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)	65-85	85-105	105-125	125-145	145-165	165-185	185-205
No. of consumers:	4	5	13	20	14	8	4

Given:

The given frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality.

To do:

We have to find the median, mean and mode of the data and compare them.

Solution:

The frequency of the given data is as given below.

Let the assumed mean be $A=135$.

We know that,

Mean $=A+\frac{\sum{f_id_i}}{\sum{f_i}}$

Therefore,

Mean $=135+\frac{140}{68}$

$=135+2.05$

$=137.05$

The mean of the given data is 137.05.

We observe that the class interval of 125-145 has the maximum frequency(20).

Therefore, it is the modal class.

Here,

$l=125, h=20, f=20, f_1=13, f_2=14$

We know that,

Mode $=l+\frac{f-f_1}{2 f-f_1-f_2} \times h$

$=125+\frac{20-13}{2 \times 20-13-14} \times 20$

$=125+\frac{7}{40-27} \times 20$

$=125+\frac{140}{13}$

$=125+10.76$

$=135.76$

The mode of the given data is 135.76.

Here,

$N=68$

This implies, $\frac{N}{2}=\frac{68}{2}=34$

Median class $=125-145$

We know that,

Median $=l+\frac{\frac{N}{2}-F}{f} \times h$

$=125+\frac{34-22}{20} \times 20$

$=125+12$

$=137$

The median of the given data is 137.

The mean, mode and median of the above data are 135.07, 135.76 and 137 respectively.

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Updated on: 10-Oct-2022

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