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
Number of consumers 4 5 13 20 14 8 4
Answers
1 ) 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
From the above we observe that the
three measures are
approximately the same in the case .
I hope this helps you.
: )
From table ( 1 ) ,
1 ) 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
From the above we observe that the
three measures are
approximately the same in the case .
I hope this helps you.