SOA Exam P Sample Question #696. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #696 — Verified Solution

Independent solution

How to solve this Normal Distribution question

Answer in brief

Independence makes the one-year positive-profit probability the square root of 0.9025, or 0.95. This determines the standard deviation, and the normal interquartile width is 205.03, selecting choice B.

Setup

Let X be one annual profit with mean 250 and standard deviation sigma. Use independence to recover the one-year probability of a positive profit.

X\sim N(250,\sigma^2)

\Pr(X_1>0,X_2>0)=\Pr(X>0)^2=0.9025

Model

The one-year probability is 0.95. Standardizing zero shows that 250 divided by sigma is the 95th standard normal percentile.

\Pr(X>0)=\sqrt{0.9025}=0.95

\Phi\!\left(\frac{250}{\sigma}\right)=0.95

Compute

Solve for sigma and multiply it by the distance between the standard normal 75th and 25th percentiles.

\sigma=\frac{250}{z_{0.95}}=\frac{250}{1.644853627}=151.98920798\ldots

\operatorname{IQR}=(z_{0.75}-z_{0.25})\sigma

\operatorname{IQR}=2(0.6744897502)(151.98920798)=205.0303258\ldots

Answer

The normal profit distribution has an interquartile range of about 205.

\boxed{\operatorname{IQR}\approx205\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 is approximately the recovered standard deviation 151.99; the interquartile range is \(1.34898\sigma\), not \(\sigma\).
CTreating 0.9025 as the one-year positive-profit probability, instead of taking its square root, gives an IQR about 260.
DThis uses a standard-normal half-width near 0.915 rather than the quartile value 0.67449, so it measures a wider central interval.
EThis similarly substitutes a half-width near 1.21 for the normal quartile, producing a much wider interval than the middle 50 percent.

Original practice · fully worked

Original variant: cleanroom inspection score

A cleanroom's hourly particle-control score is normally distributed with mean 72. Scores recorded during independently selected inspection hours have the same unknown standard deviation. The probability that both of two selected scores are below 90 is 0.9216. Calculate the interquartile range of one hourly score.

A
6.9
B
10.3
C
13.9
D
17.1
E
20.6

Variant answer in brief

One selected score is below 90 with probability 0.96. This gives a standard deviation of 10.282, and multiplying by the normal interquartile factor yields 13.870, so choice C.

Setup

Let C be one inspection score with standard deviation tau and recover its one-score probability from independence.

C\sim N(72,\tau^2)

\Pr(C<90)=\sqrt{0.9216}=0.96

Model

Standardizing the score threshold identifies the 18-point distance from the mean as the 96th standard normal percentile times tau.

\Phi\!\left(\frac{90-72}{\tau}\right)=0.96

\tau=\frac{18}{z_{0.96}}

Compute

Evaluate the standard deviation, then apply the symmetric normal quartile spacing.

\tau=\frac{18}{1.750686071}=10.28168345\ldots

\operatorname{IQR}=2z_{0.75}\tau

\operatorname{IQR}=2(0.6744897502)(10.28168345)=13.86978020\ldots

Answer

The middle half of hourly inspection scores spans about 13.9 points.

\boxed{13.9\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #696 — Verified Solution.” Verified 2026-07-30.

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