Worked solution · Exam P
This Exam P sample reference tests Sums of Independent Random Variables. Independence lets the second component inherit mean 0.8 and variance 1.35 from the stated totals. Its coefficient of variation is √(1.35)/0.8 = 1.45237, so the answer is E.
Independent solution
Setup
Recover the mean of the unobserved component by subtracting component means.
Model
For independent components, the variance of their sum is the sum of their variances.
Compute
Divide the component standard deviation by its mean.
Answer
The coefficient of variation rounds to 1.45.
Test the method on unseen problems
30 original multivariate 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
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
Original practice · fully worked
A fabrication shop models two independent random resource loads, X and Y. Their means are 3 and 5 units, and their standard deviations are 1 and 2 units, respectively. A batch cost index is defined by C=2X+Y. Calculate the coefficient of variation of C.
The weighted cost has mean 2(3)+5=11. Independence gives variance 2 squared times 1 squared plus 2 squared, or 8. Therefore its coefficient of variation is √(8)/11=0.2571297, so choice B is correct.
Setup
Apply linearity of expectation to the weighted cost index.
Model
Independence removes the covariance term, while the coefficient on X is squared in the variance.
Compute
Divide the standard deviation of the cost index by its mean.
Answer
The cost-index coefficient of variation rounds to 0.257.
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