Let the mean and variance of 12 observations | vedclass

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(C) Given $n = 12$,$\bar{x} = \frac{9}{2}$,and $\sigma^2 = 4$. $\sum x = n 	imes \bar{x} = 12 	imes \frac{9}{2} = 54$. $\sigma^2 = \frac{\sum x^2}{n

Vedclass · Accepted Answer

$317$

Vedclass · Answer

(C) Given $n = 12$,$\bar{x} = \frac{9}{2}$,and $\sigma^2 = 4$. $\sum x = n 	imes \bar{x} = 12 	imes \frac{9}{2} = 54$. $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 \implies 4 = \frac{\sum x^2}{12} - (\frac{9}{2})^2$. $\frac{\sum x^2}{12} = 4 + \frac{81}{4} = \frac{16 + 81}{4} = \frac{97}{4}$. $\sum x^2 = 12 	imes \frac{97}{4} = 3 	imes 97 = 291$. Correct sum $\sum x_{	ext{new}} = 54 - (9 + 10) + (7 + 14) = 54 - 19 + 21 = 56$. Correct sum of squares $\sum x_{	ext{new}}^2 = 291 - (9^2 + 10^2) + (7^2 + 14^2) = 291 - (81 + 100) + (49 + 196) = 291 - 181 + 245 = 355$. Correct variance $\sigma_{	ext{new}}^2 = \frac{\sum x_{	ext{new}}^2}{n} - (\frac{\sum x_{	ext{new}}}{n})^2 = \frac{355}{12} - (\frac{56}{12})^2 = \frac{355}{12} - (\frac{14}{3})^2 = \frac{355}{12} - \frac{196}{9}$. $\sigma_{	ext{new}}^2 = \frac{355 	imes 3 - 196 	imes 4}{36} = \frac{1065 - 784}{36} = \frac{281}{36}$. Since $m = 281$ and $n = 36$ are co-prime,$m + n = 281 + 36 = 317$.

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on,it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$,where $m$ and $n$ are co-prime,then $m + n$ is equal to

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