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Mean deviation for n observations x1, x2...

Mean deviation for n observations `x_1, x_2,..., x_n` from their mean x is given by

A

`sum_(i=1)^n(x_i-barx)`

B

`1/nsum_(i=1)^n|x_i-barx|`

C

`sum_(i=1)^n(x_i-barx)^2`

D

`1/nsum_(i=1)^n(x_i-barx)^2`

Text Solution

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The correct Answer is:
B
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