Question 1
The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption(in units) | No. of customers |
|---|---|
| 65-85 | 4 |
| 85-105 | 5 |
| 105-125 | 13 |
| 125- 145 | 20 |
| 145-165 | 14 |
| 165-185 | 8 |
| 185- 205 | 4 |
Solution:
Find the cumulative frequency of the given data as follows:
| Class Interval | Frequency | Cumulative frequency |
|---|---|---|
| 65-85 | 4 | 4 |
| 85-105 | 5 | 9 |
| 105- 125 | 13 | 22 |
| 125-145 | 20 | 42 |
| 145- 165 | 14 | 56 |
| 165-185 | 8 | 64 |
| 185- 205 | 4 | 68 |
| N = 68 |
From the table, it is observed that, N = 68 and hence $\dfrac{N}{2}$=34
Hence, the median class is 125-145 with cumulative frequency = 42
Where, l = 125, N = 68, cf = 22, f = 20, h = 20
Median is calculated as follows:
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
= 125 + $[\dfrac{(34 − 22)}{20}]$ × 20
= 125 + 12
= 137
Therefore, median = 137
To calculate the mode:
Modal class = 125-145,
$f_m$ or $f_1$ = 20, $f_0$ = 13, $f_2$ = 14 & h = 20
Mode formula:
| Mode = $l+ [\dfrac{(f_1 – f_0)}{(2f_1 – f_0 – f_2)}] × h$ |
Mode = 125 + $[\dfrac{(20 – 13)}{(40 – 13 – 14)}]$ × 20
= 125 + $(\dfrac{140}{13})$
= 125 + 10.77
= 135.77
Therefore, mode = 135.77
Calculate the Mean:
| Class Interval | $f_i$ | $x_i$ | $d_i=x_i-a$ | $u_i=\dfrac {d_i}{h}$ | $f_iu_i$ |
|---|---|---|---|---|---|
| 65-85 | 4 | 75 | -60 | -3 | - 12 |
| 85-105 | 5 | 95 | -40 | -2 | -10 |
| 105- 125 | 13 | 115 | -20 | -1 | -13 |
| 125-145 | 20 | 135 = a | 0 | 0 | 0 |
| 145-165 | 14 | 155 | 20 | 1 | 14 |
| 165 -185 | 8 | 175 | 40 | 2 | 16 |
| 185- 205 | 4 | 195 | 60 | 3 | 12 |
| Sum$ f_i$ = 68 | Sum $ f_iu_i$= 7 |
| $\overline{x} = a + h (\dfrac{∑ f_iu_i }{∑f_i})$ |
= 135 + 20 $(\dfrac{7}{68})$
Mean = 137.05
In this case, mean, median and mode are more/less equal in this distribution.