This Exam P sample reference tests Central Limit Theorem. The aggregate payment has mean 700,000 and standard deviation 185,768.67. Its 99th-percentile normal approximation is 1,132,162.56, so the smallest offered fund above that amount is 1,150,000, choice D.
Represent each covered life by an independent death indicator and write the aggregate claim payment as their scaled sum.
Ii∼Bernoulli(0.014),S=50000i=1∑1000Ii
Model
Model
Independence makes the means and variances additive. A normal approximation is appropriate for the large portfolio.
μS=50000(1000)(0.014)=700000
σS2=500002(1000)(0.014)(0.986)=34510000000
Compute
Compute
Use the one-sided 99th standard-normal quantile and compare the result with the available fund levels.
σS=185768.6734
q0.99≈μS+z0.99σS
q0.99=700000+2.3263479(185768.6734)=1132162.56
Answer
Answer
A fund of 1,100,000 is below the approximated percentile, while 1,150,000 is the next available level.
1150000(D)
Test the method on unseen problems
Probability Multivariate & Risk Models Practice Workbook
30 original multivariate 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.
AThe amount 750,000 pays for 15 claims. It merely rounds the expected count of 14 upward by one and includes no 99th-percentile uncertainty margin.
BThe amount 850,000 corresponds to 17 claims. It results from adding a roughly one-standard-deviation count buffer, rounded down to three deaths, rather than using the 99th-percentile multiplier.
CUsing the one-sided 95th-percentile multiplier 1.645 gives about 20.11 claims; rounding that count upward produces 21 payments, or 1,050,000. The requested confidence level is higher.
EThe amount 1,400,000 simply doubles the expected payment, equivalently funding 28 deaths. Doubling the mean is not the normal-quantile calculation.
Original practice · fully worked
Original variant: percentile across correlated weekly components
For each of 50 independent weeks, a center records workload A and support load B. Their weekly means are 8 and 5, their variances are 9 and 4, and Cov(A,B)=-3. The score for a week is A+2B. Using a normal approximation for the sum of the 50 weekly scores, calculate its 95th percentile.
A 900.0
B 925.5
C 941.9
D 950.0
E 958.2
Answer and calculation
One weekly score has mean 18 and variance 13 after accounting for the negative covariance. The 50-week total has mean 900 and variance 650, so its normal 95th percentile is 941.9 and choice C.
Setup
Setup
Let W be one weekly score and T the sum across the 50 independent weeks.
W=A+2B,T=i=1∑50Wi
Model
Model
Compute the weekly mean and variance, retaining the within-week covariance term.
E[W]=8+2(5)=18
Var(W)=9+22(4)+2(1)(2)(−3)=13
E[T]=50(18)=900,Var(T)=50(13)=650
Compute
Compute
Apply the 95th standard-normal quantile to the aggregate approximation.
