This Exam P sample reference tests Normal Distribution. The sum of n normal lifetimes has mean 3n and standard deviation √(n). Requiring the 40-month threshold two standard deviations below the mean gives n=16, choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AAt n=14 the standardized margin is only about 0.535, far below the required value 2, so 14 does not qualify.
CThe value 20 exceeds the boundary solution and qualifies, but it is not the count determined by the equality among the listed choices.
DThe value 40 is not obtained by setting the mean 3n equal to the 40-month threshold; that equation would give n=13.33.
Original practice · fully worked
Original variant: minimum battery modules for a reliability target
Independent battery modules each supply a normally distributed operating time with mean 6 hours and variance 4 hours squared. What is the smallest listed number of modules whose combined life exceeds 50 hours with probability at least 0.9332?
A 8 modules
B 9 modules
D 10 modules
C 11 modules
E 12 modules
Variant answer in brief
For n modules, the sum has mean 6n and standard deviation 2 √(n). At n=10 the standardized 50-hour threshold is -1.581, exceeding the 0.9332 target; n=9 does not.
Setup
Setup
For n independent modules, total life is normal with mean 6n and standard deviation 2 times the square root of n.
Sn∼N(6n,4n)
P(Sn>50)≥0.9332
Model
Model
The target probability 0.9332 requires a standardized safety margin of at least 1.5. Check adjacent integer counts around the boundary.
P(Sn>50)=Φ(2n6n−50)
Compute
Compute
Nine modules give margin 0.6667 and fail; ten give margin 1.5811 and pass. Thus ten is the first qualifying integer.
n=9:z=64=0.6667<1.5
n=10:z=21010=1.5811>1.5
Answer
Answer
The minimum listed module count is 10, corresponding to choice D.
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