Quartile
Quartile
A quartile is the variate value which divides the value of given frequency into four equal parts. There are three quartiles, they are: 
 Lower Quartile (Q_{1})
 Middle Quartile (Q_{2})
 Upper Quartile (Q_{3})
Quartiles can be calculated as following ways: 
Individual series
After arranging the data in ascending order, the quartiles are calculated by applying the following formula:
Lower quartile (Q_{1}) = value of (\(\frac{N+1}{4}\))^{th }item.
Upper Quartile (Q_{3}) =value of 3(\(\frac{N+1}{4}\))^{th} item.
Where, N = Total sum of frequency.
Discrete Series
When the data are arranged in ascending order, the quartile is calculated by the following formula:
Lower quartile (Q_{1}) = value of (\(\frac{N+1}{4}\))^{th }item
Upper quartile (Q_{3}) = value of 3(\(\frac{N+1}{4}\))^{th} item
Where N = Total sum of frequency
Continuous Series
For continuous series, the following formulae area used.
Lower quartile class = value of 3(\(\frac{N}{4}\))^{th} item
Exact Lower Quartile (Q_{1}) = L + \(\frac{\frac{N}{4}  C.F}{f}\)× i
where,
L = Lower limit of corresponding class
f = frequency
c.f = cumulative frequency
i = class size
N = Total sum of frequency
Upper quartile class = value of 3(\(\frac{N}{4}\))^{th} item
Exact upper quartile (Q_{3}) = L + \(\frac{\frac{3N}{4}  C.F}{f}\)× i
 Quartile is the variate value which divides the value of given frequency into four equal parts.
 There are three quartiles, they are: lower quartile, middle quartile and upper quartile.
Solution:
Now,
Arranging in ascending order
42, 52, 62, 72, 82, 90, 100
Total number (N) = 7
We know that,
or, Q_{3} = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th}item
or, Q_{3} = 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th}item
or, Q_{3} = 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}item
or, Q_{3} = 3 \(\times\) 2^{th}item
or, Q_{3} = 6^{th}item
\(\therefore\) The value of Q_{3}is 90.
Find the lower quartile and upper quartile of the following data set of scores:
18 20 23 20 23 27 24
Solution:
Arrange the values in ascending order of magnitude:
18 20 20 23 23 23 24 27 29
Here, n =7
Lower quartile = value of\(\begin{pmatrix}\frac{n+1}{4} \end{pmatrix}\)^{th}term
= value of\(\begin{pmatrix}\frac{7+1}{4} \end{pmatrix}\)^{th}term
= value of\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}term
= value of 2^{nd}item
\(\therefore\) Lower Quartile = 20
Upper Quartile = value of\(\begin{pmatrix}\frac{3(n+1)}{4} \end{pmatrix}\)^{th}term
= value of\(\begin{pmatrix}\frac{3(7+1)}{4} \end{pmatrix}\)^{th}term
= value of\(\begin{pmatrix}\frac{3 \times 8}{4} \end{pmatrix}\)^{th}term
= value of\(\begin{pmatrix}\frac{24}{4} \end{pmatrix}\)^{th}term
= value of 6^{th}term
\(\therefore\) Upper Quartile = 27
Solution:
Here,
6, x+5, 12, 14, 17, 20, 21
Q_{1}= 8
numbers = 7
We know that,
or, Q_{1} = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th} item
or, Q_{1} = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th} item
or, Q_{1} = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th} item
or, Q_{1} = 2^{th} item
or, Q_{1} =x + 5
Now,
or,Q_{1}= x + 5
or, 8= x + 5
or, x = 8  5
\(\therefore\) x = 5
Solution:
Here,
numbers = 7
Q_{3}= 60
Now,
or, Q_{3}= 3(\(\frac{N + 1}{4}\))^{th} item
or, Q_{3}= 3(\(\frac{7 + 1}{4}\))^{th} item
or, Q_{3}= 3(\(\frac{8}{4}\))^{th} item
or, Q_{3}= 3 \(\times\) 2^{th} item
or, Q_{3}= 6^{th} item
i.e.Q_{3}= x + 25
Then,
or,Q_{3}= x + 25
or, 60 = x + 25
or, x + 60  25
\(\therefore\) x = 35
Solution:
Now,
Arranging in ascending order
50, 52, 56, 61, 64, 68, 72
numbers = 7
We know that,
or, Q_{1} = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th}item
or, Q_{1} = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th}item
or, Q_{1} = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}item
or, Q_{1} = 2^{th}item
or,Q_{1} = 52
Again,
or, Q_{2}= \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)^{th}item
or, Q_{2}= \(\begin{pmatrix}\frac{7 + 1}{2} \end{pmatrix}\)^{th}item
or, Q_{2}= \(\begin{pmatrix}\frac{8}{2} \end{pmatrix}\)^{th}item
or, Q_{2}= 4^{th}item
or, Q_{2}= 61
Similarly,
or, Q_{3}= 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th}item
or, Q_{3}= 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th}item
or, Q_{3}= 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}item
or, Q_{3}= 3 \(\times\)2^{th}item
or, Q_{3}= 6^{th}item
or, Q_{3 }= 68
\(\therefore\) The value ofQ_{1}, Q_{2} and Q_{3}is 52, 61 and 68 respectively.
