Worked solution · Exam P

SOA Exam P Sample Question #614 Solution

This Exam P sample reference tests Normal Distribution. The independent two-period total is normal with standard deviation √(1100²+2640²)=2860. The positive-total probability corresponds to z=1.099844, so the unknown second-period mean is 2860z-660=2485.55 and choice C is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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

Form the distribution of the sum of the two independent normal results.

S=X_1+X_2

E[S]=660+\mu

Model

Independence makes the variances additive.

\operatorname{SD}(S)=\sqrt{1100^2+2640^2}=2860

S\sim N(660+\mu,2860^2)

Compute

Translate the positive-total probability into a standard-normal quantile and solve for the unknown mean.

\Phi\!\left(\frac{660+\mu}{2860}\right)=0.8643

\frac{660+\mu}{2860}=\Phi^{-1}(0.8643)=1.099844224\ldots

\mu=2485.5544806\ldots

Answer

The unknown mean rounds to the listed value 2486.

\boxed{2486\quad\text{(C)}}

Test the method on unseen problems

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30 original multivariate 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.

AThis understates the mean needed for an 86.43% positive-total probability by using an inadequately small spread for the two-period sum.
BThis results from using the stated probability itself in place of the required standard-normal quantile and then applying an inconsistent mean adjustment.
DThis uses a shifted or rounded cutoff equation instead of standardizing zero against the full total mean 660+mu.
EThis adds the two standard deviations to get 3740 instead of adding variances; 3740(1.099844)-660 is approximately 3454.

Original practice · fully worked

Original variant: combined production cost

Three independent production-stage costs are normally distributed. Their respective means are 100, 150, and 250, and their standard deviations are 10, 20, and 30. Calculate the 95th percentile of the total production cost.

A
438.5
B
500.0
C
537.4
D
561.5
E
598.7

Answer and calculation

The total cost is normal with mean 500 and variance 1400. Adding 1.6448536 standard deviations to the mean gives the 95th percentile 561.5448, so choice D is correct.

Setup

Add the three stage means to obtain the center of the total cost.

E[T]=100+150+250=500

Model

The sum is normal, and independence makes the stage variances additive.

\operatorname{Var}(T)=10^2+20^2+30^2=1400

T\sim N(500,1400)

Compute

Apply the 95th-percentile standard-normal quantile to the total standard deviation.

q_{0.95}=500+\Phi^{-1}(0.95)\sqrt{1400}

q_{0.95}=500+1.644853627(\sqrt{1400})

q_{0.95}=561.5447872\ldots

Answer

The 95th percentile rounds to 561.5.

\boxed{561.5\quad\text{(D)}}

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Next steps for Exam P Multivariate Random Variables

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: combined production cost

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables