If the mean and standard deviation of 5 obser | vedclass

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(D) Given,$n_1 = 5$,$\bar{x} = 10$,and $\sigma = 3$.Sum of observations $\sum x_i = n_1 	imes \bar{x} = 5 	imes 10 = 50$.Variance $\sigma^2 = 9 = \f

Vedclass · Accepted Answer

$507$

Vedclass · Answer

(D) Given,$n_1 = 5$,$\bar{x} = 10$,and $\sigma = 3$.Sum of observations $\sum x_i = n_1 	imes \bar{x} = 5 	imes 10 = 50$.Variance $\sigma^2 = 9 = \frac{\sum x_i^2}{n_1} - (\bar{x})^2$.$9 = \frac{\sum x_i^2}{5} - 100 \implies \frac{\sum x_i^2}{5} = 109 \implies \sum x_i^2 = 545$.Now,we have $6$ observations: $x_1, x_2, x_3, x_4, x_5$ and $-50$.New sum $\sum x_{new} = 50 + (-50) = 0$.New mean $\bar{x}_{new} = \frac{0}{6} = 0$.New sum of squares $\sum x_{new}^2 = \sum x_i^2 + (-50)^2 = 545 + 2500 = 3045$.New variance $\sigma_{new}^2 = \frac{\sum x_{new}^2}{n_2} - (\bar{x}_{new})^2 = \frac{3045}{6} - 0^2 = 507.5$.

If the mean and standard deviation of $5$ observations $x_1, x_2, x_3, x_4, x_5$ are $10$ and $3$,respectively,then the variance of $6$ observations $x_1, x_2, x_3, x_4, x_5$ and $-50$ is equal to: (in $.5$)

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