Worked solution · Exam P

SOA Exam P Sample Question #206 Solution

This Exam P sample reference tests Normal Distribution. The supplied cumulative probabilities correspond to z=0.72 and z=1.32. Their 0.60 separation across 1000 units gives sigma=1666.67 and mu=800; the target z-score is 0.12, so Φ(0.12)=0.5478, 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

Translate the two given cumulative probabilities into standard-normal quantiles.

\Phi(0.72)=0.7642

\Phi(1.32)=0.9066

Model

Standardization makes the observed cutoffs affine functions of their z-scores.

\frac{2000-\mu}{\sigma}=0.72

\frac{3000-\mu}{\sigma}=1.32

Compute

Subtract the equations, solve the parameters, and standardize the target.

\sigma=\frac{1000}{1.32-0.72}=1666.6667

\mu=2000-0.72\sigma=800

z_{1000}=\frac{1000-800}{1666.6667}=0.12

Answer

The standard-normal cdf at 0.12 is 0.5478.

\boxed{\Phi(0.12)=0.5478\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 probability 0.1424 corresponds to z=-1.070, not the required z=0.12. It is inconsistent with the affine spacing of the two calibrated quantiles.
BThe probability 0.3022 corresponds to z=-0.518. Substituting that score into x=mu+sigma z does not return the target cutoff.
DThe probability 0.6218 corresponds to z=0.310. This uses too small a downward z-step from 0.72 when the cutoff falls by 1000.
EThe value 0.7257 is Φ(0.60), using the z-gap between the two known cutoffs as the target z-score rather than subtracting it from 0.72.

Original practice · fully worked

Original variant: a percentile-calibrated sample mean

A package mass X is normally distributed. Its median is 80 grams, and its 90th percentile is 92 grams. Four packages are selected independently. Calculate the probability that their average mass exceeds 85 grams.

A
0.1000
B
0.1428
C
0.2967
D
0.7033
E
0.8572

Answer and calculation

The median gives mean 80, while the 90th percentile gives standard deviation 12/1.2815516. The four-package average has half that standard deviation, so its threshold z-score is 1.06796 and its upper-tail probability is 0.14277, choice B.

Setup

Use the median and upper percentile to identify the individual-package normal model.

\mu=80

z_{0.90}=1.2815516

\sigma=\frac{92-80}{1.2815516}=9.3636498

Model

Independence makes the average of four packages normal with the same mean and half the individual standard deviation.

\overline X\sim N\left(80,\frac{\sigma^2}{4}\right)

\operatorname{SD}(\overline X)=\frac{9.3636498}{2}=4.6818249

Compute

Standardize the average-mass threshold and take the upper tail.

z=\frac{85-80}{4.6818249}=1.0679596

\Pr(\overline X>85)=1-\Phi(1.0679596)=0.1427694

Answer

The four-package average exceeds 85 grams with probability about 0.1428.

\boxed{0.1428\quad\text{(B)}}

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: a percentile-calibrated sample mean

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables