The following results are obtained on the basis of daily wages(in Rs.) paid to workers
of two firms a and b.
n1=550; n2=600;X1;=60; X2=48.5;S1=10;S2=12.
(0) Which firm shows more variation in wages paid to its workers?
(1) Find combined mean.
Answers
Answer:
Number of wage earner in firm A=586
Mean of monthly wage of firm A=Rs5253
Mean of monthly age of firm A =
No. of wage earners in firm A
Total amount paid
5253=
586
Total amount paid
Total amount paid by Firm A =5253×586
Number of wage earner in firm B=648
Mean of monthly wage of firm B=Rs5253
Mean of monthly age of firm B =
No. of wage earners in firm B
Total amount paid
5253=
648
Total amount paid
Total amount paid by Firm B =5253×648
Clearly, firm B paid larger amount as monthly wage.
(ii) Variance of the distribution of wages in firm A (σ
1
2
)=100
∴ Standard deviation of the distribution of wages in firm A(σ
1
)=
100
=10
Variance of the distribution of wages in firm B(σ
1
2
)=121
∴ Standard deviation of the distribution of wages in firm B(σ
2
)=
121
=11
The mean of monthly wages of both the firms is same i.e.5253. So, the firm with greater standard deviation will have more variability.
Thus firm B has greater variability in the individual wages