Question : The mean of the following data is 42. Find the missing frequencies x and y, if the sum of the frequencies is 100.
| C.I. | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| fi | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
Doubt by Anvesha
Solution :
| C.I. | fi | xi | di=xi-a | ui=di/h | fiui |
| 0-10 | 7 | 5 | -30 | -3 | -21 |
| 10-20 | 10 | 15 | -20 | -2 | -20 |
| 20-30 | x | 25 | -10 | -1 | -x |
| 30-40 | 13 | 35=a | 0 | 0 | 0 |
| 40-50 | y | 45 | 10 | 1 | y |
| 50-60 | 10 | 55 | 20 | 2 | 20 |
| 60-70 | 14 | 65 | 30 | 3 | 42 |
| 70-80 | 9 | 75 | 40 | 4 | 36 |
| Σfi=100 |
Σfiui=-x+y+57 |
Mean = 42 (Given)
Σfi=100 (Given)
63+x+y=100
x+y=100-63
x+y=37 — (1)
Mean by Step Deviation Method
Mean (x̄) = a+[Σfiui/Σfi]×h
42=35+[-x+y+57/100]×10
42-35=[-x+y+57]/10
7×10=-x+y+57
70-57=-x+y
-x+y=13 — (2)
Adding equation (1) and (2)
x+y+(-x+y)=37+13
x+y-x+y=50
2y=50
y=50/2
y=25
Putting this value of y in equation (1)
x+25=37
x=37-25
x=12
Hence, x=12 and y=25