For data X = a, a+1, a+2, ..., a+9 (10 consecutive integers), variance:

Question

Accepted 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.

Answer

10

Answer

Cannot determine

Answer

9

For data X = a, a+1, a+2, ..., a+9 (10 consecutive integers), variance:

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