Worked solution · Exam P

SOA Exam P Sample Question #731 Solution

This Exam P sample reference tests Normal Distribution. The normal percentile identifies an input variance of 25, after which the variance condition fixes the positive scale factor at one. The required output mean then gives an intercept of 30, matching choice D.

Independent analysis by ActuaryProof · reviewed July 30, 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

Standardize the reported percentile to estimate the standard deviation of the raw-score distribution.

z_{0.80}=0.841621\ldots

\sigma_X=\frac{24.20-20}{z_{0.80}}=4.99037\ldots

Model

Use the integer-variance condition and the moment rules for a positive affine transformation.

\operatorname{Var}(X)=25

\operatorname{Var}(Y)=a^2\operatorname{Var}(X)=25

a>0\quad\Longrightarrow\quad a=1

Compute

Translate the stated relationship between the input variance and output expectation, then solve the affine mean equation.

\operatorname{Var}(X)=\frac12\mathbb{E}[Y]\quad\Longrightarrow\quad\mathbb{E}[Y]=50

50=\mathbb{E}[aX+b]=a\mathbb{E}[X]+b=20+b

b=30

Answer

The intercept is the positive value listed under choice D.

\boxed{b=30\quad\text{(D)}}

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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.

AWith the verified scale a=1, b=-34.72 would make E[Y]=20-34.72=-14.72 rather than the required 50.
BThe sign-error value b=-30 gives E[Y]=-10; solving 50=20+b requires adding 30, not subtracting it.
CSetting b=0 preserves the raw mean at 20 and ignores the condition that E[Y]=50.
EThe value b=34.72 would produce E[Y]=54.72, so it is incompatible with the mean constraint even though its sign is positive.

Original practice · fully worked

Original variant: acoustic calibration offset

An acoustic laboratory's uncalibrated noise index R is normally distributed with average 40, and 90% of readings do not exceed 52.8155. The lab publishes S=cR+d. Instrument records state that the variance of R is an integer, while calibration requires c to be positive, the standard deviation of S to be 15, and the average of S to be 180. Determine the programmed offset d.

A
-120
B
60
C
100
D
120
E
180

Answer and calculation

The 90th percentile identifies an input standard deviation of 10, so the positive output scale is 15/10=1.5. Matching the required output average then gives d=120, which is choice D.

Setup

Use the 90th-percentile z value to recover the spread of the uncalibrated index.

z_{0.90}=1.2815516\ldots

\sigma_R=\frac{52.8155-40}{z_{0.90}}=9.99999\ldots

Model

The integer variance is therefore 100, and a positive affine scale multiplies standard deviation by c.

\operatorname{Var}(R)=100

15=\operatorname{SD}(S)=c\operatorname{SD}(R)=10c

c=1.5

Compute

Apply the output-average requirement to determine the remaining offset.

180=\mathbb{E}[S]=c\mathbb{E}[R]+d

180=1.5(40)+d

d=120

Answer

The laboratory must program the offset shown in choice D.

\boxed{d=120\quad\text{(D)}}

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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: acoustic calibration offset

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables