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 (in Rs)	Number of families
1000 − 1500	24
1500 − 2000	40
2000 − 2500	33
2500 − 3000	28
3000 − 3500	30
3500 − 4000	22
4000 − 4500	16
4500 − 5000	7

Answer 1

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.

Therefore, modal class = 1500 − 2000

Lower limit (l) of modal class = 1500

Frequency (f₁) of modal class = 40

Frequency (f₀) of class preceding modal class = 24

Frequency (f₂) of class succeeding modal class = 33

Class size (h) = 500

Mode = `l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

=`1500+((40-24)/(2(40)-24-33))xx500`

=`1500+((16)/(80-57))xx500`

=`1500+8000/23`

Therefore, modal monthly expenditure was Rs 1847.83.

To find the class mark, the following relation is used.

`"class mark" =("Upper class limit + Lower class limit")/2`

Class size (h) of the given data = 500

Taking 2750 as assumed mean (a), d_i, u_i, and f_iu_iare calculated as follows

aking 2750 as assumed mean (a), d_i, u_i, and f_iu_iare calculated as follows.

Expenditure (in Rs)	families fi	xi	di = xi  − 2750	ui = di/500	fiui
1000 − 1500	24	1250	−1500	−3	−72
1500 − 2000	40	1750	−1000	−2	−80
2000 − 2500	33	2250	−500	−1	−33
2500 − 3000	28	2750	0	0	0
3000 − 3500	30	3250	500	1	30
3500 − 4000	22	3750	1000	2	44
4000 − 4500	16	4250	1500	3	48
4500 − 5000	7	4750	2000	4	28
Total	200				−35

From the table, we obtain

`sumf_i = 200`

`sumf_iu_i = -35`

Mean `barx = a+ ((sumf_iu_i)/(sumf_i))xxh`

`barx=2750+((-354)/200)xx500`

= 2750 - 87.5

= 2662.5

Therefore, mean monthly expenditure was Rs 2662.50.