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Mean deviation for n observations x(1), ...

Mean deviation for n observations `x_(1), x_(2),….., x_(n)` from their mean `bar(x)` is given by :

A

`sum_(i = 1)^(n) (x_(i) - x)`

B

`1/nsum_(i = 1)^(n) |x_(i) - x|`

C

`sum_(i = 1)^(n) (x_(i) - bar x)`

D

`1/nsum_(i = 1)^(n) (x_(i) - x)`

Text Solution

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The correct Answer is:
B
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