SOA Exam P Sample Question #670. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #670

A normal median equals its mean. The upper-tail condition places the premium at the 95th percentile, so its excess above the median is z0.95 times 500, or 822.427, which rounds to choice A.

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

Answer in brief

A normal median equals its mean. The upper-tail condition places the premium at the 95th percentile, so its excess above the median is z0.95 times 500, or 822.427, which rounds to choice A.

Setup

A normal distribution is symmetric, so its median is its mean.

m=2500

Model

The stated upper-tail probability makes the premium the 95th percentile of the loss distribution.

\Pr(L>P)=0.05\quad\Longrightarrow\quad \Pr(L\le P)=0.95

P=2500+500z_{0.95}

Compute

Subtract the median before evaluating the standard-normal quantile.

P-m=500z_{0.95}

P-m=500(1.6448536270)=822.4268135\ldots

Answer

The premium exceeds the median loss by approximately 822.

\boxed{822\quad\text{(A)}}

Error diagnosis

Why the other letters tempt

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

BThis uses the two-sided 95% critical value 1.96, giving 1.96(500)=980, instead of the one-sided 95th percentile 1.64485.
CThis reports the median loss itself, not the amount by which the premium exceeds it.
DThis is approximately the premium 2500+822, while the requested quantity is only its excess above the median.
EThis both uses the two-sided critical value 1.96 and reports the resulting premium 2500+980 rather than the excess.

Original practice · fully worked

Original variant: wind-farm reserve threshold

A wind farm's hourly output is normally distributed with mean 180 MW and standard deviation 24 MW. An operator chooses a reserve threshold R so that output exceeds R in 10% of hours. Calculate the amount by which R exceeds the median hourly output.

A
24.000 MW
B
30.757 MW
C
39.476 MW
D
180.000 MW
E
210.757 MW

Variant answer in brief

The threshold is the 90th percentile and the median is 180 MW. Their difference is 24z0.90=30.757 MW, so choice B.

Setup

Use normal symmetry to identify the median with the stated mean.

m=180\text{ MW}

Model

An upper-tail probability of 0.10 places the threshold at cumulative probability 0.90.

R=180+24z_{0.90}

Compute

Subtract the median and evaluate the 90th standard-normal quantile.

R-m=24(1.2815515655)

R-m=30.7572376\ldots\text{ MW}

Answer

The reserve threshold is 30.757 MW above the median.

\boxed{30.757\text{ MW}\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #670 — Verified Solution.” Verified 2026-07-30.

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