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For data X = a, a+1, a+2, ..., a+9 (10 consecutive integers), variance:
A10
BCannot determine
C9
D(n²-1)/12 = (100-1)/12 = 99/12 = 8.25
Answer & Solution
Correct answer: D. (n²-1)/12 = (100-1)/12 = 99/12 = 8.25
Variance of n consecutive integers (or any AP with d=1) = (n²-1)/12, independent of starting value. For n=10: (100-1)/12 = 99/12 = 8.25. Mean shifts but variance is translation-invariant.
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