This Exam P sample reference tests Sums of Independent Random Variables. This problem asks for the standard deviation of an independent sum. Squaring the four component standard deviations and adding gives variance 30, so the total standard deviation is √(30)=5.4772 and choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value (1+2+3+4)/4=2.5 averages the component standard deviations; an aggregate is not an average of its components.
CThe value (1²+2²+3²+4²)/4=7.5 averages the four variances and also omits the final square root.
DThe value 1+2+3+4=10 adds standard deviations directly. Independence supports addition of the squared standard deviations.
EA standard deviation of 12.5 would imply variance 156.25, which is incompatible with the independently summed variance 30.
Original practice · fully worked
Original variant: uncertainty in a corrected sensor reading
Three independent centered calibration errors U, V, and W have standard deviations 1.5, 2, and 4 units. A corrected reading has error R=2U-V+0.5W. Calculate the standard deviation of R.
A 3.000
B 4.123
C 7.000
D 17.000
E 49.000
Variant answer in brief
Variance scales with the square of each coefficient. Independence gives Var(R)=4(1.5²)+2²+0.5²(4²)=17, so SD(R)=√(17)=4.123 and choice B.
Setup
Setup
Write the variance rule for the stated linear combination of independent errors.
R=2U−V+0.5W
Var(R)=22Var(U)+(−1)2Var(V)+(0.5)2Var(W)
Model
Model
Convert the three supplied standard deviations to variances before applying the squared coefficients.
Var(U)=1.52,Var(V)=22,Var(W)=42
Compute
Compute
Add the three weighted variance contributions, then return to standard-deviation units.
Var(R)=4(2.25)+4+0.25(16)=17
SD(R)=17=4.1231056256
Answer
Answer
The corrected reading has standard deviation approximately 4.123 units.
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