Worked solution · Exam P

SOA Exam P Sample Question #223 Solution

This Exam P sample reference tests Normal Distribution. A 2% stockout probability places inventory at the 98th percentile. Using z0.98=2.05375 gives 20+2(2.05375)=24.1075 pounds, so the listed amount is 24, 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

Let I be the inventory level and D the normally distributed monthly demand.

D\sim N(20,2^2),\qquad \Pr(D>I)=0.02

Model

Convert the upper-tail requirement to a cumulative normal percentile.

\Pr(D\le I)=0.98

\frac{I-20}{2}=\Phi^{-1}(0.98)

Compute

Evaluate the standard-normal percentile and return to the demand scale.

\Phi^{-1}(0.98)=2.0537489

I=20+2(2.0537489)=24.1074978

Answer

The closest listed inventory level is 24 pounds.

\boxed{24\quad\text{pounds (C)}}

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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.

AUsing the lower 2nd percentile gives 20-2.05375(2)=15.89, nearest 16. That reverses the stockout tail.
BUsing the 95th percentile gives 20+1.645(2)=23.29, nearest 23. This permits about a 5% stockout rate instead of 2%.
DThe value 32 can arise from treating the variance 4 as though it were the standard deviation and adding a three-unit safety multiplier. The given standard deviation is 2 and the required multiplier is about 2.054.
EAn inventory of 43 is 11.5 standard deviations above the mean, making the normal stockout probability effectively zero. It is not compatible with a 2% target.

Original practice · fully worked

Original variant: interval probability from two calibration percentiles

The fill volume X of a production line is normally distributed with unknown mean and standard deviation. Historical data show that 15.8655% of fills are at or below 48 milliliters and 97.7250% are at or below 72 milliliters. Calculate the probability that a fill is between 40 and 64 milliliters.

A
0.682689
B
0.818595
C
0.841345
D
0.954500
E
0.977250

Answer and calculation

The two calibration percentiles correspond to z-scores -1 and 2, giving sigma=8 and mean 56. The requested endpoints standardize to -2 and 1, so the interval probability is Φ(1)-Φ(-2)=0.818595, choice B.

Setup

Translate the two stated cumulative probabilities into their standard-normal scores.

\frac{48-\mu}{\sigma}=-1,\qquad \frac{72-\mu}{\sigma}=2

Model

Solve the two calibration equations for the distribution's scale and center.

\frac{72-48}{\sigma}=2-(-1),\qquad \sigma=8

48-\mu=-8,\qquad \mu=56

Compute

Standardize the requested interval and subtract its two cumulative probabilities.

\frac{40-56}{8}=-2,\qquad \frac{64-56}{8}=1

\Pr(40<X<64)=\Phi(1)-\Phi(-2)

=0.8413447-0.0227501=0.8185946

Answer

The probability of a fill between 40 and 64 milliliters is approximately 0.818595.

\boxed{0.818595\quad\text{(B)}}

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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: interval probability from two calibration percentiles

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables