This Exam P sample reference tests Mixed Distributions. The distribution has an atom of one half at 1 and density x-1 on (1,2). Its first two moments yield variance 5/36, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
DThe value 4/3 is E[X], the first moment, rather than Var(X).
EThe value 23/12 is the second moment. Variance additionally requires subtracting the squared mean, 16/9.
Original practice · fully worked
Original variant: variance of a mixed inspection score
An inspection score S equals 1 with probability 0.4. Otherwise it is drawn uniformly from 1 to 3, independently of that mixture decision. Determine Var(S).
A 0.2000
E 0.2400
C 0.3333
D 0.4000
B 0.4400
Variant answer in brief
The mixture has mean 1.6 and second moment 3.0, so its variance is 3.0-1.6²=0.44.
Setup
Setup
Treat S as a two-component mixture: a mass at 1 with weight 0.4 and a Uniform(1,3) component with weight 0.6.
P(S=1)=0.4
S∣{S=1}∼Unif(1,3)
Model
Model
Weight the first and second moments of each component by its mixture probability before forming the variance.
E[S]=0.4(1)+0.6(2)=1.6
E[S2]=0.4(1)+0.6(312+1(3)+32)
Compute
Compute
The weighted first moment is 1.6 and the second raw moment is 3.0. Subtracting the squared mean gives variance 0.44.
E[S2]=3.0
Var(S)=3.0−(1.6)2=0.44
Answer
Answer
The inspection-score variance is 0.4400, corresponding to choice B.
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