This Exam P sample reference tests Sums of Independent Random Variables. The independent components have combined mean 250 and combined standard deviation √(40²+30²)=50. Dividing 50 by 250 gives coefficient of variation 0.20, so choice A is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis adds the component standard deviations to get 70 and divides by 250, instead of adding variances.
CThis averages the two individual coefficients of variation, 0.40 and 0.20, without weighting the independent aggregate correctly.
DThis reports the first component's coefficient of variation, 40/100, rather than that of the sum.
EThis adds the two component coefficients of variation, even though coefficients of variation are not additive.
Original practice · fully worked
Original variant: unknown component spread
Two independent laboratory costs A and B have means 40 and 60. The standard deviation of A is 6, while the coefficient of variation of A+B is 0.10. Calculate the standard deviation of B.
A 4
B 6
C 8
D 10
E 16
Variant answer in brief
The combined mean is 100, so CV 0.10 implies combined standard deviation 10 and variance 100. Independence gives Var(B)=100-36=64, hence SD(B)=8 and choice C is correct.
Setup
Setup
Use the combined mean and coefficient of variation to recover the total standard deviation.
E[A+B]=40+60=100
SD(A+B)=0.10(100)=10
Model
Model
Independent component variances add to the total variance.
102=62+Var(B)
Compute
Compute
Isolate the unknown variance and take its square root.
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