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(D) Given $n = 15$,$	ext{mean} (\bar{x}) = 8$,and $	ext{standard deviation} (\sigma) = 3$.$	ext{Variance} = \sigma^2 = 3^2 = 9$.Using the formula $

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$17$

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(D) Given $n = 15$,$	ext{mean} (\bar{x}) = 8$,and $	ext{standard deviation} (\sigma) = 3$.$	ext{Variance} = \sigma^2 = 3^2 = 9$.Using the formula $	ext{Variance} = \frac{\sum x_i^2}{n} - (\bar{x})^2$,we have $9 = \frac{\sum x_i^2}{15} - 8^2$.$\frac{\sum x_i^2}{15} = 9 + 64 = 73 \Rightarrow \sum x_i^2 = 15 	imes 73 = 1095$.Also,$\sum x_i = n 	imes \bar{x} = 15 	imes 8 = 120$.Correcting the values: $20$ was misread as $5$,so we subtract $5$ and add $20$.Corrected $\sum x_i = 120 - 5 + 20 = 135$.Corrected $\sum x_i^2 = 1095 - 5^2 + 20^2 = 1095 - 25 + 400 = 1470$.Corrected $	ext{mean} (\bar{x}_{new}) = \frac{135}{15} = 9$.Corrected $	ext{Variance} = \frac{1470}{15} - (9)^2 = 98 - 81 = 17$.

The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking,it was found that,in the observations,$20$ was misread as $5$. Then,the correct variance is equal to......

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