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  • 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.

1.  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.

        3.1

Answers (1)

best_answer

Let the assumed mean be a = 130 and h = 20

Class

Number of

consumers f_i

Cumulative

Frequency

Class mark

x_i

 d_i = x_i -a u_i = \frac{d_i}{h} f_iu_i
65-85 4 4 70 -60 -3 -12
85-105 5 9 90 -40 -2 -10
105-125 13 22 110 -20 -1 -13
125-145 20 42 130 0 0 0
145-165 14 56 150 20 1 14
165-185 8 64 170 40 2 16
185-205 4 68 190 60 3 12
 

 

\sum f_i = N

= 68

 

    

\sum f_ix_i

= 7

MEDIAN:
N= 68 \implies \frac{N}{2} = 34
\therefore Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125; 

c.f. = 22; f = 20; h = 20
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 125 + \left (\frac{34-22}{20} \right ).20 \\ \\ = 125 + 12

= 137

Thus, the median of the data is 137

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency (f_1) of the modal class = 20; frequency (f_0) of class preceding the modal class = 13, frequency (f_2) of class succeeding the modal class = 14.

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

\\ = 125 + \left(\frac{20-13}{2(20)-13-14} \right).20 \\ \\ = 125 + \frac{7}{13}.20

= 135.76

Thus, Mode of the data is 135.76

MEAN:

Mean, 
\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h 
= 130 + \frac{7}{68}\times20 = 137.05

Thus, the Mean of the data is 137.05

Posted by

HARSH KANKARIA

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