This Exam P sample reference tests Gamma Distribution. The density kernel is gamma with shape 6 and rate 0.01. Its mean is 6/0.01=600, so choice E is correct.
Match the power and exponential factor to a gamma density written in shape-rate form.
x5e−0.01x=x6−1e−0.01x
α=6,β=0.01
Model
Model
For a gamma variable with shape alpha and rate beta, the normalizing constant does not need to be found separately to use the mean formula.
E[X]=βα
Compute
Compute
Substitute the identified shape and rate.
E[X]=0.016=600
Answer
Answer
The expected value is 600.
600(E)
Test the method on unseen problems
Probability Univariate Distributions Practice Workbook
30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis reports only the gamma scale 1/0.01=100 and omits the shape multiplier 6.
BThis treats the polynomial factor as x rather than x⁵, incorrectly assigning shape 2 and obtaining 2/0.01=200.
CThis halves the identified shape from 6 to 3 and therefore computes 3/0.01=300.
DThis reports the gamma mode (alpha-1)/beta=(6-1)/0.01=500 instead of the mean.
Original practice · fully worked
Original variant: infer gamma variance from field summaries
A sensor's recalibration delay T follows a gamma distribution with shape alpha and scale θ. Field engineers report a mode of 6 minutes and a coefficient of variation of 1/2. Calculate Var(T).
A 4
B 8
C 12
D 16
E 64
Answer and calculation
The coefficient of variation gives shape 4, and the mode then gives scale 2. Thus the variance is 4(2²)=16, so choice D is correct.
Setup
Setup
Use the gamma coefficient of variation to recover the shape parameter.
CV(T)=α1=21
α=4
Model
Model
For shape greater than one, the gamma mode identifies the scale after the shape is known.
mode(T)=(α−1)θ
Compute
Compute
Recover the scale and then evaluate the gamma variance.