Worked solution · Exam P

SOA Exam P Sample Question #224 Solution

This Exam P sample reference tests Continuous Random Variables. Symmetry assigns half the mass to each side of 6.5, so the left density constant is 0.5/ln(7.5). Reflecting the 60th percentile to the 40th gives k=14-7.5⁰·⁸=8.9876, choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Continuous Random Variables question

Setup

Write the density on the left half as c divided by t+1. Symmetry makes its total probability one half.

f(t)=\frac{c}{t+1},\qquad 0\le t\le6.5

\int_0^{6.5}\frac{c}{t+1}\,dt=c\ln(7.5)=0.5

c=\frac{0.5}{\ln(7.5)}

Model

Let k be the 60th percentile. Its reflection 13-k is the 40th percentile and lies on the known left half.

F_T(13-k)=1-F_T(k)=0.40

Compute

Integrate the left density to the reflected point, then solve for k.

0.40=\frac{0.5}{\ln(7.5)}\ln(14-k)

\ln(14-k)=0.8\ln(7.5)

k=14-7.5^{0.8}=8.9875620

Answer

The student registers after approximately 8.99 elapsed days.

\boxed{8.99\quad\text{days (C)}}

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.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThe value 4.01 is the reflected point 13-8.99, which is the 40th percentile. It is the helper quantile on the known side, not the requested 60th percentile.
BThe value 7.80 is 0.60(13), the 60th percentile of a uniform distribution on the full interval. The stated density is not uniform.
DExtending the left-half formula beyond its valid interval and setting it directly equal to 0.60 gives 7.5¹·²-1=10.22. The right half must be handled by symmetry.
EAt 10.51, symmetry gives F(10.51)=1-[0.5 ln(3.49)/ln(7.5)], approximately 0.690. This is not the 60th percentile.

Original practice · fully worked

Original variant: percentile under a stepped density

A device's energy use X has density c from 0 through 2 units and density 2c from 2 through 5 units, with zero density elsewhere. Calculate the 70th percentile of X.

A
2.000 units
B
2.800 units
C
3.500 units
D
3.800 units
E
4.800 units

Answer and calculation

Normalization gives c=1/8, so the lower segment contains probability 1/4 and the upper density is 1/4. Accumulating another 0.45 above 2 requires 1.8 units, placing the 70th percentile at 3.8 and choice D.

Setup

Normalize the two density levels over their respective intervals.

\int_0^2c\,dx+\int_2^5 2c\,dx=2c+6c=1

c=\frac18

Model

Locate the percentile above the density break because the first segment holds only one quarter of the probability.

F_X(2)=2c=\frac14<0.70

F_X(q)=\frac14+2c(q-2),\qquad 2<q<5

Compute

Accumulate the remaining probability at the upper segment's density.

0.70=\frac14+\frac14(q-2)

q-2=1.8,\qquad q=3.8

Answer

The 70th percentile of energy use is 3.8 units.

\boxed{3.800\quad\text{units (D)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: percentile under a stepped density

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables