SOA Exam P Sample Question #666. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #666 — Verified Solution

Independent solution

How to solve this Normal Distribution question

Answer in brief

The permitted interval is symmetric about the normal mean, so 80% central coverage leaves 10% in each tail. Dividing the half-width by z_{0.90} gives a largest standard deviation of 0.390152, matching choice B.

Setup

Write the tolerance interval as a symmetric distance of 0.5 from the mean.

X\sim N(11,\sigma^2),\qquad \Pr(|X-11|\le0.5)\ge0.80

Model

For a centered normal distribution, the interval probability decreases as sigma increases. The largest allowable sigma therefore makes the probability exactly 0.80.

\Pr\!\left(-\frac{0.5}{\sigma}\le Z\le\frac{0.5}{\sigma}\right)=0.80

\Phi\!\left(\frac{0.5}{\sigma}\right)=0.90

Compute

Use the 90th standard-normal percentile because the two outside tails each have probability 0.10.

z_{0.90}=1.2815515655\ldots

\sigma_{\max}=\frac{0.5}{z_{0.90}}=0.3901520730\ldots

Answer

The largest standard deviation rounds to 0.39.

\boxed{\sigma_{\max}\approx0.39\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

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

AThis smaller spread would satisfy the coverage requirement, but it is unnecessarily conservative and therefore cannot be the largest permitted value.
CThis uses the 80th percentile as the upper standardized endpoint. Central 80% coverage instead requires the 90th percentile after splitting the remaining probability between two tails.
DThis results from an inverted or width-scaled standardization. Substitution gives only about 24.5% central coverage, far below the required level.
EThis treats a normal quantile as a standard deviation multiplier rather than dividing the half-width by it; its central coverage is only about 15.5%.

Original practice · fully worked

Original variant: precision pouch filling

Fill weights for a precision powder pouch are normally distributed with mean 50 grams and standard deviation s. A quality rule requires at least 90% of pouches to weigh between 48.8 and 51.2 grams. Calculate the largest allowable value of s.

A
0.466
B
0.730
C
0.936
D
1.974
E
3.948

Variant answer in brief

A central probability of 90% places 5% in each tail. Dividing the 1.2-gram half-width by z_{0.95} gives 0.729548 grams, so choice B is correct.

Setup

The permitted range extends 1.2 grams to either side of the mean.

W\sim N(50,s^2),\qquad \Pr(|W-50|\le1.2)\ge0.90

Model

At the largest scale, the central interval contains exactly 90%, leaving 5% in each tail.

\Phi\!\left(\frac{1.2}{s}\right)=0.95

Compute

Divide the half-width by the 95th standard-normal percentile.

z_{0.95}=1.6448536270\ldots

s_{\max}=\frac{1.2}{z_{0.95}}=0.7295481983\ldots

Answer

The maximum allowable standard deviation is approximately 0.730 grams.

\boxed{s_{\max}\approx0.730\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #666 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.