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Question
If `barx` is the mean of n values of x, then `sum_(i = 1)^n (x_i - barx)` is always equal to ______. If a has any value other than `barx`, then `sum_(i = 1)^n (x_i - barx)^2` is ______ than `sum(x_i - a)^2`
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Solution
If `barx` is the mean of n values of x, then `sum_(i = 1)^n (x_i - barx)` is always equal to 0. If a has any value other than `barx`, then `sum_(i = 1)^n (x_i - barx)^2` is less than `sum(x_i - a)^2`
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