Question

The following table gives production yield per hectare of wheat of 100 farms of a village:

Production yield	Frequency
40 – 45	4
45 – 50	6
50 – 55	16
55 – 60	20
60 – 65	20
65 – 70	24

Change the distribution to a 'more than' type distribution and draw its ogive.

Answer 1

Class Interval	Frequency	Cumulative frequency
0 – 100	2	2
100 – 200	5	7
200 – 300	x	7 + x
300 – 400	12	19 + x
400 – 500	17	36 + x
500 – 600	20	56 + x
600 – 700	y	56 + x + y
700 – 800	9	65 + x + y
800 – 900	7	72 + x + y
900 – 1000	4	76 + x + y
	N = 100

Also, 76 + x + y = 100

⇒ x + y = 100 – 76 = 24   ...(i)

Given, Median = 525, which lies between class 500 – 600.

⇒ Median class = 500 – 600

Now, Median = `l + (n/2 - c.f.)/f xx h`

⇒ `525 = 500 + [(100/2 - (36 + x))/20] xx 100`

⇒ 25 = (50 – 36 – x)5

⇒ `14 - x = 25/5 `

⇒ 14 – x = 5

⇒ x = 14 – 5

⇒ x = 9

Putting the value of x in equation (i), we get

y = 24 – 9

y = 15

Hence, x = 9 and y = 15.