Worked solution · Exam P

SOA Exam P Sample Question #575 Solution

This Exam P sample reference tests Normal Distribution. The four independent quarterly results sum to a normal variable with mean 4(8)=32 and variance 4(24²)=2304, so its standard deviation is 48. Standardizing zero gives -32/48=-0.67, making choice C correct.

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.

Read the solution Open official source

Independent solution

How to solve this Normal Distribution question

Setup

Let Y be the four-quarter sum. Add the four identical means.

E[Y]=4(8)=32

Model

Independence makes the quarterly variances additive, and a sum of independent normal variables remains normal.

\operatorname{Var}(Y)=4(24^2)=2304

Y\sim N(32,48^2)

Compute

Standardize the zero boundary for a positive aggregate result.

\Pr(Y>0)=\Pr\!\left(Z>\frac{0-32}{48}\right)

\Pr(Y>0)=\Pr(Z>-0.666666\ldots)

Answer

The standardized probability matches the expression using -0.67.

\boxed{\Pr(Z>-0.67)\quad\text{(C)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis reverses the one-quarter standardized ratio, using -24/8=-3.00, and also ignores aggregation over four quarters.
BThis first obtains aggregate mean 32 and standard deviation 48 but reverses the z-score ratio, using -48/32=-1.50.
DAdding the four quarterly standard deviations gives 96, so -32/96=-0.33; independent standard deviations do not add directly.
EThis uses one quarterly mean with the four-quarter standard deviation, giving -8/48=-0.17.

Original practice · fully worked

Original variant: demand-supply buffer

Independent daily demand X and supply Y are normally distributed with means 70 and 60 and standard deviations 8 and 6, respectively. Determine the buffer b for which P(X-Y≤b)=0.95.

A
-6.449
B
10.000
C
18.704
D
33.028
E
26.449

Answer and calculation

The difference X-Y is normal with mean 10 and standard deviation √(8²+6²)=10. Its 95th percentile is 10+1.64485(10)=26.449, so choice E is correct.

Setup

Form the normal distribution of the demand-supply difference.

D=X-Y

E[D]=70-60=10

\operatorname{Var}(D)=8^2+6^2=100

Model

The requested buffer is the 95th percentile of this difference.

\Pr(D\le b)=0.95

\frac{b-10}{10}=\Phi^{-1}(0.95)

Compute

Insert the standard-normal 95th percentile and return to the original scale.

\Phi^{-1}(0.95)=1.644853627\ldots

b=10+10(1.644853627\ldots)=26.44853627\ldots

Answer

The required buffer is approximately 26.449.

\boxed{b=26.449\quad\text{(E)}}

Keep practicing

Next steps for Exam P Multivariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Multivariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: demand-supply buffer

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables