For a statistical data x_1, x_2, ldots, x_{10 | vedclass

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(A) Given the initial mean $\overline{x} = 5.5$ for $n = 10$ values,the initial sum is $\sum_{i=1}^{10} x_i = 5.5 	imes 10 = 55$. Given $\sum_{i=1}^{

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$7$

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(A) Given the initial mean $\overline{x} = 5.5$ for $n = 10$ values,the initial sum is $\sum_{i=1}^{10} x_i = 5.5 	imes 10 = 55$. Given $\sum_{i=1}^{10} x_i^2 = 371$. Correcting the sum: $(\sum x_i)_{	ext{new}} = 55 - (4 + 5) + (6 + 8) = 55 - 9 + 14 = 60$. Correcting the sum of squares: $(\sum x_i^2)_{	ext{new}} = 371 - (4^2 + 5^2) + (6^2 + 8^2) = 371 - (16 + 25) + (36 + 64) = 371 - 41 + 100 = 430$. The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2$. $\sigma^2 = \frac{430}{10} - (\frac{60}{10})^2 = 43 - 6^2 = 43 - 36 = 7$.

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