Question

The following table gives production yield per hectare of pulses of 100 farms.

If the mean production is 43 Kg/hr, find the missing frequencies ‘x’ & ‘y’.

Answer with complete steps & earn 15 points...☺

Answer 1

Hi GK,

Here we use direct method formula

of mean

We know that ,

Class mark ( xi ) = avg of class limits

____________________
C. I. |. f | (xi). | fixi
___________________
25- 30| 2 |27.5 | 55
____________________
30- 35| 7 | 32.5 | 227.5
___________________
35- 40| x. | 37.5| 37.5x

40-45| 36. | 42.5| 1530
____________________
45-50| y. | 47.5| 47.5y
___________________
50-55| 5. | 52.5| 262.5
____________________
55-60| 2. | 67.5 | 135
____________________

Sigma fi = sum of frequencies

= 52 +x + y ----( 1 )

Sum of fixi = sigma fixi

= 2210 + 37.5x + 47.5y----( 2 )

According to the problem given,

Sum of frequencies = 100--( 3 )

( 1 ) = ( 3 )

52 + x + y = 100

x + y = 100 - 52

x + y = 48

x = 48 - y ------( 4 )

Mean production = 43 ----( 5 )


Mean = ( 2 ) / 100

(2210 + 37.5x + 47.5y)/100 = 43

2210+37.5x+47.5y = 4300

37.5x + 47.5 y = 2090

Put x = 48 - y from ( 4 ),

37.5( 48 - y) + 47.5y = 2090

1800-37.5y+ 47.5y = 2090

10y = 290

y = 29

Put y = 29 in ( 4 ) , we get

x = 48 - 29

x = 19

Therefore ,

x = 19 and y = 29

I hope this helps you.

:)

Answer 2

Above mentioned answer by someone else is wrong. My answer is correct