Partition Values Based on Median Principles Median is

Partition Values Based on Median Principles Median is the division of the series into two parts, sometimes we may be required to divide the series into more than two parts. The value of each is known as partition value. Generally the series may be divided in four parts called Quartiles. Ten Parts – Deciles, Hundred parts – Percentiles. (are those values of the variate which divide the total frequency into hundred, ten , four equal parts. ) Uses of P. V. – present the behavior Of the different parts of the series.

Steps for the calculation of Q 1 , Q 2 , Q 3 Steps : (1) Arrange the data in ascending order only. (2) Locate the item by finding out For Median OR(Q 2) N + 1 th ______ M= size of item 2 For Q 1 N+1 ______ 4 th For Q 3 3 ___ N+1 4 th

Example - partition Values Calculate Q 1, Q 2, Q 3 , from the following data of heights of 7 students in a class. S. No. Name of the Students Height (C. M) 1 A 140 2 B 142 3 C 144 4 D 145 5 E 147 6 F 149 7 G 151

Solution : Partition values S. No. Name of the Students Height (C. M) 1 A 140 2 B 142 3 C 144 4 D 145 5 E 147 6 F 149 7 G 151 (Q 1 First quartile or lower quartile ) (Median -Q 2 Middle Quartile) (Q 3 third quartile or upper quartile) (2) Median = = Size of N+1 ___ 2 Size of th ____ 7+1 th 2 = Size of 8 __ 2 =4 = Size of 4 th item Therefore – M = 145 N+1 th item (1) Solution – Q 1 = Size of _____ N+1 th ___ 4 (3) Q 3 = Size of 3 4 7+1 = size of ____ 8 ___ 7+1 =2 = Size of 3 ____ 4 Size of (3 X 2) th 4 4 = 06 th item 8 Therefore =149 = size of 3 __ = Size of 2 th item is =142 4 Upper Quartile

Partition Values in Three Series 1. Individual Series Formula s for individual & Discrete series 1. In case of Quartile = N+1 ____ 4 th item 2. In case of deciles = ____ N+1 10 th item 3. In case of percentile = 1. 4 th Deciles ( D 4 ) 2. 28 th Percentile ____ N+1 100 4 N+1 ____ 10 th item N+1 th item 28 _____ 100

Example in Individual Series (Partition Values) Q. Calculate Q 1, Q 3, D 4. P 60 with the help of following data : 1, 2, 3, 4, 6, 8, 9, 10, 11, 13, 15, 17, 19, 20, 21, 22, 23, 24, 25.

Partition Value in Individual series S. No. Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 6 Q 1 8 9 10 D 4 11 13 15 17 P 60 19 20 21 Q 3 22 23 24 P 90 25 1. Q 1 = size of N+1 ____ 4 ____ 19+1 4 th item th = size of 5 th item =6 Therefore Q 1 = 6 item

Partition Value in Individual series S. No. Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 6 Q 1 8 9 10 D 4 11 13 15 17 P 60 19 20 21 Q 3 22 23 24 P 90 25 2. Q 3 = size of 3 N+1 ____ 4 ____ 19+1 = size of 3 4 th item th = Size of 15 th item = 21 Therefore Q 3 = 21 item

Partition Value in Individual series S. No. Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 6 Q 1 8 9 10 D 4 11 13 15 17 P 60 19 20 21 Q 3 22 23 24 P 90 25 3. D 4 = size of 4 N+1 ____ 10 ____ 19+1 = size of 4 10 th item th = Size of 8 th item = 10 Therefore D 4 = 10 item

Partition Value in Individual series S. No. Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 6 Q 1 8 9 10 D 4 11 13 15 17 P 60 19 20 21 Q 3 22 23 24 P 90 25 4. P 60 size of 60 N+1 ____ th item 100 ____ th item 19+1 = size of 60 100 = Size of 12 th item = 17 Therefore P 60 = 17

Partition Value in Individual series S. No. Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 6 Q 1 8 9 10 D 4 11 13 15 17 P 60 19 20 21 Q 3 22 23 24 P 90 25 5. P 90 size of 90 N+1 ____ th item 100 ____ th item 19+1 = size of 90 100 = Size of 18 th item = 24 Therefore P 90 = 24

Example in Discrete Series (Partition Values) Q. Calculate Q 1, Q 3, D 6. P 10 with the help of following data : X: 10 15 20 25 30 35 40 45 50 55 60 65 70 Y: 4 6 5 7 12 5 8 9 12 10 7 8 6

Solution : In discrete Series Partition Value 1. Q 1 X F C. F. 10 15 20 25 30 35 40 45 50 55 60 65 70 4 6 5 7 12 5 8 9 12 10 7 8 6 4 10 15 22 34 39 47 56 68 78 85 93 99 N= 99 = size of N+1 ____ 4 ____ 99+1 4 th item = size of 25 th item lies in c. f = 34 Therefore Q 1 = 30

Solution : In discrete Series Partition Value 2. Q 3 = size of 3 X F C. F. 10 15 20 25 30 35 40 45 50 55 60 65 70 4 6 5 7 12 5 8 9 12 10 7 8 6 4 10 15 22 34 (Q 1) 39 47 56 68 78(Q 3) 85 93 99 N= 99 N+1 ____ 4 ____ 99+1 = size of 3 4 th item = Size of 75 th item. It lies in 78 c. f. = 55 Therefore Q 3 = 55

Solution : In discrete Series Partition Value 3. D 6 X F C. F. 10 15 20 25 30 35 40 45 50 55 60 65 70 4 6 5 7 12 5 8 9 12 10 7 8 6 4 10 15 22 34 (Q 1) 39 47 56 68(D 6) 78(Q 3) 85 93 99 N= 99 = size of 6 N+1 ____ th item 10 ____ th item 99+1 = size of 6 10 = Size of 60 th item. It lies in 68 c. f. = 50 Therefore D 6 = 50

Solution : In discrete Series Partition Value 3. P 10 X F C. F. 10 15 20 25 30 35 40 45 50 55 60 65 70 4 6 5 7 12 5 8 9 12 10 7 8 6 4 10(P 10) 15 22 34 (Q 1) 39 47 56 68(D 6) 78(Q 3) 85 93 99 N= 99 = size of 10 N+1 ____ th item 100 ____ th item 99+1 = size of 10 10 = Size of 10 th item. It lies in 10 c. f. = 15 Therefore P 10 = 15

Example in Continuous Series (Partition Values) Q. Calculate Q 1, Q 3, D 4. D 6 P 90 with the help of following data : Marks No. of students 0– 5 4 5 – 10 6 10 – 15 8 15 - 20 12 20 - 25 12 25 – 30 8 30 – 35 6 35 – 40 4

Solution : In Continuous Series Partition Value 1. Q 1 = size of __ N 4 __ 60 4 th th item Marks No. of students c. f 0– 5 4 4 = size of 15 th item lies in c. f. 18, 10 – 15 group 5 – 10 6 10 C i __ (Q 1 – C) Therefore Q 1 = L 1 + f 5 __ (15 – 10) = Q 1 = 10 + 8 10 – 15 8 18(Q 1) 15 - 20 12 30 20 - 25 12 42 25 – 30 8 50 30 – 35 6 56 35– 40 4 60 N=60 = 10 + 3. 125 = 13. 125

Solution : In Continuous Series Partition Value 1. Q 3 = size of 3 __ N 4 __ 60 4 th th item Marks No. of students c. f 0– 5 4 4 = size of 45 th item lies in c. f. 50 , 25 – 30 group 5 – 10 6 10 10 – 15 8 18 15 - 20 12 30 20 - 25 12 42 C i __ (Q 3 – C) Therefore Q 3 = L 1 + f 5 __ (45 – 42) = Q 3 = 25 + 8 25 – 30 8 50 (Q 3) 30 – 35 6 56 35– 40 4 60 N=60 = 25 + 1. 875 = 26. 875

Solution : In Continuous Series Partition Value 1. D 4 = size of 4 Marks No. of students c. f 0– 5 4 4 5 – 10 6 10 10 – 15 8 18 C 15 - 20 30(D 4) 12 20 - 25 12 42 25 – 30 8 50 30 – 35 6 56 35– 40 4 60 N=60 = size of 4 __ N th 10 item __ th 60 item 10 = size of 24 th item lies in (c. f 30) 15 – 20 group Therefore = = D 4 i __ (D 4 – C) L 1 + f 5 __ (24 – 18) = 15 + 12 = 15 + 2. 5 = 17. 5

Solution : In Continuous Series Partition Value __ N 1. P 90 = size of 90 th 100 item 60 th __ = size of 90 item 100 Marks No. of students c. f 0– 5 4 4 = size of 54 th item lies in (c. f. 56) 30 – 35 group 5 – 10 6 10 10 – 15 8 18 15 - 20 12 30 20 -25 12 42 i __ (P 90 – C) Therefore = L 1 + f 5 __ (54 – 50) = P 90= 30 + 6 25 – 30 8 50 C 30 – 35 6 56 35– 40 4 60 N=60 = 30 + 3. 33 = 33. 33

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