The standard deviation of n observations a_{1 | vedclass

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(C) Let the observations be $x_{i} = a_{i}$ for $i = 1, 2, \ldots, n$. The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{1}{n} \sum_{i

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$|\lambda| \sigma$

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(C) Let the observations be $x_{i} = a_{i}$ for $i = 1, 2, \ldots, n$. The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^2}$.Let the new observations be $y_{i} = \lambda a_{i} = \lambda x_{i}$.The mean of the new observations is $\bar{y} = \frac{1}{n} \sum_{i=1}^{n} \lambda x_{i} = \lambda \left( \frac{1}{n} \sum_{i=1}^{n} x_{i} \right) = \lambda \bar{x}$.The standard deviation of the new observations $\sigma_{y}$ is:$\sigma_{y} = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (y_{i} - \bar{y})^2}$$= \sqrt{\frac{1}{n} \sum_{i=1}^{n} (\lambda x_{i} - \lambda \bar{x})^2}$$= \sqrt{\frac{1}{n} \sum_{i=1}^{n} \lambda^2 (x_{i} - \bar{x})^2}$$= \sqrt{\lambda^2 \cdot \frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^2}$$= \sqrt{\lambda^2} \cdot \sigma$$= |\lambda| \sigma$.

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

Size $(x_i)$	$3.5$	$4.5$	$5.5$	$6.5$	$7.5$	$8.5$	$9.5$
Frequency $(f_i)$	$3$	$7$	$22$	$60$	$85$	$32$	$8$

The standard deviation of $n$ observations $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ is $\sigma$. Then,the standard deviation of the observations $\lambda a_{1}, \lambda a_{2}, \ldots, \lambda a_{n}$ is

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$C$.$I$. $0$ - $6$ $6$ - $12$ $12$ - $18$
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Let $x_1, x_2, \dots, x_n$ be $n$ observations,$\bar{x}$ be their mean,and $\sigma^2$ be their variance.
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