If $\sum\limits_{i = 1}^{18} {(x_i - 8) = 9}$ and $\sum\limits_{i = 1}^{18} {(x_i - 8)^2 = 45}$,then the standard deviation of $x_1, x_2, \dots, x_{18}$ is:

  • A
    $4/9$
  • B
    $9/4$
  • C
    $3/2$
  • D
    None of these

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Two teams $A$ and $B$ have the same mean and their coefficients of variation are $4$ and $2$,respectively. If $\sigma_A$ and $\sigma_B$ are the standard deviations of teams $A$ and $B$ respectively,then the relation between them is

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