Worked solution · Exam P

SOA Exam P Sample Question #378 Solution

This Exam P sample reference tests Normal Distribution. This problem transfers a known normal percentile to a higher percentile while retaining the same mean and standard deviation. The 80th-percentile information implies standard deviation about 1.996, and the resulting 90th percentile is 9.278, which rounds to choice C.

Independent analysis by ActuaryProof · reviewed July 31, 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

Represent each percentile through its standard-normal score.

z_{0.80}=0.8416212336,\qquad z_{0.90}=1.2815515655

q_p=\mu+\sigma z_p

Model

Use the supplied mean and 80th percentile to recover the scale.

8.40=6.72+\sigma z_{0.80}

\sigma=\frac{8.40-6.72}{0.8416212336}=1.9961473558

Compute

Apply the 90th-percentile standard-normal score using the recovered standard deviation.

q_{0.90}=6.72+(1.9961473558)(1.2815515655)

q_{0.90}=9.2781657689

Answer

The 90th percentile rounds to 9.28.

\boxed{9.28\quad\text{(C)}}

Test the method on unseen problems

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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.

AScaling only the 1.68 displacement above the mean by the percentile-number ratio 0.90/0.80 gives 6.72+1.68(0.90/0.80)=8.61. Normal quantiles are not linear in p.
BLinear extrapolation in percentile rank from the median at 6.72 to the 80th percentile gives 8.40+[0.10/0.30](1.68)=8.96. The normal quantile function is curved.
DMultiplying the raw 80th-percentile value by 0.90/0.80 gives 8.40(1.125)=9.45. A percentile level is not a scale factor for observed values.
ETreating the mean as zero and applying rounded z-score ratio 1.28/0.84 to the full value gives 8.40(1.28/0.84)=12.80. The ratio applies only to deviations from the mean.

Original practice · fully worked

Original variant: transfer complementary normal percentiles

Calibration score A follows a normal model centered at 100. Score B follows another normal model with center 100 and a standard deviation 1.5 times as large as A's. The 90th percentile of A is 112. Calculate the 10th percentile of B.

A
82
B
88
C
92
D
100
E
118

Answer and calculation

Complementary normal percentiles have opposite standardized displacements. B's displacement is 1.5 times A's 12-point displacement, so its 10th percentile is 100-18=82 and choice A.

Setup

Write A's upper-percentile displacement from the shared mean.

q_{A,0.90}-\mu=112-100=12

12=z_{0.90}\sigma_A

Model

Scale that displacement for B and use normal symmetry between the 90th and 10th percentiles.

\sigma_B=1.5\sigma_A

z_{0.10}=-z_{0.90}

Compute

B's lower-percentile displacement has magnitude 1.5 times twelve and points below the mean.

q_{B,0.10}=100-z_{0.90}(1.5\sigma_A)

q_{B,0.10}=100-1.5(12)=82

Answer

The 10th percentile of B is 82.

\boxed{82\quad\text{(A)}}

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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: transfer complementary normal percentiles

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables