This Exam P sample reference tests Exponential Distribution. An exponential lifetime has standard deviation equal to its mean. Solving the supplied five-year failure probability gives a mean of 9.788 years, so choice D is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis computes the rate, -ln(0.60)/5=0.1022, but reports that rate as the standard deviation.
BThis takes an extra square root, √(9.788)=3.128, as though the exponential mean were a variance.
CThis treats 0.40 as the survival probability instead of the failure probability, giving -5/ln(0.40)=5.46.
EThis divides 5 by 0.40 to get 12.50, incorrectly replacing the exponential distribution function by a linear proportion.
Original practice · fully worked
Original variant: calibration survival after inspection
The time until a monitoring unit loses calibration is exponentially distributed. Its 75th percentile is 10 days. A unit has already remained calibrated for 6 days. Calculate the probability that it remains calibrated for at least 5 additional days.
A 0.2176
B 0.2500
C 0.3750
D 0.5000
E 0.7500
Variant answer in brief
The percentile statement gives S(10)=0.25. Memorylessness reduces the requested conditional probability to S(5)=√(0.25)=0.50, so choice D is correct.
Setup
Setup
Translate the 75th percentile into a survival probability.
F(10)=0.75,S(10)=0.25
Model
Model
Exponential memorylessness makes the additional five-day survival probability independent of the six days already observed.
Pr(T>11∣T>6)=Pr(T>5)=S(5)
Compute
Compute
The exponential survival over half as much time is the square root of the ten-day survival.
S(5)=S(10)5/10=0.251/2=0.50
Answer
Answer
The unit has a one-half chance of lasting at least five more days.
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