If the sum of squares of the deviations from the mean of the data $x_i, (i=1, 2, \ldots, n)$ is $n\bar{x}^2$,where $\bar{x}$ is the mean of $x_i$'s,then the sum of squares of $x_i$'s is

  • A
    $4n\bar{x}^2$
  • B
    $3n\bar{x}^2$
  • C
    $n\bar{x}^2$
  • D
    $2n\bar{x}^2$

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