This Exam P sample reference tests Covariance. One person's combined hours have mean 70 and variance 100 after including covariance. For 100 people, 7100 is one standard deviation above the total mean, giving 0.8413 and choice B.
For one person, let Z=X+Y be the combined hours. Its mean is 70, and its variance must include the covariance term.
Z=X+Y
E[Z]=50+20=70
Model
Model
The per-person variance is 50+30+2(10)=100. Independence across 100 people then gives an aggregate standard deviation of 100.
Var(Z)=50+30+2(10)=100
T=i=1∑100Zi
Compute
Compute
The aggregate mean is 7000, so 7100 is exactly one aggregate standard deviation above the mean. The corresponding normal cumulative probability is 0.8413447.
E[T]=7000,SD(T)=100(100)=100
P(T<7100)≈Φ(1)=0.8413447
Answer
Answer
The approximation rounds to 0.84, corresponding to choice B.
0.84(B)
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.
COmitting the covariance gives per-person variance 80 and z about 1.118, producing approximately 0.87.
Original practice · fully worked
Original variant: combined task time across a work crew
For one technician, inspection time X and repair time Y have means 4 and 6 hours, variances 1 and 4, and covariance -0.5. Twenty-five technicians work independently. Approximate the probability that their total X+Y time is below 255 hours.
E 0.6915
B 0.7734
C 0.8413
D 0.8944
A 0.9332
Answer and calculation
A technician's combined time has mean 10 and variance 4. Across 25 workers the mean is 250 and standard deviation 10, so 255 corresponds to z=0.5 and probability 0.6915.
Setup
Setup
For each technician, combine inspection and repair time as Z=X+Y. The negative covariance reduces the variance of that sum.
Z=X+Y
E[Z]=10
Model
Model
The per-technician mean is 10 and variance is 4. Independence across 25 technicians gives total mean 250 and standard deviation 10.
Var(Z)=1+4+2(−0.5)=4
S=i=1∑25Zi
Compute
Compute
The threshold 255 is 0.5 aggregate standard deviations above the mean, giving normal cumulative probability 0.691462.
E[S]=250,SD(S)=25(4)=10
P(S<255)≈Φ(0.5)=0.691462
Answer
Answer
The approximate probability is 0.6915, corresponding to choice E.
