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Question

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
Firm A Firm B
No.of wage earners 586 648
Mean of monthly wages Rs.5253 Rs.5253
Variance of the distribution of wages 100 121

(i.) Which firm A or B pays larger amount as monthly wages?

(ii.) Which firm, A or B, shows greater variability in individual wages?

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Solution

Mean monthly wages of firm A=Rs.5253
and Number of wage earners in firm A=586
Then,
Total amount paid in firm A=586×5253
Total amount paid in firm A=Rs.3078258

Mean monthly wages of firm B=Rs.5253
Number of wage earners in firm B=648
Then,
Total amount paid in firm B
=648×5253=Rs.3403944
Thus, firm B pays larger amount as monthly wages than firm A.

For firm A:
Variance (σ2A)=100 (Given)
Standard deviation (σA)=Variance
(σA)=100
(σA)=10
For firm B:
Variance (σ2B)=121 (Given)
Standard deviation (σB)=Variance
(σB)=121=11

Since, mean in both the firms is same i.e., 5253, the plant with greater standard deviation will have more variability.

Therefore, the plant B has greater variability in the individual wages.

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