Worked solution · Exam P

SOA Exam P Sample Question #140 Solution

This Exam P sample reference tests Normal Distribution. The relevant linear combination is normal with mean 1 and variance 50.49. Standardizing zero gives a probability of 0.55596, which selects 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 Normal Distribution question

Setup

Represent the comparison by moving every lifetime to the same side of the inequality.

W=X_1+X_2-1.9X_3

\Pr(X_1+X_2>1.9X_3)=\Pr(W>0)

Model

Independence makes the covariance terms vanish, so the mean and variance of the normal linear combination follow directly.

E[W]=10+10-1.9(10)=1

\operatorname{Var}(W)=3^2+3^2+(1.9)^2 3^2=50.49

Compute

Standardize the zero threshold with the resulting standard deviation.

\sigma_W=\sqrt{50.49}=7.1056316

\Pr(W>0)=1-\Phi\!\left(\frac{0-1}{7.1056316}\right)=0.5559597

Answer

The probability rounds to 0.556.

\boxed{0.556\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.

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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.

AThe value 0.407 is the complement of 0.593. It results from both treating the third lifetime as fixed and reversing the requested comparison.
BThe value 0.444 is the complementary normal probability P(W<0), approximately 0.44404.
DThe value 0.593 is obtained by treating 1.9 times the third lifetime as the fixed value 19. That leaves variance 18 for the first two lifetimes and gives Φ(1/√(18))=0.593.
EThe value 0.604 corresponds to a still smaller standard error. All three independent lifetime variances contribute to W.

Original practice · fully worked

Original variant: percentile of a calibrated Gaussian score

A laboratory models two independent calibration offsets: U has a Gaussian distribution with mean 1 and variance 4, while V has a Gaussian distribution with mean -2 and variance 9. The reported composite score is T=3U-2V. Determine its 95th percentile.

A
7.000
B
16.009
C
20.957
D
23.631
E
26.738

Answer and calculation

The composite score has mean 7 and variance 72. Adding 1.64485 standard deviations to its mean gives the 95th percentile 20.957, choice C.

Setup

Compute the center of the reported linear score.

E[T]=3(1)-2(-2)=7

Model

Square the two coefficients when propagating independent variances.

\operatorname{Var}(T)=3^2(2^2)+(-2)^2(3^2)=72

T\sim N(7,72)

Compute

Use the one-sided standard-normal 95th percentile.

t_{0.95}=7+\Phi^{-1}(0.95)\sqrt{72}

t_{0.95}=7+1.6448536(8.4852814)=20.9570458

Answer

The requested percentile is approximately 20.957.

\boxed{20.957\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: percentile of a calibrated Gaussian score

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables