean 90 and variance 12,900. The aggregate mean is 9,000 and standard deviation 1,135.78, so the interval probability is about 0.0392.
Setup
Setup
The device payout distribution has values 0, 100, and 300 with probabilities 0.5, 0.3, and 0.2.
E[Y]=0.3(100)+0.2(300)=90
E[Y2]=0.3(10000)+0.2(90000)=21000
Model
Model
The single-device mean is 90 and variance is 12,900, giving aggregate mean 9,000 and standard deviation 1,135.78 for 100 devices.
Var(Y)=21000−902=12900
E[S]=9000,SD(S)=1290000=1135.78
Compute
Compute
The interval endpoints standardize to 1.761 and 5.283. The normal area between them is approximately 0.0392.
zL=1135.7811000−9000=1.761
zU=1135.7815000−9000=5.283
P≈Φ(5.283)−Φ(1.761)=0.0392
Answer
Answer
Thus the approximate interval probability is 0.0392, corresponding to choice B.
