Worked solution · Exam P
This Exam P sample reference tests Binomial Distribution. This is the standard deviation of a binomial count. With n=500 and p=0.12, the variance is 500(0.12)(0.88)=52.8, so the standard deviation is √(52.8)≈7.27 and choice A.
Independent solution
Setup
Independent returns with a common probability produce a binomial count.
Model
Use the binomial variance formula and then take its square root.
Compute
Substitute the item count and return probability.
Answer
The standard deviation rounds to 7.27.
Test the method on unseen problems
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.
Error diagnosis
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Original practice · fully worked
An audit runs 240 independent checks, each passing with probability 0.35. The audit score adds 5 points for every pass and subtracts 2 points for every failed check. Calculate the standard deviation of the total audit score.
If N is the pass count, the total score is 5N-2(240-N)=7N-480. The pass-count standard deviation is sqrt[240(0.35)(0.65)], and scaling by 7 gives 51.724, choice B.
Setup
Represent the number of passing checks by a binomial variable.
Model
Rewrite the total score as an affine function of the pass count.
Compute
Calculate the pass-count variance and apply the score scale.
Answer
The standard deviation of the total score is approximately 51.724 points.
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