SOA Exam P Sample Question #671. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #671

The average of the two independent normals is normal with mean 75 and variance 149. Standardizing 80 gives z=0.4096 and an upper tail of 0.341044, so choice C.

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 Linear Combinations of Independent Random Variables question

Answer in brief

The average of the two independent normals is normal with mean 75 and variance 149. Standardizing 80 gives z=0.4096 and an upper tail of 0.341044, so choice C.

Setup

Let A be the arithmetic average of the two independent normal lifetimes.

A=\frac{X+Y}{2}

Model

A linear combination of independent normal variables is normal. Means combine linearly and variances use squared coefficients.

\operatorname{E}[A]=\frac{70+80}{2}=75

\operatorname{Var}(A)=\frac{14^2+20^2}{4}=149

Compute

Standardize the threshold using the standard deviation of the average and evaluate the upper tail.

z=\frac{80-75}{\sqrt{149}}=0.40961596\ldots

\Pr(A>80)=1-\Phi(0.40961596)=0.34104384\ldots

Answer

The probability that the average exceeds 80 is approximately 0.341.

\boxed{0.341\quad\text{(C)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis is approximately Pr(X>80)Pr(Y>80), which incorrectly replaces the event that the average exceeds 80 by the stricter event that both values exceed 80.
BThis is approximately Pr(X>80) for the first distribution alone and ignores the second value in the average.
DThis is close to the result of averaging the component standard deviations directly; independent variances must be combined with squared coefficients.
EThis is the lower-tail complement Pr(A less than or equal to 80), rather than the requested upper tail.

Original practice · fully worked

Original variant: blended laboratory score

Two independent laboratory instruments produce normally distributed quality scores. Instrument A has mean 10 and standard deviation 6; instrument B has mean 14 and standard deviation 8. A blended score is W=0.25A+0.75B. Calculate the probability that W exceeds 15.

A
0.091
B
0.373
C
0.395
D
0.450
E
0.627

Variant answer in brief

The weighted normal score has mean 13 and variance 38.25. Its threshold z-score is 0.32338, giving upper-tail probability 0.373203 and choice B.

Setup

Treat the reported score as a linear combination of the two independent readings.

W=0.25A+0.75B

Model

Combine the normal means and independent variances using the linear weights.

\operatorname{E}[W]=0.25(10)+0.75(14)=13

\operatorname{Var}(W)=0.25^2(6^2)+0.75^2(8^2)=38.25

Compute

Standardize 15 under the resulting normal distribution.

z=\frac{15-13}{\sqrt{38.25}}=0.32338083\ldots

\Pr(W>15)=1-\Phi(0.32338083)=0.37320342\ldots

Answer

The blended score exceeds 15 with probability approximately 0.373.

\boxed{0.373\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #671 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.