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

Updated On: Sep 10, 2024
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Solution and Explanation

The cumulative frequencies with their respective class intervals are as follows.

Monthly consumption 
(in units)

 Number of consumers

Cumulative frequency

60 - 85 

4

4

85 - 105

5

4 + 5 = 9

105 - 125

13

9 + 13 = 22

125 - 145

20

22 + 20 = 42

145 - 165

14

42 + 14 = 56

165 - 185

8

56 + 8 = 64

185 - 205

4

64 + 4 = 68

Total(n)

68

 

From the table, we obtain 
n = 68  
Cumulative frequency just greater \(\frac{n}2 ( i.e., \frac{68}2 = 34)\) than is 42, belonging to class interval 125 - 145.
Median class = 125 - 145
Lower limit (\(l\)) of median class = 125
Frequency (\(f\)) of median class = 20
Cumulative frequency (\(cf\)) of median class = 22
Class size (\(h\)) = 20

 Median = \(l + (\frac{\frac{n}2 - cf}f \times h)\)

Median =  \(125 + (\frac{34 - 22}{20} \times 20)\)

Median = 125 +12
Median = 137


To find the class mark (xi) for each interval, the following relation is used.  

Class mark  \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)

Taking 11.5 as assumed mean (a), \(d_i\)\(u_i\), and \(f_iu_i\) are calculated according to step deviation method as follows.

Monthly consumption 
(in units)

 Number of consumers

         \(\bf{x_i}\)       

\(\bf{d_i = x_i -11.5}\)

\(\bf{u_i = \frac{d_i}{3}}\)

       \(\bf{f_iu_i}\)          

60 - 85 

4

75-60-3-12

85 - 105

5

95-40-2-10

105 - 125

13

115-20-1-13

125 - 145

20

135000

145 - 165

14

15520114

165 - 185

8

17540216

185 - 205

4

19560312

Total 

68

   7

From the table, it can be observed that  

\(\sum f_i = 68\)
\(\sum f_iu_i = 7\)

Mean, \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i})\times h\) 

\(\overset{-}{x}\) = \(135 + (\frac{7 }{68})\times 20\)

\(\overset{-}{x}\) = 135 + \(\frac{140}{68}\)
Mean, \(\overset{-}{x}\) = 137.058


The data in the given table can be written as 
 

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


From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.  

Therefore, Modal class = 125 − 145 
Lower limit (\(l\)) of modal class = 125  
Frequency (\(f_1\)) of modal class = 40
Frequency (\(f_0\)) of class preceding the modal class = 13
Frequency (\(f_2\)) of class succeeding the modal class = 14
Class size (\(h\)) = 20  

Mode = \(l\) + \((\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h\)

Mode = \(125 + (\frac{20 - 13 }{ 2(20) - 13 - 14}) \times 20\)

Mode =\(125+ [\frac{7}{13}] \times 20\)

Mode = \(125 +( \frac{ 140}{ 13})\)

Mode = 135.76

Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.  The three measures are approximately the same in this case. 

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