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.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis results from mixing the first component's mean with the second component's variance rather than using a consistent standard-deviation-to-mean ratio.
BThis uses an incorrect variance subtraction or divides by a total-scale quantity instead of the mean of the requested component.
CThis is the standard deviation √(1.35) by itself, before division by the component mean.
DThis is the variance 1.35, not the coefficient of variation.
Original practice · fully worked
Original variant: weighted fabrication cost
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.
A 0.223
B 0.257
C 0.354
D 0.364
E 0.727
Variant answer in brief
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
Setup
Apply linearity of expectation to the weighted cost index.
E[C]=2E[X]+E[Y]=2(3)+5=11
Model
Model
Independence removes the covariance term, while the coefficient on X is squared in the variance.
Var(C)=22Var(X)+Var(Y)
Var(C)=4(12)+22=8
Compute
Compute
Divide the standard deviation of the cost index by its mean.
CV(C)=E[C]Var(C)
CV(C)=118=0.2571297386…
Answer
Answer
The cost-index coefficient of variation rounds to 0.257.
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