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$.

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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