Question
A factory independently produces two trial batches. Let \(X\) be the number of defects in its 30-item morning batch and \(Y\) the number of defects in its 20-item afternoon batch. The batch totals are independent, with
\[
\operatorname{Var}(X)=9,\qquad \operatorname{Var}(Y)=4.
\]
The quality supervisor compares the two defect rates, using the random difference
\[
D=\frac{X}{30}-\frac{Y}{20}.
\]
What is \(\operatorname{Var}(D)\)? How would your calculation change if the two batch totals were correlated, and why does the minus sign not make the second variance contribution negative?
Get hint
Hint 1
For independent random variables, the variance of a weighted sum uses the squares of the fixed weights.
Check your answer
Solution
For any square-integrable \(X,Y\) and fixed constants \(a,b\),
\[
\operatorname{Var}(aX+bY)
=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y).
\]
Here \(a=1/30\) and \(b=-1/20\). The batch totals are independent, so \(\operatorname{Cov}(X,Y)=0\). Consequently,
\[
\operatorname{Var}(D)
=\left(\frac1{30}\right)^2 9
+\left(-\frac1{20}\right)^2 4
=\frac9{900}+\frac4{400}
=0.01+0.01=\boxed{0.02}.
\]
The standard deviation of the defect-rate difference is therefore \(\sqrt{0.02}\approx0.1414\), although only the variance was requested.
If the batch totals were correlated, the full answer would be
\[
\operatorname{Var}(D)=0.02-\frac{2}{30\cdot20}\operatorname{Cov}(X,Y).
\]
The negative weight for \(Y\) reverses its direction in the difference and in its mean; it does not create negative uncertainty. Since \((-1/20)^2=1/400\), \(Y\) contributes positively to variance. Dependence could add either a positive or a negative cross term, but independence makes that term vanish here.
Assumptions
\(X\) and \(Y\) are random defect counts for separate batches, and are independent across batches.
Their individual count variances are exactly 9 and 4; no assumptions about within-batch defect independence are required.
\(D\) compares fraction-defective rates for the two listed batch sizes, so its coefficients are exactly \(+1/30\) and \(-1/20\).
Final Answer
\(\operatorname{Var}(D)=9/30^2+4/20^2=1/50=0.02\) (squared defect-rate units). Without independence an additional term \(-2\operatorname{Cov}(X,Y)/(30\cdot20)\) would appear.