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(A) Given that the mean square deviation about $c = -2$ is $18$:$\frac{1}{n} \sum (x_i + 2)^2 = 18 \implies \sum (x_i^2 + 4x_i + 4) = 18n \implies \su

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$3$

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(A) Given that the mean square deviation about $c = -2$ is $18$:$\frac{1}{n} \sum (x_i + 2)^2 = 18 \implies \sum (x_i^2 + 4x_i + 4) = 18n \implies \sum x_i^2 + 4 \sum x_i + 4n = 18n \implies \sum x_i^2 + 4 \sum x_i = 14n$  $(1)$Given that the mean square deviation about $c = 2$ is $10$:$\frac{1}{n} \sum (x_i - 2)^2 = 10 \implies \sum (x_i^2 - 4x_i + 4) = 10n \implies \sum x_i^2 - 4 \sum x_i + 4n = 10n \implies \sum x_i^2 - 4 \sum x_i = 6n$  $(2)$Adding $(1)$ and $(2)$:$2 \sum x_i^2 = 20n \implies \frac{\sum x_i^2}{n} = 10$Subtracting $(2)$ from $(1)$:$8 \sum x_i = 8n \implies \frac{\sum x_i}{n} = 1$The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2 = 10 - (1)^2 = 9$.Therefore,the standard deviation $\sigma = \sqrt{9} = 3$.

\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}	$A: 81, B: 100$

The mean square deviation of a set of observations $x_1, x_2, \dots, x_n$ about a point $c$ is defined as $\frac{1}{n} \sum_{i=1}^n (x_i - c)^2$. If the mean square deviations about $-2$ and $2$ are $18$ and $10$ respectively,find the standard deviation of this set of observations.

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