Verified solution · Exam P
This Exam P sample reference tests Binomial Distribution. The count is binomial. Its mean fixes the common success probability at 3/4, after which np(1-p)=12(3/4)(1/4)=2.25, so choice C is correct.
Independent solution
Setup
Represent each person's qualifying outcome by an independent Bernoulli indicator and let N be the sum of the twelve indicators.
Model
First recover the Bernoulli probability from the supplied mean, then use the binomial variance formula.
Compute
Substitute n=12, p=3/4, and 1-p=1/4.
Answer
The variance of the count is 2.25.
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Original practice · fully worked
A dashboard totals the pass indicators from 16 independent devices, all with the same pass probability p. The variance of the displayed total is 3.84, and engineers know p is greater than 0.50. Calculate the expected number of passing devices.
The variance equation gives p(1-p)=0.24, whose roots are 0.40 and 0.60. The condition p>0.50 selects 0.60, so the expected count is 16(0.60)=9.60 and choice D.
Setup
Let T be the total number of passing devices. Equal independent pass probabilities make T binomial.
Model
Solve the variance equation for its two symmetric probability roots and use the directional condition to choose one.
Compute
The quadratic roots are 0.40 and 0.60; only the larger root satisfies the engineering information.
Answer
The expected number of passing devices is 9.60.
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These references share syllabus concepts or sit nearby in the official sequence. All 388 Exam P univariate random variables practice problems.
Drill the method: Exam P Variance problems
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ActuaryProof. “SOA Exam P Sample Question #494 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
