This Exam P sample reference tests Exponential Distribution. This is an exponential memorylessness problem. Conditioning on survival to the stated age shifts the lifetime but leaves the residual variance equal to the square of the nine-year mean, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 14 is the conditional mean, obtained by adding the five elapsed years to the nine-year expected residual lifetime; it is not a variance.
BThe value 16 comes from subtracting the elapsed age from the original mean and then squaring. Memorylessness does not shorten the residual mean from nine to four.
DThe value 106 adds the square of the five-year shift to 81. A deterministic shift contributes zero variance, so that extra 25 must not be included.
EThe value 196 is the square of the conditional mean, 14 squared. Variance is the second central moment, not the squared mean.
Original practice · fully worked
Original variant: archive-drive inspection
A cold-storage archive drive has an exponentially distributed operating lifetime with mean six months. A technician inspects a drive that is still running after four months. Calculate the conditional variance, in square months, of its total operating lifetime.
A 4
B 10
C 16
D 36
E 100
Variant answer in brief
The four completed months are a fixed shift, while the remaining lifetime is exponential with mean six months. Its variance is therefore 6 squared, or 36, selecting choice D.
Setup
Setup
Let L be the total operating lifetime. The exponential mean of six corresponds to rate 1/6.
L∼Exp(1/6)
Model
Model
After conditioning on operation through month four, memorylessness supplies a fresh residual variable R with the same exponential law.
L∣(L≥4)=d4+R,R∼Exp(1/6)
Compute
Compute
The constant four has no variance. An exponential variable's variance is the square of its mean.
Var(L∣L≥4)=Var(R)=62=36
Answer
Answer
The requested conditional variance is 36 square months.
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