This Exam P sample reference tests Limited Random Variables. The replacement age is min(X,4). Its first two moments are 2.4 and 7.4667, giving variance 1.7067 and official choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 1.3 is the rounded variance of a uniform variable on the interval from 0 to 4. It incorrectly spreads the capped mass continuously over that interval.
EThe value 7.5 approximates the second moment 7.4667 without subtracting the squared mean 5.76.
Original practice · fully worked
Original variant: variance of a capped discrete delay score
A delivery delay T equals one hour on 20% of trips, four hours on 50%, and seven hours on the other 30%. A dashboard records Y=min(T,5), so the longest delay is entered as five hours. Find Var(Y).
A 1.21
B 1.70
C 2.01
D 2.49
E 3.30
Variant answer in brief
The recorded values are 1, 4, and 5. Their mean is 3.7 and second moment 15.7, so the variance is 2.01.
Setup
Setup
Capping the seven-hour delay at five changes the three possible recorded values to 1, 4, and 5 while preserving their probabilities.
Y∈{1,4,5}with probabilities 0.2,0.5,0.3
Model
Model
Compute the first two moments directly from the three-point distribution of the recorded value Y.
E[Y]=0.2(1)+0.5(4)+0.3(5)=3.7
E[Y2]=0.2(1)+0.5(16)+0.3(25)=15.7
Compute
Compute
The first moment is 3.7 and the second raw moment is 15.7, so subtracting the squared mean gives variance 2.01.
Var(Y)=15.7−(3.7)2=2.01
Answer
Answer
Thus the dashboard-recorded delivery delay has variance 2.01, corresponding to choice C.
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