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Question
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | Number of consumers |
| 65 - 85 | 4 |
| 85 - 105 | 5 |
| 105 - 125 | 13 |
| 125 - 145 | 20 |
| 145 - 165 | 14 |
| 165 - 185 | 8 |
| 185 - 205 | 4 |
Solution
The given data is shown below.
| Monthly Consumption (in units) | No. of consumers (fi) | xi | fixi | C.f. |
| 65−85 | 4 | 75 | 300 | 4 |
| 85−105 | 5 | 95 | 475 | 9 |
| 105−125 | 13 | 115 | 1495 | 22 |
| 125−145 | 20 | 135 | 2700 | 42 |
| 145−165 | 14 | 155 | 2170 | 56 |
| 165−185 | 8 | 175 | 1400 | 64 |
| 185−205 | 4 | 195 | 780 | 68 |
| Total | `sumf=68` | `sumf_1x_1=9320` |
Here, the maximum frequency is 20 so the modal class is 125 − 145.
Therefore,
l = 125, h = 20, f = 20, f1 = 13 and f2 = 14
Mode `=l+(f-f_1)/(2f-f_1-f_2)xxh`
`=125+7/13xx20`
`=125+140/13`
= 125 + 10.76
= 135.76
Mode = 135.76 units
Thus, the mode of the monthly consumption of electricity is 135.76 units.
Mean `=(sumf_1x_1)/(sumf)`
`=9320/68=137.05`
Mean = 137.05
Thus, the mean of the monthly consumption of electricity is 137.05 units.
Here,
Total number of consumers, N = 68 (even)
Then, `N/2=34`
Median `=l+(N/2_f)/fxxh`
`=125+(68/2_22)/20xx20`
`=125+(34-22)/20xx20`
`=125+12/20xx20`
= 125 + 12
= 137
Thus, the median of the monthly consumption of electricity is 137 units.
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