This Exam P sample reference tests Continuous Distributions. Normalizing the power density gives a CDF proportional to x raised to a+1. The supplied CDF value calibrates that exponent near 2.5, and the upper tail beyond 4 is 0.4276, choice B.
How to solve this Continuous Distributions question
Setup
Setup
Normalize the density over its finite support before using the probability statement.
1=∫05cxadx=a+1c5a+1
Model
Model
After normalization, the CDF depends only on the ratio of the point to the upper endpoint.
F(x)=(5x)a+1,0<x<5
0.4871=(0.75)a+1
Compute
Compute
Recover the power from logarithms and evaluate the complementary CDF at 4.
a+1=log(0.75)log(0.4871)=2.5002804
Pr(X>4)=1−(0.8)2.5002804=0.4276024
Answer
Answer
The requested upper-tail probability rounds to 0.428.
0.428(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 0.404 corresponds to using an exponent near 2.31 rather than the exponent fixed by the supplied CDF point. The log calibration gives 2.5003.
CThe value 0.500 treats 4 as a median without deriving that claim from the density. The calibrated CDF at 4 is about 0.5724.
DThe value 0.572 is F(4), the probability below 4. The question requires its complement.
EThe value 0.596 is the complement of distractor A and inherits the same miscalibrated power before also selecting the wrong side of the cutoff.
Original practice · fully worked
Original variant: infer a bounded power law from a conditional fraction
A quality score X lies between 0 and 12 and has cumulative distribution F(x)=(x/12)ᵏ on that interval, where k is positive. Among scores no greater than 9, the fraction no greater than 6 is 4/9. Determine the median score.
A 6.000
B 8.000
C 8.485
D 9.000
E 10.392
Answer and calculation
The conditional fraction gives (2/3)ᵏ=4/9, so k=2. Solving F(m)=1/2 then gives m=12/√(2)=8.485, choice C.
Setup
Setup
Express the supplied conditional fraction as a ratio of CDF values.
Pr(X≤6∣X≤9)=F(9)F(6)=94
Model
Model
Substitute the power-form CDF; the common endpoint cancels.
(9/12)k(6/12)k=(32)k=94
Compute
Compute
The calibration gives k=2, after which the median equation is immediate.