This Exam P sample reference tests Covariance. This is a variance-of-a-sum calculation with nonzero covariance. The two marginal variances contribute 5 and 8, while covariance enters twice, so the total variance is 5+8+2(3)=19 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 7 is 5+8-2(3). It uses the covariance contribution with the wrong sign.
BThe value 10 is 5+8-3. It both subtracts the covariance and includes it only once.
CThe value 13 is 5+8. It treats the two counts as uncorrelated and drops the supplied covariance.
DThe value 16 is 5+8+3. Covariance occurs twice in the variance of a sum, not once.
Original practice · fully worked
Original variant: combined count from exclusive inspection classes
Twenty components are inspected independently. Each is classified as cosmetic rework with probability 0.30, functional rework with probability 0.20, or acceptable otherwise. Let X and Y be the counts in the two rework classes. Calculate Var(X+Y).
A 1.20
B 3.20
C 5.00
D 7.40
E 9.80
Variant answer in brief
For each component, membership in either rework class has probability 0.50. Therefore X+Y is Binomial(20,0.50) and has variance 20(0.50)(0.50)=5, so choice C.
Setup
Setup
Combine the two mutually exclusive rework labels into one indicator for whether a component requires any rework.
prework=0.30+0.20=0.50
Model
Model
The combined count is binomial because the twenty component classifications are independent.
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