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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 $x_{1}, x_{2} \dots x_{18}$ is

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detailed solution

Correct option is C

Let di=xi−8 but σx2=σd2=118∑di2−118∑di2=118×45−9182=52−14=94Therefore σx=3/2.

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If ∑i=118 xi−8=9 and ∑i=118 xi−82=45then 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 $x_{1}, x_{2} \dots x_{18}$ is