The variance of 20 observations is 5. If each | vedclass

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(B) Let the observations be $x_{1}, x_{2}, \ldots, x_{20}$ and $\bar{x}$ be their mean. Given that variance $= 5$ and $n = 20$.We know that variance $

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$20$

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(B) Let the observations be $x_{1}, x_{2}, \ldots, x_{20}$ and $\bar{x}$ be their mean. Given that variance $= 5$ and $n = 20$.We know that variance $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.Thus,$5 = \frac{1}{20} \sum_{i=1}^{20} (x_i - \bar{x})^2$,which implies $\sum_{i=1}^{20} (x_i - \bar{x})^2 = 100$.If each observation is multiplied by a constant $k = 2$,the new observations are $y_i = 2x_i$.The new mean is $\bar{y} = 2\bar{x}$.The new variance is $\sigma_{new}^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2 = \frac{1}{20} \sum_{i=1}^{20} (2x_i - 2\bar{x})^2$.$\sigma_{new}^2 = \frac{1}{20} \sum_{i=1}^{20} 4(x_i - \bar{x})^2 = 4 	imes \left( \frac{1}{20} \sum_{i=1}^{20} (x_i - \bar{x})^2 ight) = 4 	imes 5 = 20$.Therefore,the new variance is $20$.

The variance of $20$ observations is $5$. If each observation is multiplied by $2$,find the new variance of the resulting observations.

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