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 wages 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?
(i) From the given table,
Mean monthly wages of firm A = Rs 5253
and Number of wage earners = 586
Then,
Total amount paid = 586 × 5253
= Rs 3078258
Mean monthly wages of firm B = Rs 5253
Number of wage earners = 648
Then,
Total amount paid = 648 × 5253
= Rs 34,03,944
So, firm B pays larger amount as monthly wages.
(ii) Variance of firm A = 100
We know that, standard deviation (σ)= \( \sqrt{100}\)
=10
Variance of firm B = 121
Then,
Standard deviation (σ)=\( \sqrt{121}\)
=11
Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.
Answered by Pragya Singh | 1 year agoFind the mean deviation from the mean and from a median of the following distribution:
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No. of students | 5 | 8 | 15 | 16 | 6 |
The age distribution of 100 life-insurance policy holders is as follows
Age (on nearest birthday) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |
No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |
Compute mean deviation from mean of the following distribution:
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |
Find the mean deviation from the mean for the following data:
Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |
Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |
Find the mean deviation from the mean for the following data:
Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |
Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |