9. Given the following data on monthly wages of 1000 workers, frequency of two class interval are missing. Locate these if, Median of the series is 50.
| Wages (in Rs) | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| No. of workers | 60 | 60 | ? | 200 | ? | 285 |
10. Calculate mode in the following distribution using grouping method:-
| Marks | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| f | 12 | 3 | 24 | 20 | 12 | 2 |
Dear Student,
9)
605 + f1 + f2 = 1000
f1 + f2 = 395 --------------------- (i)
As median is 56 median class is 50-60
Median = l + n/2 - cf / f * i
56 = 50 + 1000/2 - ( 120 + f1 ) / 200 * 10
56 = 50 + 500 - 120 - f1 / 20
56 = 1000 + 380 - f1 / 20
56 * 20 = 1380 - f1
1120 = 1380 - f1
f1 = 260
substituting value in (i)
f1 + f2 = 395
260 + f2 = 395
f2 = 135
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9)
| Wages | frequency | cumulative frequency |
| 20-30 | 60 | 60 |
| 30-40 | 60 | 120 |
| 40-50 | f1 | 120+ f1 |
| 50-60 | 200 | 320 + f1 |
| 60-70 | f2 | 320 + f1 +f2 |
| 70-80 | 285 | 605 + f1 +f2 |
| Total | 1000 | 1000 |
605 + f1 + f2 = 1000
f1 + f2 = 395 --------------------- (i)
As median is 56 median class is 50-60
Median = l + n/2 - cf / f * i
56 = 50 + 1000/2 - ( 120 + f1 ) / 200 * 10
56 = 50 + 500 - 120 - f1 / 20
56 = 1000 + 380 - f1 / 20
56 * 20 = 1380 - f1
1120 = 1380 - f1
f1 = 260
substituting value in (i)
f1 + f2 = 395
260 + f2 = 395
f2 = 135
You are requested not to pile up questions in different threads use different threads for different videos
Regards