Question

In the frequency distribution of 100 families
given below, the number of families corresponding
to expenditure groups
missing
the median
from the table
12 known to
'be le. 50. find the missing frequency
No. of families
21-40
However
Expenditure (in Re.
0-20
14
20-40
-
40-60
27
60-80
-
80-100
15​

Answer 1

Solution :-

Expenditure(in Rs.) Frequency cumulative freq

0-20 14 14

20-40 x 14+x

40-60 28 42+x

60-80 y 42+x+y

80-100 15 57+x+y

100

given that,

→ Total no of families = 100 = N

so,

→ 57 + x + y = 100

→ (x + y) = 100 - 57

→ (x + y) = 43 ------------------ Eqn.(1)

now, we know that,

• Median = Lower limit of median class + [{(N/2) - Cumutative frequency}/f] * Class length

given that,

• Median = 50 .

then, we get,

• Median class = 40 - 60
• Lower limit = 40
• N/2 = 100/2 = 50
• CF = (14 + x)
• Class length = 20
• f = 28

Putting all values we get,

→ 40 + [{50 - (14+x)}/28] * 20 = 50

→ {(50 - 14 - x)/28 } * 20 = 50 - 40

→ {(36 - x)/28} * 20 = 10

→ (36 - x) / 28 = (1/2)

→ 36 - x / 14 = 1

→ 36 - x = 14

→ x = 36 - 14

→ x = 22 (Ans.)

putting value of x in Eqn.(1), we get,

→ x + y = 43

→ 22 + y = 43

→ y = 43 - 22

→ y = 21 (Ans.)