Question

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
Expenditure Frequency
1000-1500 24
1500-2000 40
2000-2500 33
2500-3000 28
Expenditure Frequency
3000-3500 30
3500-4000 22
4000-4500 16
4500-5000 7

Answer 1

SOLUTION :  

FREQUENCY DISTRIBUTION TABLE is in the attachment

For MODE :

Here the maximum frequency is 40, and the class corresponding to this frequency is 1500 – 2000. So the modal class is 1500 – 2000.

Therefore, l = 1500, h = 500,  f1= 40,  f0= 24 , f2 = 33

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

= 1500 + [(40 - 24)/(2 × 40 - 24 – 33) ] ×500

= 1500 + [16 × 500)/(80 - 67)]

= 1500 + [8000/ 23]

= 1500 + 347.83

= 1847.83  

MODE = 1847.83

Hence, the modal monthly expenditure of the families is ₹ 1847.83

MEAN :  

From the table : Σfi = 200 , Σfixi = 532500

Mean = Σfixi /Σfi

Mean =  532500/200 = 5325/2 = 2,662.5

Hence, the Mean monthly expenditure of the families is ₹ 2,662.5.

★★ 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.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Step-by-step explanation:

1 ) From Table ( 1 )

Here ,

The maximum class frequency is 40

Now ,

lower limit ( l ) = 1500

modal class = 1500 - 2000

class size ( h ) = 500

frequency of the modal class ( f1 ) = 40

frequency of class preceding the

modal class ( f0 ) = 24 ,

frequency of class succeeding the

modal class ( f2 ) = 33 ,

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

z = 1500 + [(40-24)/(2×40-24-33)]×500

z = 1500 + ( 16 × 500 )/( 23 )

z ≈ 18847.83

2 ) From table ( 2 ),

Mean = 2750 +[ ( -35 )/200 ]× 500

Mean = 2662.5

I hope this helps you.

: )