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

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Solution

Class
(Production in

Thousand rupees)
Class Mark
x_i d_i = x_i− A Frequency
(Number of farm owners)
f_i Frequency × deviation
f_i× d_i

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

$\underset{}{\sum f_{i} = 80}$ $\sum_{} f_{i} d_{i} = - 175$

Required Mean = $A + \frac{\sum_{} f_{i} d_{i}}{\sum_{} f_{i}}$
$37.5 - \frac{175}{80}$
= 37.5 − 2.19
= 35.31 thousand rupees
= Rs 35310
Hence, the mean production of oranges is Rs 35310.

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Calculation (i) Arithmetic Mean and (ii) Standard Deviation for the following frequency distribution.

Class Interval

Frequency

20 – 25

25 – 30

30 – 35

35 – 40

40 – 45

45 – 50

N = 50

Marks	10−20	20−30	30−40	40−50	50−60
Frequency	12	35	45	25	13

Age	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	50	70	100	180	150	120

Draw a histogram to represent the following frequency distribution.

Class interval	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	30	24	52	28	46	10

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

Class (Production in Thousand rupees)	Class Mark x_i	d_i = x_i− A	Frequency (Number of farm owners) f_i	Frequency × deviation f_i× d_i
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
			$\underset{}{\sum f_{i} = 80}$	$\sum_{} f_{i} d_{i} = - 175$