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(B) Given: $\sum_{i=1}^n (x_i + 1)^2 = 9n$  $(1)$$\sum_{i=1}^n (x_i - 1)^2 = 5n$  $(2)$Expanding both equations:$\sum (x_i^2 + 2x_i + 1) = 9n$ $\Right

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$\sqrt{5}$

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(B) Given: $\sum_{i=1}^n (x_i + 1)^2 = 9n$  $(1)$$\sum_{i=1}^n (x_i - 1)^2 = 5n$  $(2)$Expanding both equations:$\sum (x_i^2 + 2x_i + 1) = 9n$ $\Rightarrow \sum x_i^2 + 2\sum x_i + n = 9n$ $\Rightarrow \sum x_i^2 + 2\sum x_i = 8n$  $(3)$$\sum (x_i^2 - 2x_i + 1) = 5n$ $\Rightarrow \sum x_i^2 - 2\sum x_i + n = 5n$ $\Rightarrow \sum x_i^2 - 2\sum x_i = 4n$  $(4)$Adding $(3)$ and $(4)$:$2\sum x_i^2 = 12n \Rightarrow \frac{\sum x_i^2}{n} = 6$Subtracting $(4)$ from $(3)$:$4\sum x_i = 4n \Rightarrow \frac{\sum x_i}{n} = 1$Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2 = 6 - (1)^2 = 5$Standard deviation $\sigma = \sqrt{5}$

$A$ data consists of $n$ observations $x_1, x_2, ......, x_n$. If $\sum_{i=1}^n (x_i + 1)^2 = 9n$ and $\sum_{i=1}^n (x_i - 1)^2 = 5n$,then the standard deviation of this data is

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