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
Step-by-Step Solution
Key Concept: For grouped data, the median is obtained using the formula \(L+\frac{\frac{N}{2}-c_f}{f_m}\times h\). The mean is calculated by taking class‑midpoints as representative values: \(\bar{x}=\frac{\sum f_i x_i}{N}\). The mode for grouped data uses \(L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\), where \(f_1\) is the highest frequency, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively.
1. Tabulate the data
| Class (units) | Frequency (f) | Mid‑point (x) |
|---------------|--------------|--------------|
| 65 – 85 | 4 | 75 |
| 85 – 105 | 5 | 95 |
| 105 – 125 | 13 | 115 |
| 125 – 145 | 20 | 135 |
| 145 – 165 | 14 | 155 |
| 165 – 185 | 8 | 175 |
| 185 – 205 | 4 | 195 |
Total \(N = 68\).
2. Median
- Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68.
- \(\frac{N}{2}=34\). The first cumulative frequency ≥ 34 is 42, so the median class is 125 – 145.
- Lower limit \(L = 125\), class width \(h = 20\), frequency of median class \(f_m = 20\), cumulative frequency before median class \(c_f = 22\).
- \[\text{Median}=L+\frac{\frac{N}{2}-c_f}{f_m}\times h =125+\frac{34-22}{20}\times20 =125+\frac{12}{20}\times20 =125+12 = \mathbf{137\;units}\]
3. Mean
- Compute \(\sum f_i x_i\):
\(75\times4=300\)
\(95\times5=475\)
\(115\times13=1495\)
\(135\times20=2700\)
\(155\times14=2170\)
\(175\times8=1400\)
\(195\times4=780\)
- Sum = \(300+475+1495+2700+2170+1400+780 = 9320\).
- \[\bar{x}=\frac{\sum f_i x_i}{N}=\frac{9320}{68}=\mathbf{137.06\;units\;(≈137)}\]
4. Mode
- Highest frequency = 20 (class 125 – 145). Hence modal class = 125 – 145.
- \(f_1 = 20\), \(f_0 = 13\) (preceding class), \(f_2 = 14\) (succeeding class).
- \[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h =125+\frac{20-13}{2\times20-13-14}\times20 =125+\frac{7}{13}\times20\]
- \(\frac{7}{13}=0.5385\). Hence \[\text{Mode}=125+0.5385\times20 =125+10.77 = \mathbf{135.77\;units\;(≈136)}\]
5. Comparison
- Mean \(\approx 137\), Median \(=137\), Mode \(\approx 136\).
- All three measures are very close, indicating that the distribution of electricity consumption is nearly symmetric about the central value.
- Hence, the data set is fairly balanced with no extreme skewness.
Correct Answer: Median = 137 units, Mean ≈ 137.06 units (≈137), Mode ≈ 135.77 units (≈136). The three measures are almost equal, showing a nearly symmetric distribution.