This Exam P sample reference tests Poisson Distribution. The aggregate count has mean and variance 5000. Standardizing 5100 under its normal approximation gives z=√(2), whose upper-tail probability is 0.078650, so choice A is correct.
Let T denote the aggregate count. Independent Poisson counts add to another Poisson count, with parameter equal to the sum of the component parameters.
T∼Poisson(100⋅50)=Poisson(5000)
E[T]=Var(T)=5000
Model
Model
Because the aggregate mean is large, approximate T by a normal random variable with the same mean and variance.
T∼˙N(5000,5000)
Z=5000T−5000
Compute
Compute
Standardize the stated threshold and evaluate the standard normal upper tail.
z=50005100−5000=2=1.4142136…
Pr(T>5100)≈1−Φ(2)=0.0786496…
Answer
Answer
The normal approximation gives an upper-tail probability of about 0.08.
Pr(T>5100)≈0.08(A)
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.
BA probability of 0.23 would require a standardized gap near 0.74, but the actual gap is 100/√(5000)=1.414.
CThis is approximately the probability between the mean and the threshold, 0.5-0.07865=0.42135, rather than the probability above the threshold.
DAdding the 100 component standard deviations gives the incorrect aggregate standard deviation 100sqrt(50), producing an upper tail near 0.44; independent variances, not standard deviations, add.
EUsing the aggregate variance 5000 as though it were the standard deviation gives an upper tail near 0.49 and ignores the required square root.
Original practice · fully worked
Original variant: migration-night audio uploads
A wildlife-monitoring network has 80 independent acoustic sensors. On a migration night, each sensor uploads a Poisson number of audio clips with mean 30. Using a normal approximation, calculate the probability that the network uploads more than 2490 clips in total.
A 0.0331
B 0.4186
C 0.4669
D 0.5000
E 0.9669
Answer and calculation
The total is approximately normal with mean and variance 2400. Its standardized threshold is 1.8371, giving an upper tail of 0.033096 and choice A.
Setup
Setup
Write U for the total number of uploaded clips and combine the independent Poisson parameters.
U∼Poisson(80⋅30)=Poisson(2400)
E[U]=Var(U)=2400
Model
Model
Use a normal distribution with mean 2400 and standard deviation √(2400) for the large aggregate count.
U∼˙N(2400,2400)
Compute
Compute
Convert the excess over the mean to standard deviation units and evaluate the upper tail.
z=24002490−2400=1.8371173…
Pr(U>2490)≈1−Φ(1.8371173)=0.0330963…
Answer
Answer
The probability of exceeding 2490 uploads is approximately 3.31 percent.
