# 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) - Algebra

Sum

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

#### Solution

 Class (Production in Thousand rupees) Class Markxi di = xi − A Frequency(Number of farm owners)fi Frequency × deviationfi × 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 $\sum f_i = 80_{}$ $\sum_{} f_i d_i = - 175$

Required Mean = bard = (∑f_i d_i)/N

= (-35)/16

= - 2.19

Mean (barX) = A + bard

= 37.5 - 2.19

= 35.31

= 35.31 × 1000

= Rs 35,310
Hence, the mean production of oranges is Rs 35,310.

Is there an error in this question or solution?
Chapter 6: Statistics - Practice Set 6.1 [Page 138]

#### APPEARS IN

Balbharati Maths 1 Algebra 10th Standard SSC Maharashtra State Board
Chapter 6 Statistics
Practice Set 6.1 | Q 4 | Page 138

Share