Verified 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 August 4, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

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)}}

Continue without hunting through PDFs

All 718 Exam P sample solutions in syllabus order

The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.

See inside the Exam P study manual PDF

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

Variant answer in brief

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

Turn this solution into a complete study sequence

The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.

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

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #380 — 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: joint signs of orthogonal normal scores

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution