Constract of series when mid value are given
Mid value. 2.5 7.5 12.5 17.5
Frequency 4 6 7 10
Answers
Explanation:
The difference between any two consecutive mid values is 5, the width of class interval = 5
∴ Class interval with mid-value 2.5 is 0 - 5 Class interval with mid-value 7.5 is 5 - 10, etc.
We construct the less than cumulative frequency table as given below:
Class Interval Frequency
(f) Less than cumulative frequency
(c.f.)
0 - 5 7 7
5 - 10 18 25
10 - 15 25 50 ← D4, P48
15 - 20 30 80
20 - 25 20 100
Total 100
Here, N = 100
D4 class = class containing
(4N10)th observation
∴
4N10=4×10010 = 40
Cumulative frequency which is just greater than (or equal to) 40 is 50.
∴ D4 lies in the class 10 - 15
∴ L = 10, f = 25, c.f. = 25, h = 5
D4 =
L+hf(4N10-c.f.)
=
10+525(40-25)
=
10+15(15)
= 10 + 3
∴ D4 = 13
P48 class = class containing
(48N100)th observation
∴
48N100=48×100100 = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P48 lies in the class 10 - 15
∴ L=10, f = 25, c.f. = 25, h = 5
P48 =
L+hf(48N100-c.f.)
=
10+525(48-25)
=
10+15(23)
= 10 + 4.6
∴ P48 = 14.6
Concept: Relations Among Quartiles, Deciles and Percentiles
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Chapter 1: Partition Values - Exercise 1.2 [Page 15]
Q 6Q 5
Answer:
14.6
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