This Exam P sample reference tests Exponential Distribution. For an exponential variable, the mean and standard deviation are the same scale parameter. The scale is √(7225)=85 and the median is 85 ln(2), so median minus mean is -26.0825 and choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 7225(ln2-1)=-2217.0 uses the variance itself as the exponential scale instead of taking its square root.
CThe value zero treats the exponential distribution as symmetric with equal mean and median. Its right skew makes the median smaller.
DThe positive value 26 reverses the requested subtraction and computes mean minus median.
EThe value 2217 both uses the variance as the scale and reverses the subtraction order.
Original practice · fully worked
Original variant: recover a competing alarm's mean
Two independent alarms can end a laboratory run. The first alarm has an exponential waiting time with mean 4 minutes. The second alarm also has an exponential waiting time, with unknown mean m. The median time until the first alarm of either kind is 1 minute. Calculate m.
A 1.060
B 1.443
C 2.000
D 2.257
E 4.000
Variant answer in brief
The minimum has rate 1/4+1/m and median ln(2)/(1/4+1/m). Setting that median to one gives m=1/(ln2-1/4)=2.25659, so choice D.
Setup
Setup
Convert the two exponential means to rates and add them for the minimum waiting time M.
λM=41+m1
M∼Exponential(λM)
Model
Model
A median of one minute means the minimum survives one minute with probability one half.
Pr(M>1)=e−λM=21
λM=ln2
Compute
Compute
Remove the known alarm rate and invert the remaining rate.
m1=ln2−41
m=ln2−1/41=2.2565866237
Answer
Answer
The second alarm's mean waiting time is approximately 2.257 minutes.
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