Worked solution · Exam P

SOA Exam P Sample Question #327 Solution

This Exam P sample reference tests Sampling Distributions. The sample mean has standard deviation 50. Its 99th percentile is therefore 2500 plus 2.32635 × 50, or 2616.317, which corresponds to choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Sampling Distributions question

Setup

Convert the individual-claim standard deviation to the standard deviation of the mean of 100 independent claims.

\mu_{\bar X}=2500,\qquad \sigma_{\bar X}=\frac{500}{\sqrt{100}}=50

Model

An upper-tail probability of 0.01 places the threshold at the 0.99 standard-normal quantile.

\Pr(\bar X>K)=0.01\quad\Longleftrightarrow\quad \Phi\!\left(\frac{K-2500}{50}\right)=0.99

\Phi^{-1}(0.99)=2.326347874\ldots

Compute

Rescale the standard-normal quantile to the sampling distribution.

K=2500+50(2.326347874\ldots)=2616.3173937\ldots

Answer

The listed threshold is 2616.

\boxed{2616\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThe value 2505 adds 500/100 to the mean. A sample mean uses 500 divided by the square root of 100, and the result must also be multiplied by the 0.99 normal quantile.
BThe value 2512 is obtained from 2500 plus 2.326 × 500/100. It uses the correct quantile but divides the individual standard deviation by 100 instead of by the square root of 100.
DThe value 3663 is approximately 2500 plus 2.326 × 500. It treats the sample mean as though it retained the standard deviation of one claim.
EThe value 4950 equals 2500 × 1.98, a direct scaling of the mean by twice 0.99. Tail probabilities determine a standard-normal quantile; they do not multiply the population mean.

Original practice · fully worked

Original variant: spread from a three-team contrast

Three laboratories independently sample readings from the same normal process. Their sample sizes are 4, 9, and 16, and their averages are A, B, and C. Individual readings have an unknown standard deviation sigma. For the contrast L=2A-B-C, the probability that L exceeds 6.5 is 0.0668072. Calculate sigma.

A
2.737
B
3.692
C
4.000
D
4.333
E
6.658

Answer and calculation

The common process mean cancels from 2A-B-C. Independence gives contrast variance sigma squared times 4/4+1/9+1/16=169/144, so its standard deviation is 13sigma/12. The supplied tail corresponds to z=1.5; solving 6.5/(13sigma/12)=1.5 gives sigma=4, choice C.

Setup

Write the stated contrast of the three sample averages. Its coefficients sum to zero, so their common process mean cancels.

L=2A-B-C,\qquad E[L]=(2-1-1)\mu=0

Model

Independence makes the three coefficient-squared sample-mean variances additive.

\operatorname{Var}(L)=4\frac{\sigma^2}{4}+\frac{\sigma^2}{9}+\frac{\sigma^2}{16}=\frac{169\sigma^2}{144}

\operatorname{SD}(L)=\frac{13\sigma}{12}

Compute

Convert the supplied tail to its standard-normal boundary and solve for the process standard deviation.

1-\Phi(1.5)=0.0668072\ldots

\frac{6.5}{13\sigma/12}=1.5

\sigma=4

Answer

The individual-reading standard deviation is 4 units.

\boxed{4.000\quad\text{(C)}}

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How to solve this Sampling Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: spread from a three-team contrast

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables