Worked solution · Exam P

SOA Exam P Sample Question #380 Solution

This Exam P sample reference tests Normal Distribution. This problem forms a linear combination of three independent normal variables. The result has mean 1203 and variance 6, so the requested upper-tail probability is 0.889664 and 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 Normal Distribution question

Setup

Apply the constants and signs in the stated linear combination to its mean.

\mu_L=1205-2-2+2=1203

Model

A linear combination of independent normal variables is normal. Signs disappear after coefficients are squared in the variance.

\sigma_L^2=5.0+0.5+0.5=6

L\sim N(1203,6)

Compute

Standardize the lower boundary of the requested upper tail.

\Pr(L\ge1200)=\Pr\!\left(Z\ge\frac{1200-1203}{\sqrt6}\right)

\Pr(L\ge1200)=\Phi\!\left(\frac3{\sqrt6}\right)=0.8896643190\ldots

Answer

The probability rounds to 0.89.

\boxed{0.89\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.

AThis uses the variance 6 itself as the standard deviation: Φ(3/6)=0.691462, which rounds to 0.69.
CThis ignores the two subtracted variables' variance contributions and uses only variance 5, giving Φ(3/√(5))=0.910144.
DThis subtracts variances because the variables are subtracted, producing variance 5-0.5-0.5=4 and Φ(3/2)=0.933193.
EThis computes the tail for X alone and ignores the entire adjustment: P(X≥1200)=Φ(5/√(5))=0.987326, which rounds to 0.99.

Original practice · fully worked

Original variant: joint signs of orthogonal normal scores

Two independent standard-normal process shocks U and V drive scores R=3U+4V and S=4U-3V. Calculate the probability that both scores are positive.

A
0
B
0.25
C
0.50
D
0.75
E
1.00

Answer and calculation

The two scores are jointly normal and have covariance 3(4)+4(-3)=0. They are therefore independent and symmetric, so their joint positive probability is 0.25 and choice B.

Setup

Write the two scores as linear combinations of the independent standard-normal inputs.

R=3U+4V,\qquad S=4U-3V

E[R]=E[S]=0

Model

Any pair of linear combinations of a jointly normal vector is jointly normal. Compute their covariance from the shared inputs.

\operatorname{Cov}(R,S)=3(4)\operatorname{Var}(U)+4(-3)\operatorname{Var}(V)

\operatorname{Cov}(R,S)=12-12=0

Compute

Zero covariance makes this jointly normal pair independent. Each centered normal score is positive with probability one half.

\Pr(R>0)=\Pr(S>0)=\frac12

\Pr(R>0,S>0)=\frac12\cdot\frac12=\frac14

Answer

Both scores are positive with probability 0.25.

\boxed{0.25\quad\text{(B)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: joint signs of orthogonal normal scores

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables