This Exam P sample reference tests Cumulative Distribution Functions. This is a policy-limit quantile obtained after normalizing the stated cumulative distribution at its upper endpoint. A partial-payment probability of 0.56 makes the required CDF value 0.44 and gives a limit of approximately 5.40, which selects choice A.
How to solve this Cumulative Distribution Functions question
Setup
Setup
The distribution has no mass beyond its finite upper endpoint, so the CDF must equal one there. This determines the unknown scale constant.
1=c(1510)4/3
c=(1015)4/3
Model
Model
After substituting the normalization constant, the CDF on the support has a simpler endpoint-scaled form. A loss is only partially reimbursed precisely when it exceeds the policy limit.
F(x)=(10x)4/3,0≤x≤10
Pr(X>m)=0.56⟹F(m)=0.44
Compute
Compute
Invert the power CDF using the reciprocal exponent.
(10m)4/3=0.44
m=10(0.44)3/4=5.4024344656…
Answer
Answer
The reimbursement limit is approximately 5.40.
5.40(A)
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.
BUsing 0.56 as the CDF value rather than as the survival probability, then applying the inverse power, gives 6.47.
CAt m=7.03 the normalized model gives P(X>m)=1-(0.703)⁽⁴⁄³⁾=0.375, so this choice cannot produce the stated 0.56 partial-payment rate.
DAt m=7.80 the remaining tail probability is only 1-(0.780)⁽⁴⁄³⁾=0.282. This is the result of moving in the wrong direction when converting the tail probability to a quantile.
EOmitting endpoint normalization and retaining the original scale 15 instead of the normalized endpoint 10 gives 8.10.
Original practice · fully worked
Original variant: interquartile range from a logarithmic CDF
A normalized queue-load index T has cumulative distribution function F(t)=a ln(1+t) for 0≤t≤3, with F(t)=1 above that range. The constant a is positive. Calculate the interquartile range of T.
A 0.414
B 1.414
C 1.500
D 1.828
E 3.000
Answer and calculation
Endpoint normalization gives a=1/ln(4). The lower and upper quartiles are √(2)-1 and 2sqrt(2)-1, so the interquartile range is √(2), approximately 1.414, and choice B.
Setup
Setup
Use the upper endpoint of the support to determine the CDF constant.
1=F(3)=aln4
a=ln41
Model
Model
Invert the normalized CDF at a general percentile level p.
p=ln4ln(1+qp)
qp=4p−1
Compute
Compute
Evaluate the two quartiles and subtract the lower from the upper.
