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If $\sum_{i = 1}^{18} (x_{i} - 18) = 9 a n d \sum_{i = 1}^{18} {(x_{i} - 18)}^{2} = 45$ then the standard deviatioin of $x 1, x 2 … … … x 18$ is

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49

32

23

94

answer is C.

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Detailed Solution

Let

d i = x i − 8 but σ x 2 = σ d 2 = 118 ∑ d i 2 − 118 ∑ d i 2 = 118 × 45 − 9182 = 52 − 14 = 94

∴ σ x = 32

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