Worked solution · Exam P

SOA Exam P Sample Question #66 Solution

This Exam P sample reference tests Normal Distribution. The mean of 25 normal claims has standard deviation 1000. Standardizing 20,000 gives z=0.6 and upper-tail probability 0.2743, choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Normal Distribution question

Setup

The mean of 25 independent normal claims is normal with mean 19,400 and standard deviation 5,000 divided by 5, or 1,000.

\bar X\sim N\left(19400,\frac{5000^2}{25}\right)

Model

Standardize the threshold 20,000 using the sample-mean standard deviation, not the individual-claim standard deviation.

z=\frac{20000-19400}{5000/\sqrt{25}}=0.6

Compute

The standardized threshold is 0.6, so the required upper tail is one minus the standard normal cumulative probability at 0.6, equal to 0.274253.

P(\bar X>20000)=1-\Phi(0.6)=0.274253

Answer

The probability rounds to 0.27, corresponding to choice C.

\boxed{0.27\quad\text{(C)}}

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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.

DThe value 0.33 is a rough approximation not obtained from the standard normal upper tail at z=0.6.
EThe value 0.45 results approximately from using the individual standard deviation 5,000, which gives z=0.12, instead of the sample-mean standard deviation 1,000.

Original practice · fully worked

Original variant: sample-average fill volume

Bottle fill volume is normal with mean 502 milliliters and standard deviation 8 milliliters. For a random sample of 16 bottles, find the probability that the sample mean is below 499 milliliters.

A
0.0228
B
0.0668
C
0.1587
D
0.3085
E
0.6915

Answer and calculation

The sample-mean standard deviation is 2. Standardizing 499 gives -1.5, whose normal lower tail is 0.0668.

Setup

For 16 normal bottle fills, the sample mean remains normal with mean 502 ml and standard deviation 8 divided by 4, or 2 ml.

\bar X\sim N\left(502,\frac{8^2}{16}\right)

Model

Standardize 499 relative to the sample-mean distribution, giving a lower-tail cutoff of -1.5.

z=\frac{499-502}{8/4}=-1.5

Compute

The standard normal lower-tail probability at -1.5 is 0.0668072.

P(\bar X<499)=\Phi(-1.5)=0.0668072

Answer

Thus the probability that the sample mean is below 499 ml is 0.0668, corresponding to choice B.

\boxed{0.0668\quad\text{(B)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: sample-average fill volume

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables