In a data set with 15 observations x_1, x_2, | vedclass

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(A) Given,$\sum_{i=1}^{15} x_i^2 = 3600$ and $\sum_{i=1}^{15} x_i = 175$ for $n = 15$. When the incorrect observation $20$ is replaced by the correct

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$151$

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(A) Given,$\sum_{i=1}^{15} x_i^2 = 3600$ and $\sum_{i=1}^{15} x_i = 175$ for $n = 15$. When the incorrect observation $20$ is replaced by the correct value $40$,the new sums are: $\sum x_i = 175 - 20 + 40 = 195$ $\sum x_i^2 = 3600 - (20)^2 + (40)^2 = 3600 - 400 + 1600 = 4800$ The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$. Substituting the corrected values: $\sigma^2 = \frac{4800}{15} - \left(\frac{195}{15}\right)^2$ $\sigma^2 = 320 - (13)^2$ $\sigma^2 = 320 - 169 = 151$.

\text{Parameter}	\text{Plant } $A$ \text{ and } $B$ \text{ Data}
\text{No. of workers}	$A: 500, B: 6000$
\text{Average monthly wages}	$A: Rs. 2500, B: Rs. 2500$
\text{Variance of distribution of wages}	$A: 81, B: 100$

In a data set with $15$ observations $x_1, x_2, x_3, \ldots, x_{15}$,we are given $\sum_{i=1}^{15} x_i^2 = 3600$ and $\sum_{i=1}^{15} x_i = 175$. If the value of one observation $20$ was found to be incorrect and was replaced by its correct value $40$,then the corrected variance of the data is:

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