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Question
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
Options
`sum_(i = 1)^n (x_i - barx)`
`1/n sum_(i = 1)^n |x_i - barx|`
`sum_(i = 1)^n (x_i - barx)^2`
`1/n sum_(i = 1)^n (x_i - barx)^2`
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Solution
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by `1/n sum_(i = 1)^n |x_i - barx|`.
Explanation:
M.D. = `1/n sum_(i = 1)^n |x_i - barx|`
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