This Exam P sample reference tests Compound Distributions. A recruit produces 0, 1, or 2 pensions with probabilities 0.6, 0.1, and 0.3. The total's normal approximation with continuity correction gives 0.9887, choice E.
A single recruit generates 0, 1, or 2 pensions with probabilities 0.6, 0.1, and 0.3.
P(W=0)=0.6,P(W=1)=0.1,P(W=2)=0.3
Model
Model
Compute the first two moments of the three-point count, giving mean 0.7 and variance 0.81 per recruit.
E[W]=0.7
E[W2]=1.3
Var(W)=1.3−0.72=0.81
Compute
Compute
For 100 recruits the total has mean 70 and standard deviation 9. Applying the 90.5 continuity-corrected boundary gives z=2.2778 and probability 0.9886.
E[S]=70,SD(S)=9
P(S≤90)≈Φ(990.5−70)=Φ(2.2778)=0.9886
Answer
Answer
The normal approximation rounds to 0.99, corresponding to choice E.
0.99(E)
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.
DThe value 0.93 corresponds to a substantially smaller standardized boundary than the computed z=2.2778.
Original practice · fully worked
Original variant: capacity needed for subscriber rewards
Each of 200 independent subscribers earns zero rewards with probability 0.50, one reward with probability 0.30, and two rewards with probability 0.20. Approximate the probability that no more than 160 rewards are issued, using a continuity correction.
A 0.6915
E 0.7734
C 0.8413
D 0.8944
B 0.9682
Answer and calculation
One subscriber has mean 0.7 and variance 0.61. The total has mean 140 and standard deviation 11.045; 160.5 corresponds to z=1.856, giving about 0.9682.
Setup
Setup
One subscriber earns 0, 1, or 2 rewards with probabilities 0.50, 0.30, and 0.20.
P(W=0,1,2)=(0.5,0.3,0.2)
Model
Model
The single-subscriber mean is 0.7 and variance is 0.61, so for 200 subscribers the total has mean 140 and standard deviation equal to the square root of 122.
E[W]=0.7
E[W2]=1.1
Var(W)=0.61
Compute
Compute
Use 160.5 for the inclusive upper boundary. It gives z=1.85598 and normal cumulative probability 0.9682.
E[S]=140,SD(S)=122=11.0454
z=11.0454160.5−140=1.85598
P(S≤160)≈Φ(1.85598)=0.9682
Answer
Answer
The approximate probability of no more than 160 rewards is 0.9682, corresponding to choice B.
