This Exam P sample reference tests Exponential Distribution. This problem uses variance additivity for two independent exponential intervals. Removing the first interval's variance leaves 2.01 for the second, whose exponential mean is √(2.01)=1.4177, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.83 is approximately √(2.65)-0.80. That incorrectly adds standard deviations; independence makes variances additive.
BSquaring 0.96 and restoring the first variance gives 0.96²+0.80²=1.5616, which fails to reproduce the stated total variance 2.65.
DThe value 1.85 comes from 2.65-0.80, mixing a variance with a mean instead of subtracting the first variance 0.80².
EThe value 2.01 is the later interval's variance. An exponential mean is the positive square root of that variance.
Original practice · fully worked
Original variant: recover a signal mean from calibration noise
A telescope reports R=S-E, where the nonnegative signal S and the centered calibration error E are independent. The error has standard deviation 3 units. The signal is gamma distributed with shape 4 and unknown scale θ. Across repeated readings, R has standard deviation 5 units. Calculate the mean of S.
A 4
B 5
C 8
D 16
E 20
Variant answer in brief
The sign on an independent component does not change its variance. Thus Var(S)=25-9=16; gamma shape 4 gives θ=2 and E[S]=4(2)=8, selecting choice C.
Setup
Setup
Translate the two reported standard deviations into variances and write the gamma moment formulas.
Var(R)=25
Var(E)=9
Var(S)=4θ2,E[S]=4θ
Model
Model
Because S and E are independent, subtraction still contributes the error variance with a positive coefficient.
Var(S−E)=Var(S)+Var(E)
25=4θ2+9
Compute
Compute
Solve for the positive gamma scale, then multiply it by the shape to obtain the signal mean.
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