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# 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

 Class Interval 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 N=68

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 $$u_i \ = \ \frac{x_i \ - \ A}{h} \ = \ \frac{x_i \ - \ 135}{20}$$

 Class Interval $$f_i$$ $$x_i$$ $$d_i \ = \ x_i \ - \ A$$ $$u_i \ = \ \frac{d_i}{h}$$ $$f_i u_i$$ 65-85 4 75 -60 -3 -12 85-105 5 95 -40 -2 -10 105-125 13 115 -20 -1 -13 125-145 20 135 0 0 0 145-165 14 155 20 1 14 165-185 8 175 40 2 16 185-205 4 195 60 3 12 $$\sum f_i \ = \ 68$$ $$\sum f_i u_i \ = \ 7$$

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

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

Mean = 137.05

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