Worked solution · Exam P

SOA Exam P Sample Question #362 Solution

This Exam P sample reference tests Binomial Distribution. This is a continuity-corrected normal approximation to a binomial upper tail. The approximating normal has mean 30 and standard deviation 5. Replacing at least 40 by a boundary of 39.5 gives a standardized boundary of 1.9 and upper-tail probability 0.0287, choice E.

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 Binomial Distribution question

Setup

Compute the binomial mean and standard deviation.

\mu=np=180\left(\frac16\right)=30

\sigma=\sqrt{np(1-p)}=\sqrt{180\left(\frac16\right)\left(\frac56\right)}=5

Model

Translate the inclusive integer threshold to the continuous normal boundary.

\Pr(N\ge40)\approx\Pr(Y>39.5),\qquad Y\sim N(30,25)

Compute

Standardize the corrected boundary and evaluate the upper tail.

z=\frac{39.5-30}{5}=1.9

\Pr(Y>39.5)=1-\Phi(1.9)=0.0287165598

Answer

The normal approximation is 0.029 after rounding.

\boxed{0.029\quad\text{(E)}}

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 value 0.011 corresponds approximately to an upper tail at z=2.3, produced by moving the boundary to 41.5 rather than applying a half-unit correction at 40.
BThe value 0.014 corresponds approximately to z=2.2, using an incorrect boundary of 41.
CThe value 0.018 corresponds approximately to z=2.1, applying the continuity correction in the wrong direction and using 40.5.
DThe value 0.023 is approximately 1-Φ(2), obtained from the uncorrected threshold 40 instead of 39.5.

Original practice · fully worked

Original variant: central range of balanced inspections

A batch contains 100 independently inspected components, and each component has probability 0.50 of passing. Using a normal approximation with continuity correction, calculate the probability that the number of passing components is between 45 and 55, inclusive.

A
0.1357
B
0.6827
C
0.7287
D
0.8413
E
0.8643

Answer and calculation

The approximating normal has mean 50 and standard deviation 5. Continuity correction changes the inclusive range to 44.5 through 55.5, whose standardized endpoints are −1.1 and 1.1. The resulting central probability is 0.7287, choice C.

Setup

Find the mean and standard deviation of the balanced binomial count.

\mu=100(0.5)=50,\qquad \sigma=\sqrt{100(0.5)(0.5)}=5

Model

Expand both inclusive integer endpoints by one half.

\Pr(45\le N\le55)\approx\Pr(44.5<Y<55.5)

Compute

Standardize the symmetric bounds and subtract the normal CDF values.

\frac{44.5-50}{5}=-1.1,\qquad \frac{55.5-50}{5}=1.1

\Phi(1.1)-\Phi(-1.1)=2\Phi(1.1)-1=0.7286678781

Answer

The continuity-corrected approximation is 0.7287.

\boxed{0.7287\quad\text{(C)}}

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How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: central range of balanced inspections

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables