If  \(\sum_{i=1}^{18}(x_i\ -\ 8)\ =\ 3\) and  \(\sum_{i=1}^{18}(x_i\ -\ 8)^2\ =\ 45\)  then standard deviation of x1, x2, x3 ...x18 is:

1
\(\frac{3}{2}\)
2
​​\(\sqrt{{\frac{89}{36}}}\)
3
\(\frac{5}{2}\)
4
√5 

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