Worked solution · Exam P

SOA Exam P Sample Question #452 Solution

This Exam P sample reference tests Normal Distribution. This problem aggregates 25 independent normal losses into one normal total. Standardizing the threshold gives z=0.80 and an upper-tail probability of 0.2119, which selects choice D.

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

Let S be the sum of the 25 independent individual amounts.

S=\sum_{i=1}^{25}X_i

Model

A sum of independent normal variables is normal. Means add, while variances add.

E[S]=25(1000)=25{,}000

\operatorname{SD}(S)=\sqrt{25(625^2)}=3{,}125

Compute

Standardize the loss threshold and take the standard-normal upper tail.

z=\frac{27{,}500-25{,}000}{3{,}125}=0.80

\Pr(S>27{,}500)=1-\Phi(0.80)=0.211855\ldots

Answer

The loss probability rounds to 0.21.

\boxed{0.21\quad\text{(D)}}

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.

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.

AA tail near 0.07 corresponds to z about 1.48, which would require an aggregate standard deviation near 2,500/1.48=1,689 instead of 3,125.
BA tail near 0.10 corresponds to z about 1.28, which would require an aggregate standard deviation near 1,953 instead of 3,125.
CA tail near 0.14 corresponds to z about 1.08, which would require an aggregate standard deviation near 2,315 instead of 3,125.
EA tail near 0.44 corresponds to z about 0.15, which would require an aggregate standard deviation near 16,667. Using the individual standard deviation of 625 would instead give z=4 and a tail near zero.

Original practice · fully worked

Original variant: two-sided range for an average reading

Sixteen independent instrument readings are normally distributed with mean 50 and standard deviation 12. Calculate the probability that their sample average is greater than 47 but less than 54.

A
0.5918
B
0.6827
C
0.7501
D
0.8413
E
0.9088

Answer and calculation

The sample average has standard deviation 12/√(16)=3. The bounds standardize to -1 and 4/3, giving Φ(4/3)-Φ(-1)=0.7501 and choice C.

Setup

Use the normal distribution of the sample average.

\bar X\sim N\left(50,\left(\frac{12}{\sqrt{16}}\right)^2\right)=N(50,3^2)

Model

Standardize both endpoints because the requested event is a finite interval.

z_L=\frac{47-50}{3}=-1

z_U=\frac{54-50}{3}=\frac43

Compute

Subtract the lower cumulative probability from the upper cumulative probability.

\Pr(47<\bar X<54)=\Phi(4/3)-\Phi(-1)=0.7501335263\ldots

Answer

The interval probability rounds to 0.7501.

\boxed{0.7501\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate 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 Univariate 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 Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: two-sided range for an average reading

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables