Verified solution · Exam P
This Exam P sample reference tests Sums of Independent Random Variables. Let l be the low-variability count. Independence makes total variance 0.5²l+5.5²(76-l), and setting this equal to 43² gives l=15. Therefore choice B is correct.
Independent solution
Setup
Let l denote the count in the lower-standard-deviation class. The other class then has 76-l members.
Model
For independent individual amounts, variances add. Square each class standard deviation and the total standard deviation.
Compute
Expand the variance equation and solve for the requested class count.
Answer
There are 15 members in the lower-variability class.
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Original practice · fully worked
A monitoring array contains 12 independent low-noise channels and 8 independent high-noise channels. Each low-noise reading has standard deviation 2, while each high-noise reading has the same unknown standard deviation s. The standard deviation of the sum of all 20 readings is 10. Calculate s.
Independent variances add, so 10²=12(2²)+8s². This gives s²=6.5 and s=√(6.5)=2.5495, making choice B correct.
Setup
Translate each standard deviation into a variance contribution.
Model
Independence makes the variance of the full sum equal the sum of all channel variances.
Compute
Remove the low-noise contribution, divide by eight, and take the positive square root.
Answer
The high-noise channel standard deviation rounds to 2.550.
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ActuaryProof. “SOA Exam P Sample Question #542 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
