Introduction
To
Quartiles, Deciles & Percentiles
Lecture 10
Quartiles
A set of three values that split a sorted dataset into four equal parts, each part having an equal number of observations. These partitioning values are represented by Q1, Q2, and Q3.
In grouped data, quartiles can be obtained as:
Example 3.35: Calculate lower and upper quartiles for the
following data.
|
4 |
9 |
7 |
6 |
15 |
11 |
18 |
24 |
29 |
25 |
27 |
16 |
Solution: Arrange the data in ascending order of magnitude and
assign ranks.
|
Rank |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
|
Data |
4 |
6 |
7 |
9 |
11 |
15 |
16 |
18 |
24 |
25 |
27 |
29 |
The third item is 7.
So
Q1 = 7
The 9th item
is 24.
So
Q3 = 24
Example 3.36: Find the quartiles of the marks obtained by first
years in the annual examination.
|
497 |
495 |
480 |
465 |
440 |
490 |
443 |
398 |
390 |
365 |
400 |
412 |
432 |
416 |
389 |
Solution:
|
Rank
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
|
Data
|
365 |
389 |
390 |
398 |
400 |
412 |
416 |
432 |
440 |
443 |
465 |
480 |
480 |
495 |
497 |
Round up 3.75 to the 4th item.
So
Q1 = 398
Round up 11.25 to the 12th item.
So
Q3 = 480
Example 3.37: Find the lower and upper quartiles for the
following frequency distribution.
|
Class |
40
– 50 |
50- 60 |
60–70 |
70–80 |
80–90 |
90
- 100 |
100–110 |
|
Frequency |
20 |
17 |
44 |
57 |
68 |
55 |
34 |
Solution:
|
Class |
Frequency |
Cumulative
frequency |
|
40
– 50 |
20 |
20 |
|
50- 60 |
17 |
37 |
|
60–70 |
44 |
81 |
|
70–80 |
57 |
138 |
|
80–90 |
68 |
206 |
|
90–100 |
55 |
261 |
|
100
- 110 |
34 |
295 |
|
|
295 |
|
For Q1, we divided the total observation by 4.
We select the class that lies against 73.75; 73.75 is not
available in the cumulative frequency column. We select the class lies against
81, which is 60–70.
For Q3, we divided the 3 times of total observation by 4.
Which is lies against 90–100,
L1 = 90, L2 = 100, f = 55, c = 206
Where:
L1 and L2 are the lower
and upper class boundaries, and c is the cumulative frequency of the preceding
class.
PERCENTILES
Where:
L1 and L2 are the lower and upper class boundaries, and c is the cumulative frequency of the preceding class.
Example 3.38: Find the 5th, 9th deciles,
and 50th, 75th percentiles for the following frequency
distribution.
|
Group |
10 – 20 |
20- 30 |
30 – 40 |
40 – 50 |
50- 60 |
60–70 |
70 - 80 |
|
No, of person |
20 |
35 |
78 |
60 |
75 |
84 |
98 |
|
Group |
80–90 |
90 - 100 |
100 - 110 |
110 - 120 |
|
||
|
No, of person |
125 |
87 |
99 |
102 |
|||
Solution:
|
Group |
Frequency |
Cumulative frequency |
|
|
10 – 20 |
20 |
20 |
|
|
20- 30 |
35 |
55 |
|
|
30 – 40 |
78 |
133 |
|
|
40 – 50 |
60 |
193 |
|
|
50- 60 |
75 |
268 |
|
|
60–70 |
84 |
352 |
|
|
70–80 |
98 |
450 |
|
|
80–90 |
125 |
575 |
|
|
90–100 |
87 |
662 |
|
|
100–110 |
99 |
761 |
|
|
110 – 120 |
102 |
863 |
|
|
|
863 |
|
|
|
|
|||
For 5th decile:
Which lies against 70-80 group, so our L1 = 70, L2 = 80, f = 98, c = 352
For 9th decile:
Which lies against 110-120 group, so our L1 = 110, L2 = 120, f = 102, c = 761
For 50th percentiles:
For 75th percentiles:
- Read: Measure of Dispersion






















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