# The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median and mode of the data and compare them.

## Solution :

Since $$68\over 2$$ belongs to the cumulative frequency (42) of the class interval (125 – 145) therefore, it is the median class interval.

Lower limit of the median class interval = l = 125

Width of the class interval = h = 20

Total frequency = N = 68

Cumulative frequency preceding median class frequency = C = 22

Frequency of the median class = f = 20

Median = l + ($${N\over 2} – C\over f$$)(h) = 125 + $$12\times 20\over 20$$ = 125 + 12 = 137

The frequency of class (125 – 145) is maximum i.e. 20, this is the modal class.

$$x_k$$ = 125, $$f_k$$ = 20, $$f_{k – 1}$$ = 13, $$f_{k + 1}$$ = 14, h = 20

Mode = $$x_k$$ + $$f – f_{k – 1}\over 2f – f_{k – 1} – f_{k + 1}$$ = 125 + $$20 – 13\over 40 – 13 – 14$$$$\times$$ 20

= 120 + $$7\over 13$$ $$\times$$ 20 = 125 + 10.77 = 135.77