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

Calculation of median:
Preparing the table to compute median :

Monthly consumption (in units)	Number of consumers (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
Total n=68
We have, n=68
n/2=34
The cumulative frequency just greater than n/2 is 42 and the corresponding class is 125–145.
Thus, 125-145 is the median class such that n/2=34, l=125, f=20, cf=22, and h=20
Substituting these values in the formula
Median, M = l + \frac{\frac{n}{2} - cf}{f} \times h
M = 125 + (\frac{34 - 22}{20}) \times 20
M = 125 + 12 = 137 units

For calculation of mean:

Class-Interval	Mid-Value(xi)	Frequency(fi)	ui = \frac{xi - 135}{20}	fiui
65–85	75	4	-3	-12
85–105	95	5	-2	-10
105–125	105	13	-1.5	-19.5
125–145	135	20	0	0
145–165	155	14	1	14
165–185	175	8	2	16
185–205	195	4	3	12
Total		68		7
Let the assumed mean, A = 135 and class interval, h = 20
So, ui = \frac{xi - A}{h} = \frac{xi - 135}{20}
Mean, \bar{x} = A + h \times \frac{\sum fiui}{\sum fi}
\bar{x} = 135 + 20 \times \frac{7}{68} \implies \bar{x} = 135 + 2.05 = 137.05
Therefore, Mean = 137.05 units

Calculation of Mode:
The class 125–145 has the maximum frequency, therefore, this is the modal class.
Here, l = 125, h = 20, f1 = 20, f0 = 13 and f2 = 14
Now, let us substitute these values in the formula
Mode = l + (\frac{f1 - f0}{2f1 - f0 - f2}) \times h
= 125 + \frac{20 - 13}{40 - 13 - 14} \times 20
= 125 + \frac{7}{13} \times 20
= 125 + 10.76 = 135.76 units
It can be seen that the three measures are approximately the same in this case.

Eduprobe

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