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$If \sum_{i = 1}^{18} (x_{i} - 8) = 9 and \sum_{i = 1}^{18} {(x_{i} - 8)}^{2} = 45, then the standard deviation of x_{1}, x_{2}, \dots \dots \dots, x_{18} is$

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Correct option is C

[SolutionStep1]: Given that ∑i=118xi-8=9 and ∑i=118xi-82=45 , Let di=xi-8 But σx2=σd2=118∑di2-118∑di2=118×45−9182=52−14=94 Standard deviation, σx=32

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If ∑i=118xi-8=9 and ∑i=118xi-82=45, then the standard deviation of x1,x2,………,x18 is

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$If \sum_{i = 1}^{18} (x_{i} - 8) = 9 and \sum_{i = 1}^{18} {(x_{i} - 8)}^{2} = 45, then the standard deviation of x_{1}, x_{2}, \dots \dots \dots, x_{18} is$