Verified solution · Exam P

SOA Exam P Sample Question #250 Solution

This Exam P sample reference tests Normal Distribution. The daily cost is the weighted normal sum 0.2X+0.1Y, with mean 7.5 and variance 0.45. Standardizing the cost threshold gives z=-0.745356 and probability 0.228028, which rounds to choice B.

Independent analysis by ActuaryProof · reviewed August 4, 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 Normal Distribution question

Setup

Convert each mileage total into fuel cost and add the two components.

C=3\left(\frac{X}{15}+\frac{Y}{30}\right)=0.2X+0.1Y

Model

A linear combination of independent normal variables is normal; apply the weights to its mean and variance.

E[C]=0.2(25)+0.1(25)=7.5

\operatorname{Var}(C)=0.2^2(3^2)+0.1^2(3^2)=0.45

Compute

Standardize the cost boundary using the square root of the variance.

z=\frac{7-7.5}{\sqrt{0.45}}=-0.7453559925\ldots

\Pr(C<7)=\Phi(z)=0.2280282701\ldots

Answer

The required probability rounds to 0.23.

\boxed{0.23\quad\text{(B)}}

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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.

AUsing the variance 0.45 directly as the standard deviation gives Φ((7-7.5)/0.45)=0.1333, which rounds to 0.13.
CAdding the weighted standard deviations gives 0.2(3)+0.1(3)=0.9. Using that as the aggregate standard deviation gives Φ(-0.5/0.9)=0.2893.
DWeighting variances linearly instead of squaring the weights gives 0.2(9)+0.1(9)=2.7. Standardizing with √(2.7) gives 0.3805.
EDividing the deviation -0.5 by the mean cost 7.5 instead of by the standard deviation gives Φ(-0.5/7.5)=0.4734.

Original practice · fully worked

Original variant: comparing two scores with a shared normal input

Three independent normal calibration inputs U, V, and W have means 40, 25, and 10 and standard deviations 4, 3, and 2, respectively. Algorithm A reports U+V, while Algorithm B reports 1.5U+W. Calculate the probability that Algorithm B reports the larger value.

A
0.733
B
0.763
C
0.862
D
0.887
E
0.994

Variant answer in brief

Subtracting the two scores leaves D=0.5U+W-V, a normal variable with mean 5 and variance 17. Therefore P(D>0)=Φ(5/√(17))=0.887374, so choice D.

Setup

Express the score comparison as one linear contrast of the independent inputs.

D=B-A=(1.5U+W)-(U+V)=0.5U+W-V

Model

Compute the contrast mean and add its independent variance contributions.

E[D]=0.5(40)+10-25=5

\operatorname{Var}(D)=0.5^2(4^2)+2^2+3^2=17

Compute

Standardize zero under the contrast distribution.

\Pr(B>A)=\Pr(D>0)=\Phi\left(\frac5{\sqrt{17}}\right)

=0.8873735468\ldots

Answer

Algorithm B reports the larger value with probability approximately 0.887.

\boxed{0.887\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #250 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: comparing two scores with a shared normal input

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution