The mean and the standard deviation of a data | vedclass

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(A) Given: $n_1 = 8$,$\bar{x}_1 = 25$,$\sigma_1 = 5$. Sum of items: $\Sigma x_i = n_1 	imes \bar{x}_1 = 8 	imes 25 = 200$. Variance: $\sigma_1^2 = \

Vedclass · Accepted Answer

$29$

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(A) Given: $n_1 = 8$,$\bar{x}_1 = 25$,$\sigma_1 = 5$. Sum of items: $\Sigma x_i = n_1 	imes \bar{x}_1 = 8 	imes 25 = 200$. Variance: $\sigma_1^2 = \frac{\Sigma x_i^2}{n_1} - (\bar{x}_1)^2 = 25$. $\frac{\Sigma x_i^2}{8} - 625 = 25 \Rightarrow \Sigma x_i^2 = 8 	imes 650 = 5200$. New data set: $n_2 = 8 + 2 = 10$. New sum: $\Sigma x_{new} = 200 + 15 + 25 = 240$. New mean: $\bar{x}_2 = \frac{240}{10} = 24$. New sum of squares: $\Sigma x_{new}^2 = 5200 + 15^2 + 25^2 = 5200 + 225 + 625 = 6050$. New variance: $\sigma_2^2 = \frac{\Sigma x_{new}^2}{n_2} - (\bar{x}_2)^2 = \frac{6050}{10} - (24)^2 = 605 - 576 = 29$.

The mean and the standard deviation of a data set of $8$ items are $25$ and $5$ respectively. If two items $15$ and $25$ are added to this data,then the variance of the new data is:

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