Question

If the median of the following frequency distribution is 32.5, find the values of `f_1 and f_2`.

Class	0 – 10	10 – 20	20 – 30	30 -40	40 – 50	50 – 60	60 – 70	Total
Frequency	`f_1`	5	9	12	`f_2`	3	2	40

 

Answer 1

Class	Frequency (f)	Cumulative Frequency (cf)
0 – 10	`f_1`	`f_1`
10 – 20	5	`f_1`+5
20 – 30	9	`f_1`+14
30 – 40	12	`f_1`+26
40 – 50	`f_2`	`f_1`+`f_2`+26
50 – 60	3	`f_1`+`f_2`+29
60 – 70	2	`f_1`+`f_2`+31
	N = Σ𝑓 = 40

Now,` f_1 + f_2 + 31 = 40`
⇒` f_1 + f_2 = 9`
⇒ `f_2 = 9 - f_1`
The median is 32.5 which lies in 30 – 40.
Hence, median class = 30 – 40
Here, `l = 30, N/2 = 40/2 = 20, f = 12 and cf = 14 + f_1`
Now, median = 32.5
`⇒ l + ((N/2−Cf)/f) × h = 32.5`

`⇒ 30 + ((20 −(14 + f1))/ 12)× 10 = 32.5`
`⇒ (6 − f_1)/ 12` × 10 = 2.5
`⇒ (60 − 10f_1)/12 = 2.5`
`⇒ 60 – 10f_1 = 30`
`⇒ 10f_1 = 30`
`⇒ f_1 = 3`
From equation (i), we have:
`f_2 = 9 – 3`
`⇒ f_2 = 6`