This Exam P sample reference tests Joint Distributions. Sum the joint table over warranty count to get the marginal probabilities of X. Its two moments yield variance 83/144 = 0.5764, choice B.
Obtain the marginal distribution of the shipment count by summing the joint probabilities across every warranty-count value for each fixed shipment count.
P(X=0)=61
P(X=1)=121+61=41
P(X=2)=127
Model
Model
Use the marginal probabilities to compute the first and second raw moments.
E[X]=41+2127=1217
E[X2]=41+4127=1231
Compute
Compute
Subtracting the squared mean from the second raw moment gives variance approximately 0.576389.
Var(X)=1231−(1217)2=14483=0.576389
Answer
Answer
The marginal shipment count has variance approximately 0.58, selecting choice B.
0.58(B)
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AChoice A uses only one joint cell for each shipment-count value instead of summing the entire corresponding row or column.
CChoice C reports a probability-weighted quantity that does not subtract the squared mean.
DChoice D reports the mean of the shipment count rather than its variance.
EChoice E reports the second raw moment without centering it by the squared mean.
Original practice · fully worked
Original variant: marginal spread from a joint shipping table
A warehouse records packages shipped X and returns Y. Their joint probabilities are P(0,0)=0.20, P(1,0)=0.15, P(1,1)=0.25, P(2,0)=0.10, P(2,1)=0.20, and P(2,2)=0.10. Find Var(X).
A 0.36
C 0.49
B 0.56
D 0.64
E 0.80
Variant answer in brief
The marginal probabilities for X are 0.20, 0.40, and 0.40. Thus E[X]=1.2, E[X²]=2.0, and Var(X)=0.56.
Setup
Setup
Sum the joint cells for each shipment count. The resulting marginal probabilities at zero, one, and two are 0.20, 0.40, and 0.40.
PX(0)=0.20,PX(1)=0.40,PX(2)=0.40
Model
Model
The marginal distribution gives mean 1.2 and second raw moment 2.0.
E[X]=0.4+0.8=1.2
E[X2]=0.4+1.6=2.0
Compute
Compute
Subtracting the squared mean from the second raw moment gives variance 0.56.
Var(X)=2.0−(1.2)2=0.56
Answer
Answer
The shipment count has variance 0.56, selecting choice B.
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