Question

The first quartile of the following data is given to be 21.5 marks
10 15
Marks: f
15 20 24
2023 ?
25-30 90
30 35 122
35 40 ?
45 50 56
40 45 20
45 50 33
Find the missing frequencies when N=460.
[Ans. fi = 64 and 2 = 51​

Answer 1

Solution :-

Marks ------------ Fi --------------- CF

10 - 15 ------------ 24 -------------- 24

15 - 20 ------------ x -------------- (24 + x)

20 - 25 ------------90 ------------(114 + x)

25 - 30 ------------122 ------------(236 + x)

30 - 35 ------------ y --------------(236 + x + y)

35 - 40 ------------ 56 ------------(292 + x + y)

40 - 45 ------------ 20 ------------(312 + x + y)

45 - 50 ------------ 33 -------------(345 + x + y)

so,

→ 345 + x + y = 460

→ x + y = 460 - 345 = 115 --------- Eqn.(1)

now,

→ N/4 = 460/4 = 115

then, cumulative frequency greater than 115 is correspondence to class 20 - 25 . so, it is quartile class .

now,

• L = Lower limit of quartile class = 20 .
• N = Total frequency = 460 .
• f = frequency of quartile class = 90 .
• CF = cumulative frequency of the class previous to quartile class = (24 + x) .
• h = size of class = 5 .

then,

→ first quartile = L + [(N/4 - CF)/F] * H

→ 21.5 = 20 + [(115 - 24 - x)/90] * 5

→ 21.5 - 20 = (91 - x)/18

→ 91 - x = 27

→ x = 91 - 27

→ x = 64 .

therefore, putting this value in Eqn.(1),

→ 64 + y = 115

→ y = 115 - 64

→ y = 51 .

Hence, the missing frequencies are 64 and 51 respectively.

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