Worked solution · Exam P

SOA Exam P Sample Question #667 Solution

This Exam P sample reference tests Normal Distribution. Match the exponential 80th percentile, 500 ln(5), to the normal percentile 500+z0.80 sigma. This gives sigma=362.062 and second moment 500 squared plus sigma squared, or 381,089, which rounds to choice C.

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

Write the exponential distribution through its survival function and solve for its 80th percentile q.

0.20=e^{-q/500}

q=500\log 5=804.7189562\ldots

Model

For the normal model, the same percentile is the mean plus the 80th standard-normal quantile times the unknown standard deviation.

q=500+z_{0.80}\sigma

z_{0.80}=0.8416212336\ldots

Compute

Solve for the normal standard deviation, then use the identity relating the raw second moment to variance and squared mean.

\sigma=\frac{500\log 5-500}{0.8416212336}=362.0618683\ldots

\operatorname{E}[X^2]=\sigma^2+\mu^2=(362.0618683)^2+500^2

\operatorname{E}[X^2]=381088.7965\ldots

Answer

Rounding the second moment to the nearest thousand gives 381,000.

\boxed{381{,}000\quad\text{(C)}}

Test the method on unseen problems

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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.

AThis is the normal variance, approximately 131,089, and omits the squared mean 500 squared required for the raw second moment.
BThis treats the percentile probability 0.80 as the normal z-quantile and then reports the resulting variance, approximately 145,084.
DThis uses 0.80 in place of z0.80=0.841621 but does add the squared mean, producing approximately 395,084.
EThis is the exponential model's second moment, 2(500 squared), rather than the requested second moment under the normal model.

Original practice · fully worked

Original variant: parcel-handling duration models

A logistics team models a parcel-handling duration by a normal distribution with mean 40 minutes. A separate benchmark uses an exponential distribution with mean 50 minutes. The 75th percentiles of the two models coincide. Calculate the second raw moment of the duration under the normal model.

A
1600
B
1889
C
3489
D
4805
E
5000

Answer and calculation

The shared percentile is 50 ln(4). Matching it to 40+z0.75 sigma gives variance 1888.951, and adding 40 squared gives second moment 3488.951, so choice C.

Setup

Find the benchmark's 75th percentile from its exponential survival probability.

q=50\log 4=69.3147181\ldots

Model

Express the same percentile in the normal model using the 75th standard-normal quantile.

q=40+z_{0.75}\sigma

z_{0.75}=0.6744897502\ldots

Compute

Recover the variance and then the raw second moment.

\sigma=\frac{50\log4-40}{0.6744897502}=43.4620660\ldots

\operatorname{E}[Y^2]=40^2+\sigma^2=3488.9511831\ldots

Answer

The normal model's second raw moment rounds to 3489 square minutes.

\boxed{3489\quad\text{(C)}}

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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: parcel-handling duration models

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables