This Exam P sample reference tests Continuous Random Variables. Integrating after the substitution u=ln x gives F(x)=(ln x/ln 80)². Inverting F(x)=0.75 yields x=80 to the power (√(0.75))=44.4759, corresponding to choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is approximately 80 to the power (√(0.50)), so it computes the median instead of the 75th percentile.
BThis uses ln(x)/ln(80)=0.75 and forgets that the CDF is the square of that ratio.
DThis multiplies the support length 79 by 0.75 and forgets to add the lower endpoint; linear interpolation is not valid for this density.
EThis takes roughly 75% of the upper endpoint, again treating the distribution as though its quantiles were linear in x.
Original practice · fully worked
Original variant: compressed archive size
A compressed research archive has size S, measured in gigabytes, between 1 and 64. Its cumulative distribution function is F(s)=[log_2(s)/6]³ on that interval. Calculate the 12.5th percentile of S.
A 1.414
B 1.682
C 4.351
D 8.000
E 53.401
Variant answer in brief
Set the supplied CDF equal to 0.125. Taking the cube root gives log_2(s)/6=0.5, so s=2³=8 and choice D is correct.
Setup
Setup
A percentile is found by equating the CDF to its target cumulative probability.
F(s0.125)=0.125=81
Model
Model
Insert the given CDF and take the cube root.
(6log2s)3=81
6log2s=21
Compute
Compute
Undo the logarithm after restoring the scale factor.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.