This Exam P sample reference tests Pareto Distribution. Invert the Pareto distribution at probabilities 0.30 and 0.70. Their quantiles differ by about 93, corresponding to choice B.
The distribution is Pareto with scale 200 and shape 2.5. The target is the 70th percentile minus the 30th percentile.
F(x)=1−(x200)2.5,x>200
Model
Model
Set the cumulative probability at the target quantile equal to p and invert the corresponding Pareto survival probability.
xp=(1−p)0.4200
Compute
Compute
Evaluate the two quantiles using survival probabilities 0.30 and 0.70, respectively. Their difference is 93.06.
x0.70−x0.30=0.300.4200−0.700.4200
x0.70−x0.30=93.06
Answer
Answer
The requested percentile spread rounds to 93, which is choice B.
93(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.
Original variant: spread between two latency quantiles
A network’s peak latency L has survival function P(L>x)=(100/x)² for x at least 100 milliseconds. Measure the gap from the lower quartile to the upper quartile.
A 50.00 ms
B 73.21 ms
E 84.53 ms
D 100.00 ms
C 115.47 ms
Answer and calculation
The quantile is 100/√(1-p). The 75th and 25th percentiles are 200 and 115.470, a difference of 84.530 ms.
Setup
Setup
Write the latency distribution from its survival function and denote the percentile corresponding to probability p.
FL(x)=1−(x100)2
Model
Model
Solve the survival equation for the percentile, then apply the resulting reciprocal-square-root formula at probabilities 0.25 and 0.75.
ℓp=1−p100
Compute
Compute
The upper quartile is 200 ms and the lower quartile is 115.470 ms, so their difference is 84.530 ms.
ℓ0.75−ℓ0.25=200−0.75100=84.529946
Answer
Answer
Therefore the interquartile latency spread is 84.53 ms, corresponding to choice E.
