A data consists of 20 observations x_1, x_2, | vedclass

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(B) Let $S_1 = \sum_{i=1}^{20} (x_i + 5)^2 = \sum x_i^2 + 10\sum x_i + 500 = 2500$. Let $S_2 = \sum_{i=1}^{20} (x_i - 5)^2 = \sum x_i^2 - 10\sum x_i +

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$3:1$

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(B) Let $S_1 = \sum_{i=1}^{20} (x_i + 5)^2 = \sum x_i^2 + 10\sum x_i + 500 = 2500$. Let $S_2 = \sum_{i=1}^{20} (x_i - 5)^2 = \sum x_i^2 - 10\sum x_i + 500 = 100$. Subtracting $S_2$ from $S_1$: $20\sum x_i = 2400 \Rightarrow \sum x_i = 120$. The mean $\bar{x} = \frac{\sum x_i}{n} = \frac{120}{20} = 6$. Adding $S_1$ and $S_2$: $2\sum x_i^2 + 1000 = 2600 \Rightarrow 2\sum x_i^2 = 1600 \Rightarrow \sum x_i^2 = 800$. The variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{800}{20} - 6^2 = 40 - 36 = 4$. The standard deviation $\sigma = \sqrt{4} = 2$. The ratio of mean to standard deviation is $\frac{\bar{x}}{\sigma} = \frac{6}{2} = 3:1$.

$C$.$I$.	$0$ - $6$	$6$ - $12$	$12$ - $18$
f_i	$2$	$4$	$6$

$A$ data consists of $20$ observations $x_1, x_2, ..., x_{20}$. If $\sum_{i=1}^{20} (x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20} (x_i - 5)^2 = 100$, then the ratio of mean to standard deviation of this data is:

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