This Exam P sample reference tests Exponential Distribution. This is an exponential interval probability centered at the mean. The distribution has mean and standard deviation 20, so the interval is from 10 to 30; subtracting the two exponential tails gives e to the minus one-half minus e to the minus three-halves, or 0.3834 and choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.34 is approximately Φ(1)-Φ(0). It substitutes a normal model and uses a one-sided interval one full standard deviation wide, rather than the stated exponential half-standard-deviation band.
CThe value 0.50 assumes that a mean-centered interval automatically contains half the probability. An exponential distribution is right-skewed, so its mass is not symmetric about its mean.
DThe value 0.68 is the familiar normal probability within one standard deviation of a mean. Both the distribution and the stated half-width differ here.
EThe value 0.95 invokes the normal two-standard-deviation rule. It does not evaluate the exponential probability between 10 and 30.
Original practice · fully worked
Original variant: mean absolute deviation of a self-check time
A device's self-check time X, measured in minutes, follows an exponential distribution with mean 12. Calculate the expected absolute difference between the self-check time and its mean, E[absolute value of X-12].
A 4.4146 minutes
B 8.8291 minutes
C 12.0000 minutes
D 17.6582 minutes
E 24.0000 minutes
Variant answer in brief
For an exponential time with mean θ, the expected excess above θ is the integrated survival tail θ/e. Centering implies the expected shortfall below the mean is equal, so the mean absolute deviation is 2theta/e=24/e=8.8291 minutes and choice B.
Setup
Setup
Center the exponential time at its mean and split absolute deviation into positive and negative parts.
Y=X−12
∣Y∣=Y++(−Y)+
Model
Model
Because E[Y]=0, the positive and negative deviation parts have equal expectations.
E[Y+]=E[(−Y)+]
E[∣Y∣]=2E[Y+]
Compute
Compute
Evaluate the expected excess above the mean with the exponential survival function.
E[(X−12)+]=∫12∞Pr(X>t)dt
=∫12∞e−t/12dt=12e−1
E[∣X−12∣]=e24=8.829106588
Answer
Answer
The mean absolute deviation is approximately 8.8291 minutes.
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