Question

A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.

Production (Thousand rupees)	25-30	30-35	35-40	40-45	45-50
No. of Customers	20	25	15	10	10

Answer 1

Let us take 37.5 as the assumed mean. Then A = 37.5 and deviation d_i = x_i − A = x_i − 37.5

Class Production (Thousand rupees)	Class Mark xi	Deviations di = xi − A = xi − 37.5	Frequency (Farm owners) fi	Frequency × deviation fi × di
25-30	27.5	−10	20	−200
30-35	32.5	−5	25	−125
35-40	37.5 → A	0	15	0
40-45	42.5	5	10	50
45-50	47.5	10	10	100
Total	-	-	∑fi = 80	∑fidi = −175

Here, ∑f_id_i = −175, ∑f_i = 80

Required Mean `(bar d) = (sum f_i d_i)/(sum f_i)`

= `(-175)/80`

= −2.1875

≈ −2.19

Mean `(bar X) = A + bar d`

= 37.5 + (−2.19)

= 37.5 + 2.19

= 35.31 thousands

= 35.31 × 1000

= 35310

Hence, the mean production of oranges is Rs 35310.