Question

Following distribution shows the marks of 230 students in a particular subject. If the median marks are 46, then find the values of x and y.

Marks	Number of students
10 – 20	12
20 – 30	30
30 – 40	x
40 – 50	65
50 – 60	y
60 – 70	25
70 – 80	18

Answer 1

Median = 46

Total = 230 = n

Marks	xi	f	c.f.
10 – 20	15	12	12
20 – 30	25	30	42
30 – 40	35	x	42 + x → c.f.
40 – 50	45	65	107 + x
50 – 60	55	y	107 + x + y
60 – 70	65	25	132 + x + y
70 – 80	75	18	150 + x + y
		`sumf = 230` = 150 + x + y

150 + x + y = 230

x + y = 230 − 150

x + y = 80  .....(i)

Median class = 40 − 50

l = 40, h = 50 − 40 = 10, f₁ = 65, f₀ = 42 + x

Median `=l+[n/2 - (f_0))/f_1 xx h`

`46=40+(230/2-(42+x))/65xx10`

`46=40+(115-(42+x))/65xx10`

`46 - 40 = (115-42-x)/65xx10`

`6=(73-x)/65xx10`

`(6xx65)/10=73-x`

`390/10=73-x`

39 = 73 − x

x = 73 − 39

x = 34

Now put value of x in equation (i).

x + y = 80

34 + y = 80

y = 80 − 34

y = 46