For 10 observations x_1, x_2, ..., x_{10}, if | vedclass

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(D) Let $\sum x_i^2 = S_2$ and $\sum x_i = S_1$. Expanding the given equations: $S_2 + 4S_1 + 40 = 180 \implies S_2 + 4S_1 = 140$ $S_2 - 2S_1 + 10 = 9

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$3$

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(D) Let $\sum x_i^2 = S_2$ and $\sum x_i = S_1$. Expanding the given equations: $S_2 + 4S_1 + 40 = 180 \implies S_2 + 4S_1 = 140$ $S_2 - 2S_1 + 10 = 90 \implies S_2 - 2S_1 = 80$ Subtracting the two equations: $(S_2 + 4S_1) - (S_2 - 2S_1) = 140 - 80 \implies 6S_1 = 60 \implies S_1 = 10$. Substituting $S_1 = 10$ into $S_2 - 2S_1 = 80$: $S_2 - 20 = 80 \implies S_2 = 100$. Variance $\sigma^2 = \frac{S_2}{n} - (\frac{S_1}{n})^2 = \frac{100}{10} - (\frac{10}{10})^2 = 10 - 1 = 9$. Standard deviation $\sigma = \sqrt{9} = 3$.

$x_i$	$0$	$1$	$5$	$6$	$10$	$12$	$17$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$k^2-1$	$k^2-1$	$k^2-1$	$k-3$

For $10$ observations $x_1, x_2, ..., x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:

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