Question

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

Answer 1

SOLUTION :  

FREQUENCY DISTRIBUTION TABLE is in the attachment

For MODE :  

Here the maximum frequency is 20, and the class corresponding to this frequency is 125 – 145. So the modal class is 125 - 145.

Therefore,l = 125, h = 20,  f1= 20,  f0= 13 , f2 = 14

Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h

= 125 + [(20 - 13)/(2 × 20 - 13 – 14) ] ×20

= 125 + [7/ (40 - 27)]×20

= 125 + [(7×20)/13]

= 125 + [140/13]

= 125 + 10.76

= 135.76

MODE = 135.76

Hence, the mode of the data is 135.76.

MEAN :  

From the table : Σfi = 68 , Σfixi = 9320

Mean = Σfixi /Σfi

Mean = 9320/68 = 137.05

Hence, the Mean of the data is 137.05

For MEDIAN :  

Here, n = 68

n/2 = 34

Since, the Cumulative frequency just greater than 34  is 42 and the corresponding class is 125 - 145.  Therefore 125 - 145  is the median class.

Here, l = 125 , f = 20 , c.f = 22,  h = 20

MEDIAN = l + [(n/2 - cf )/f ] ×h

= 125 + [(34 - 22)/20] × 20

= 125 + [(12 × 20)/20]

= 125 + 12

Median = 137

Hence, the Median is 137 .

★★ Mode = l + (f1-f0/2f1-f0-f2) ×h

l = lower limit of the modal class

h = size of the class intervals

f1 = frequency of the modal class

f0 = frequency of the class preceding the modal class

f2 = frequency of the class succeed in the modal class.

★★ MEDIAN = l + [(n/2 - cf )/f ] ×h

Where,

l = lower limit of the median class

n = number of observations  

cf = cumulative frequency  of class interval preceding the  median class

f = frequency  of median class

h = class  size

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Median :

In the above distribution , n = 68

n/2 = 34

Now , 125 - 145 is the class whose

cumulative frequency 42 is greater

than ( and nearest to ) n/2 , i.e 34

Therefore ,

125 - 145 is the median class .

l = 125 , cf = 22 ; f = 20 , h = 20 , n = 68

Median ( M ) = l + [ ( n/2 - cf )/f ] × h

M = 125 + [ ( 34 - 22 )/20 ] × 20

M = 137 units

2 ) From Table ( 2 ) ,

Here the maximum class frequency is 20

The modal class is 125 - 145 .

l = 125 , f1 = 20 ; f0 = 13 ; f2 = 14 ; h = 20

mode ( Z ) = l + [(f1 - f0)/(2f1-f2-f0)] × h

Z = 125 + [ ( 20 - 13 )/(2×20-13-14)]×20

Z = 125 + ( 7/13 )× 20

Z = 125 + 140/13

z = 125 + 10.76

z = 135.76 units

3 ) From tanble ( 3 ),

mean = 137.05

Step-by-step explanation: