Worked solution · Exam P

SOA Exam P Sample Question #420 Solution

This Exam P sample reference tests Continuous Random Variables. This problem integrates a gamma-shaped density over the interval matching the time units. Evaluating the antiderivative at 1 and 2 gives 2/e-3/exp(2)=0.329753, so 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

Express the two calendar-time endpoints in the units used by the random variable. They correspond to density coordinates 1 and 2.

\Pr(1<X<2)=\int_1^2 xe^{-x}\,dx

Model

An integration-by-parts antiderivative collects the polynomial and exponential factors.

\int xe^{-x}\,dx=-(x+1)e^{-x}+C

Compute

Evaluate the antiderivative at the two scaled endpoints.

\Pr(1<X<2)=\left[-(x+1)e^{-x}\right]_1^2

=\frac{2}{e}-\frac{3}{e^2}=0.3297530326

Answer

The interval probability rounds to 0.3298.

\boxed{0.3298\quad\text{(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.

AIntegrating from 10 to 20 without converting to the random variable's time units gives 11 exp(−10)-21 exp(−20)=0.0004994, which rounds to 0.0005.
BThe density decreases from exp(−1)=0.3679 to 2 exp(−2)=0.2707 over this unit-width interval. Therefore its integral must exceed 0.2707, so 0.0867 fails a direct magnitude check.
DThe value 0.6702 is 1-0.3298. It gives the probability outside the target interval rather than the probability between its endpoints.
EThe value 0.9995 is the complement of the 0.0005 produced by using unscaled bounds 10 and 20, combining a unit-conversion error with an unnecessary complement.

Original practice · fully worked

Original variant: normalize a symmetric tolerance density

A fabrication process records a centered tolerance score Z between -1 and 1. On that range its density is f(z)=c(1-z²), with no mass outside the range. The coefficient c normalizes the model. Find Pr(|Z|<1/2).

A
0.3438
B
0.5000
C
0.6875
D
0.7500
E
0.9167

Answer and calculation

Normalization gives c=3/4. Symmetry then reduces the central probability to twice the integral from 0 to 1/2, producing 11/16=0.6875 and choice C.

Setup

First determine the unknown density constant from total probability one.

1=c\int_{-1}^{1}(1-z^2)\,dz

1=c\left[ z-\frac{z^3}{3}\right]_{-1}^{1}=c\frac43

Model

Solve for c and use symmetry to express the central interval probability as twice a one-sided integral.

c=\frac34

\Pr\left(|Z|<\frac12\right)=2\left(\frac34\right)\int_0^{1/2}(1-z^2)\,dz

Compute

Evaluate the polynomial antiderivative at the positive cutoff.

\Pr\left(|Z|<\frac12\right)=\frac32\left[z-\frac{z^3}{3}\right]_0^{1/2}

=\frac32\left(\frac12-\frac1{24}\right)=\frac{11}{16}=0.6875

Answer

The probability of a score within one half of the center is 0.6875.

\boxed{0.6875\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: normalize a symmetric tolerance density

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables