This Exam P sample reference tests Variance. A fixed addition has no effect on variance, while multiplication by 1.03 multiplies variance by 1.03 squared. The result is 53.045, which corresponds to choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 52 is obtained by using 1.03 × 50, which rounds 51.5 upward. Variance scales by the square of 1.03, not by 1.03 itself.
CThe value 54 comes from 50 plus 3 percent of 50 plus the fixed 2.5, rounded to an integer. The fixed maintenance amount shifts the mean but contributes nothing to variance.
DThe value 56 is obtained by first computing the correct scaled variance 53.045 and then adding the fixed 2.5. A deterministic addition cannot create dispersion.
EThe value 59 is produced by 50+3+(2.5)²=59.25, treating the percentage as three variance units and the fixed charge as a random contribution. Neither operation follows the affine-variance rule.
Original practice · fully worked
Original variant: coefficient of variation after two reporting stages
A raw measurement X has mean 40 and variance 100. An intermediate report is Y=1.5X+12, and a final dashboard value is Z=(Y-12)/3+5. Calculate the coefficient of variation of Z, defined as its standard deviation divided by its mean.
A 0.125
B 0.20
C 0.40
D 0.25
E 1.00
Variant answer in brief
Composing the two reporting stages gives Z=0.5X+5. Thus Z has mean 25 and standard deviation 5, so its coefficient of variation is 5/25=0.20 and choice B.
Setup
Setup
First compose the two affine rules so the final value is written directly in terms of the raw measurement.
Z=3(1.5X+12)−12+5=0.5X+5
Model
Model
Track location and spread separately through the composite transformation.
E[Z]=0.5E[X]+5
Var(Z)=(0.5)2Var(X)
Compute
Compute
Evaluate the final mean, standard deviation, and their ratio.
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