Worked solution · Exam P

SOA Exam P Sample Question #432 Solution

This Exam P sample reference tests Normal Distribution. This problem infers the original normal standard deviation from a lower-tail probability, then applies the stated scale change. The future interval is ±1.8 standard deviations about its mean, giving probability 0.9281 and choice B.

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.

Read the solution Open official source

Independent solution

How to solve this Normal Distribution question

Setup

Translate the historical lower-tail probability into its standard-normal quantile.

\Phi^{-1}(0.1056)\approx-1.25

\frac{400-500}{\sigma}=-1.25

Model

Solve for the historical spread and apply the future standard-deviation multiplier.

\sigma=80

\sigma_{\mathrm{future}}=1.25(80)=100

Compute

Both future endpoints are equally distant from the future mean.

\frac{370-550}{100}=-1.8,\qquad \frac{730-550}{100}=1.8

\Pr(370<N<730)=\Phi(1.8)-\Phi(-1.8)=0.928139362\ldots

Answer

The interval probability rounds to 0.928.

\boxed{0.928\quad\text{(B)}}

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.

AA central probability of 0.903 corresponds to bounds about ±1.6596 and an implied future standard deviation 180/1.6596=108.46 rather than 100.
CA central probability of 0.970 corresponds to bounds about ±2.1701 and an implied future standard deviation 82.95 rather than 100.
DKeeping the historical standard deviation 80 gives standardized bounds ±2.25 and probability 0.9756, near 0.976; it omits the 1.25 scale increase.
EA central probability of 0.985 corresponds to bounds about ±2.4324 and an implied future standard deviation 74.00, far below 100.

Original practice · fully worked

Original variant: minimum sample size for a normal average

Package weights are independently normally distributed with mean 100 grams and standard deviation 12 grams. Determine the smallest sample size n for which the sample average is within 4 grams of 100 with probability at least 0.95.

A
9
B
25
C
34
D
35
E
36

Answer and calculation

The sample-average standard deviation is 12/√(n). Requiring the symmetric standardized bound √(n)/3 to reach z_0.975 gives n≥34.573, so the smallest integer is 35 and choice D.

Setup

Write the exact distribution of the sample average.

\bar X\sim N\left(100,\frac{12^2}{n}\right)

Model

Standardize the symmetric four-gram margin.

\Pr(|\bar X-100|<4)=2\Phi\left(\frac{4\sqrt n}{12}\right)-1

=2\Phi\left(\frac{\sqrt n}{3}\right)-1

Compute

Require each tail to be at most 0.025 and solve for n.

\frac{\sqrt n}{3}\ge\Phi^{-1}(0.975)=1.959963985\ldots

n\ge(3\times1.959963985)^2=34.57312939\ldots

Answer

Round the required sample size upward to the smallest integer.

\boxed{35\quad\text{(D)}}

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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: minimum sample size for a normal average

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables