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

We have n = 68
\( \Rightarrow \ \frac{n}{2} \ = \ \frac{68}{2} \ = \ 34 \)

The cumulative frequency just greater than \( \frac{n}{2} \) , is 42 and the corresponding class is 125-145.

Thus, 125-145 is the median class such that \( \frac{n}{2} \ = \ 34 , \)
l = 125 ,
c.f. = 22 ,
f = 20 , h = 20.

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

Median \( = \ 125 \ + \ ( \frac{34 \ - \ 22}{20}) \ × \ 20 \ \)
\( = \ 125 \ + \ 12 \ = \ 137 \) units

To calculate the mode :

Modal class = \(125 - 145 , \)
\( f_1 = 20 , \)
\( f_0 = 13 ,\)
\( f_2 = 14, \) and
\( \ h = 20 \)

Mode formula :

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

Mode = \( 125 + \frac{(20 -13)}{(40 -13 -14)} × 20 \)

\( = 125 + \frac{140}{13} \)

\( = 125 + 10.77 = 135.77 \)

Therefore, mode = 135.77

Calculation of mean :

Let the assumed mean, A = 135, class interval, h = 20

so \(u_i \ = \ \frac{x_i \ - \ A}{h} \ = \ \frac{x_i \ - \ 135}{20} \)

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ 135 \ + \ 20( \frac{7}{68}) \)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.