Question :
| Monthly Consumption | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 – 205 |
| Number of Consumers | 4 | 5 | 13 | 20 | 14 | 8 | 4 |
Solution :
| Monthly Consumption of Electricity | No. of Consumers | Less than type cumulative frequency |
| 65 – 85 | 4 | 4 |
| 85 – 105 | 5 | 9 |
| 105 – 125 | 13 | 22 = C |
| 125 – 145 | 20 = f | 42 |
| 145 – 165 | 14 | 56 |
| 165 – 185 | 8 | 64 |
| 185 – 205 | 4 | 68 |
Since \(68\over 2\) belongs to the cumulative frequency (42) of the class interval (125 – 145) therefore, it is the median class interval.
Lower limit of the median class interval = l = 125
Width of the class interval = h = 20
Total frequency = N = 68
Cumulative frequency preceding median class frequency = C = 22
Frequency of the median class = f = 20
Median = l + (\({N\over 2} – C\over f\))(h) = 125 + \(12\times 20\over 20\) = 125 + 12 = 137
The frequency of class (125 – 145) is maximum i.e. 20, this is the modal class.
\(x_k\) = 125, \(f_k\) = 20, \(f_{k – 1}\) = 13, \(f_{k + 1}\) = 14, h = 20
Mode = \(x_k\) + \(f – f_{k – 1}\over 2f – f_{k – 1} – f_{k + 1}\) = 125 + \(20 – 13\over 40 – 13 – 14\)\(\times\) 20
= 120 + \(7\over 13\) \(\times\) 20 = 125 + 10.77 = 135.77
