Two plants A and B of a factory show the foll | vedclass

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(B) To determine the variability in individual wages,we compare the standard deviations of the two plants since their mean wages are equal. The varian

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Plant $B$

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(B) To determine the variability in individual wages,we compare the standard deviations of the two plants since their mean wages are equal. The variance of the distribution of wages in plant $A$ is $\sigma_{1}^{2} = 81$. Therefore,the standard deviation of the distribution of wages in plant $A$ is $\sigma_{1} = \sqrt{81} = 9$. The variance of the distribution of wages in plant $B$ is $\sigma_{2}^{2} = 100$. Therefore,the standard deviation of the distribution of wages in plant $B$ is $\sigma_{2} = \sqrt{100} = 10$. Since the average monthly wages in both plants are the same $(Rs. 2500)$,the plant with the greater standard deviation will have greater variability. Since $\sigma_{2} > \sigma_{1}$ $(10 > 9)$,plant $B$ has greater variability in individual wages.

\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}	$A: 81, B: 100$

\text{Particulars}	\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}	$A: 586, B: 648$
\text{Mean of monthly wages}	$A: Rs. 5253, B: Rs. 5253$
\text{Variance of wages}	$A: 100, B: 121$

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