SOA Exam P Sample Question #624. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #624

This is a sum of independent normal variables. Their aggregate has mean 300 and standard deviation sqrt(290000), so its positive-tail probability is 0.71127 and choice E is correct.

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

Answer in brief

This is a sum of independent normal variables. Their aggregate has mean 300 and standard deviation sqrt(290000), so its positive-tail probability is 0.71127 and choice E is correct.

Setup

Let X and Y denote the two independent normal contributions and define the combined result S.

S=X+Y

\operatorname{E}[S]=400+(-100)=300

Model

An independent sum of normal variables is normal. Independence also makes the component variances additive.

\operatorname{Var}(S)=200^2+500^2=290000

S\sim N\!\left(300,290000\right)

Compute

Standardize zero relative to the aggregate distribution and take the probability above that boundary.

\operatorname{SD}(S)=\sqrt{290000}=538.5164807135\ldots

\Pr(S>0)=\Phi\!\left(\frac{300}{538.5164807135}\right)

\Pr(S>0)=\Phi(0.5570860145\ldots)=0.7112656688\ldots

Answer

The requested probability rounds to 0.71.

\boxed{0.71\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

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

AA value below one half reverses the sign or tail despite the aggregate mean being positive. The probability of a positive result must exceed one half here.
BThis understates the signal-to-noise ratio, often by retaining only the magnitude 100 when combining the two means. Signed means must be added before standardization.
CThis is consistent with using an overly large scale after blending means and standard deviations. Only variances, not unlike quantities, can be combined in the variance calculation.
DAdding the standard deviations directly gives 700 and Phi(300/700)=0.6666, which rounds to 0.67. Independent standard deviations combine through their squared values.

Original practice · fully worked

Original variant: paired sensor correction

A materials lab applies two independent normally distributed sensor corrections. Correction A has mean 12 units and standard deviation 3 units; correction B has mean -5 units and standard deviation 4 units. The final adjustment is D=A+2B. Find the probability that D is negative.

A
0.081
B
0.407
C
0.428
D
0.500
E
0.593

Variant answer in brief

The linear combination has mean 2 and variance 73. Standardizing zero gives Phi(-2/sqrt(73))=0.40746, so choice B is correct.

Setup

Represent the final adjustment as the stated linear combination of the two independent inputs.

D=A+2B

Model

A linear combination of independent normals is normal. Apply the coefficient to both the mean and the variance contribution.

\operatorname{E}[D]=12+2(-5)=2

\operatorname{Var}(D)=3^2+2^2(4^2)=73

Compute

Standardize the zero threshold under the resulting normal distribution.

\Pr(D<0)=\Phi\!\left(\frac{0-2}{\sqrt{73}}\right)

\Pr(D<0)=\Phi(-0.2340822944\ldots)=0.4074605456\ldots

Answer

The probability is approximately 0.407.

\boxed{0.407\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #624 — Verified Solution.” Verified 2026-07-30.

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