This Exam P sample reference tests Poisson Distribution. The summed Poisson count has mean and variance 2500. Under the requested normal approximation, the standardized bounds are -1 and 2, giving probability about 0.8185 and choice B.
The aggregate of the stated independent Poisson counts is Poisson with mean 2500 and therefore standard deviation 50.
N∼Poisson(1250⋅2)=Poisson(2500)
Model
Model
Under the requested normal approximation, convert the lower and upper count bounds to z-values using mean 2500 and standard deviation 50.
E[N]=2500,SD(N)=50
Compute
Compute
The bounds become -1 and 2. The standard normal probability between them is 0.97725-0.15866=0.81859.
P(2450<N<2600)≈Φ(2)−Φ(−1)
0.97725−0.15866=0.81859
Answer
Answer
The approximate interval probability rounds to 0.82, corresponding to choice B.
0.82(B)
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.
AThe value 0.68 is the familiar central area within one standard deviation. The actual standardized interval is asymmetric, from -1 to 2.
DNo standard one-sided normal probability from either displayed bound was identified that produces 0.95.
EThe value 1.00 treats the finite interval as certain and ignores both normal tails.
Original practice · fully worked
Original variant: normal approximation for accepted quality checks
A batch contains five hundred unrelated quality checks. Each succeeds 40% of the time. Use a normal approximation without continuity correction to estimate the probability that the successful count lies between 180 and 220.
A 0.6827
B 0.7745
E 0.8664
D 0.8944
C 0.9321
Answer and calculation
The binomial count has mean 200 and standard deviation √(120). The symmetric bounds are about 1.826 standard deviations away, giving probability 0.9321.
Setup
Setup
Let N be the successful count in 500 independent checks, each with success probability 0.4.
N∼Bin(500,0.4)
Model
Model
The binomial mean is 200 and variance is 120, so the bounds 180 and 220 are symmetric around the mean.
E[N]=200,SD(N)=500(0.4)(0.6)=120
Compute
Compute
Each bound is 1.825742 standard deviations from the mean. The central normal area between them is 0.932111.
z=12020=1.825742
P(180<N<220)≈2Φ(1.825742)−1=0.932111
Answer
Answer
The approximate probability is 0.9321, corresponding to choice C.
