Worked solution · Exam P

SOA Exam P Sample Question #341 Solution

This Exam P sample reference tests Continuous Random Variables. Normalizing the inverse-square density on its finite support gives constant 35/6. Integrating from year 5 through year 10 then gives 7/36, approximately 0.194. This is choice B.

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

Set the integral of the density over its support equal to one.

1=k\int_0^{30}\frac{1}{(x+5)^2}\,dx

Model

Evaluate the normalization integral and solve for the constant.

1=k\left(\frac15-\frac1{35}\right)=k\frac6{35}

k=\frac{35}{6}

Compute

Integrate the normalized density over the required five-year interval.

\Pr(5<X<10)=\frac{35}{6}\int_5^{10}\frac{1}{(x+5)^2}\,dx

=\frac{35}{6}\left(\frac1{10}-\frac1{15}\right)=\frac7{36}=0.194444\ldots

Answer

The probability of death during that interval is approximately 0.194.

\boxed{0.194\quad\text{(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.

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Error diagnosis

Why the other letters tempt

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

AThe density throughout the five-year interval exceeds 0.025, so its area must exceed 0.125. A probability of 0.004 is incompatible with the stated density.
CThe value 0.333 treats the interval length as a fraction of the first 15 years, as though the lifetime density were uniform.
DThe value 0.583 is the probability of death during the first five years, not during years five through ten.
EThe value 0.778 is the cumulative probability of death by year ten. It includes deaths during the first five years.

Original practice · fully worked

Original variant: median of a rising calibration density

A calibration score X ranges from zero to two. Its density at score x is a constant multiple of one plus x. Determine the 50th percentile of X.

A
0.732
B
1.000
C
1.236
D
1.414
E
1.500

Answer and calculation

The integral of 1+x over the support is 4, so the density constant is 1/4. Setting the resulting cumulative probability equal to one half gives the positive root of a quadratic, approximately 1.236. This is choice C.

Setup

Normalize the rising linear density.

1=c\int_0^2(1+x)\,dx=4c

c=\frac14

Model

Write the cumulative distribution within the support.

F(x)=\frac14\left(x+\frac{x^2}{2}\right),\qquad 0\le x\le2

Compute

Set the cumulative probability to one half and retain the root inside the support.

\frac14\left(m+\frac{m^2}{2}\right)=\frac12

m^2+2m-4=0,\qquad m=\sqrt5-1=1.2360679\ldots

Answer

The median calibration score is approximately 1.236.

\boxed{1.236\quad\text{(C)}}

Keep practicing

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: median of a rising calibration density

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables