Verified solution · Exam P

SOA Exam P Sample Question #670 Solution

This Exam P sample reference tests Normal Distribution. 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 × 500, or 822.427, which rounds to choice A.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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

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)}}

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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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Cite this solution

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.

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: wind-farm reserve threshold

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution