An analysis of monthly wages paid to workers in two firms $A$ and $B$,belonging to the same industry,gives the following results:
\text{Parameter}
\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}
$A: 586, B: 648$
\text{Mean of monthly wages}
$Rs. 5253$
\text{Variance of wages}
$A: 100, B: 121$
Which firm,$A$ or $B$,pays a larger total amount as monthly wages?
| \text{Parameter} | \text{Firm } $A$ \text{ and Firm } $B$ |
| \text{No. of wage earners} | $A: 586, B: 648$ |
| \text{Mean of monthly wages} | $Rs. 5253$ |
| \text{Variance of wages} | $A: 100, B: 121$ |
Which firm,$A$ or $B$,pays a larger total amount as monthly wages?
(B) The total monthly wages paid by a firm is calculated as the product of the mean monthly wage and the number of wage earners.
For firm $A$:
Total amount $= 5253 \times 586 = 3,078,258$.
For firm $B$:
Total amount $= 5253 \times 648 = 3,403,944$.
Comparing the two,$3,403,944 > 3,078,258$.
Therefore,firm $B$ pays a larger total amount as monthly wages.
For firm $A$:
Total amount $= 5253 \times 586 = 3,078,258$.
For firm $B$:
Total amount $= 5253 \times 648 = 3,403,944$.
Comparing the two,$3,403,944 > 3,078,258$.
Therefore,firm $B$ pays a larger total amount as monthly wages.
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Consider the given data with frequency distribution:
$x_{i} = \{3, 8, 11, 10, 5, 4\}$
$f_{i} = \{5, 2, 3, 2, 4, 4\}$
Match each entry in List-$I$ to the correct entries in List-$II$.
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The correct option is :
$x_{i} = \{3, 8, 11, 10, 5, 4\}$
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Match each entry in List-$I$ to the correct entries in List-$II$.
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| $(P)$ The mean of the above data is | $(1) 2.5$ |
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| $(R)$ The mean deviation about the mean of the above data is | $(3) 6$ |
| $(S)$ The mean deviation about the median of the above data is | $(4) 2.7$ |
| $(5) 2.4$ |
The correct option is :
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