Compute Q_{1}, Q_{2} and Q_{3 }from the following data:
Marks  10  20  30  40  50  60  70 
No. of Students  2  4  6  12  8  6  1 
Solution:
Marks  No. of students(f)  cumulative frequency (c.f.) 
10  2  2 
20  4  6 
30  6  12 
40  12  24 
50  8  32 
60  6  38 
70  1  39 
N = 39 
We have,
Q1= value of \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th} term
= value of \(\begin{pmatrix}\frac{39 + 1}{4} \end{pmatrix}\)^{th} term
= value of 10^{th} term
In c.f. column just greater than 10 is 12 and its coressponding value is 30.
\(\therefore\) Q_{1} = 30
Q_{2 }=value of \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)^{th} term
= value of \(\begin{pmatrix}\frac{39 + 1}{2} \end{pmatrix}\)^{th} term
= value of 20^{th} term
In c.f. column just greater than 20 is 24and its coressponding value is 40.
\(\therefore\) Q_{2}=40
Q_{3 }=value of \(\begin{pmatrix}\frac{3(n + 1)}{4} \end{pmatrix}\)^{th} term
=value of \(\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}\)^{th} term
=value of \(\begin{pmatrix}\frac{3 * 40}{4} \end{pmatrix}\)^{th} term
=value of \(\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}\)^{th} term
=value of 30^{th} term
In c.f. column just greater than 30 is 32 and its coressponding value is 50.
\(\therefore\) Q_{3}= 50
Calculate Q_{1} and Q_{}_{3} from the following data.
marks  15  25  35  45  55 
No. of students  5  10  16  7  5 
Solution:
Marks  Frequency (f)  c.f 
15  5  5 
25  10  15 
35  16  31 
45  7  38 
55  5  43 
N = 43 
Now,
or,Q_{1}= \(\begin{pmatrix}\frac{N + 1}{4} \end{pmatrix}\)^{th} item
or,Q_{1}= \(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)^{th} item
or,Q_{1}= \(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)^{th} item
or,Q_{1}= 11^{th} item
In c.f column, c.f is just greater than 11 is 15 and its corresponding value is 25. So, Q_{1} = 25
Again,
or, Q_{3} = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)
or, Q_{3} = 3\(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)
or, Q_{3} = 3\(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)
or, Q_{3} = 3 \(\times\) 11^{th}item
or, Q_{3} = 33^{th} item
In c.f column, c.f is just greater than 33 is 38 and its corresponding value is 45.
So, Q_{3} is 45.
Calculate the first quartile (lower quartile) from the given data.
Class  0 10  10 20  20  30  30  40  40  50 
frequency  6  8  10  7  5 
Solution:
x  f  c.f 
0 10  6  6 
10  20  8  14 
20  30  10  24 
30  40  7  31 
40  50  5  36 
N = 36 
Now,
or, Q_{1} = \(\begin{pmatrix}\frac{N}{4} \end{pmatrix}\)^{th} item
or, Q_{1} = \(\begin{pmatrix}\frac{36}{4} \end{pmatrix}\)^{th} item
or, Q_{1} = 9^{th} item
In c.f column, c.f is just greater than 9 is 14 and its corresponding value is 10  20.
So, Q_{1} lies in class 10  20
Then,
l = 10, c.f = 6, f = 8, and i = 10
or, Q_{1} =L + \(\frac{\frac{N}{4}C.F}{f}\) × i
or,Q_{1} = 10 + \(\frac{\frac{36}{4}6}{8}\) × 10
or,Q_{1} = 10 + \(\frac{9  6}{8}\) × 10
or,Q_{1} = 10 + \(\frac{3}{8}\) × 10
or,Q_{1} = 10 + \(\frac{3}{4}\) × 5
or,Q_{1} = 10 + 0.75× 5
or,Q_{1} = 10 +3.75
or,Q_{1} = 13.75
\(\therefore\) The value of Q_{1}is 13.75.
Calculate the first quartile and third quartile from the given data:
Class  05  510  1015  1520  2025  2530 
Frequency  6  8  10  7  5  3 
Solution:
Class  Frequency(f)  c.f. 
05  6  6 
510  8  14 
1015  10  24 
1520  7  31 
2025  5  36 
2530  4  40 
N =40 
Now,
Q1 Class = value of (\(\frac{N}{4}\))^{th} item
= value of ( \(\frac{40}{4}\))^{th} item
= value of 10^{th} item
In c.f. column, c.f. just greater than 10 is 14 and its corresponding class is (510).
\(\therefore\) Q_{1} lies in the class 510
Here,
\(\frac{N}{4}\) = 10,l= 5, c.f. = 6, f = 8 and i = 5
\(\therefore\)Q_{1}= l + \(\frac{\frac{N}{4}− c.f}{f}\) × i
= 5 + \(\frac{10− 6}{8}\) ×5
=5 + \(\frac{4}{8}\) ×5
= 5 + 2.5
= 7.5
Again,
Q_{3} Class = value of(\(\frac{3N}{4}\))^{th} item
= value of(\(\frac{3 * 40}{4}\))^{th} item
=value of30^{th} item
In c.f. column, c.f.just greater than 30 is 31 and its corresponding class is 1520.
Here,
\(\frac{3N}{4}\) = 30,l= 5, c.f. = 24, f = 7and i = 5
\(\therefore\)Q_{3}= l + \(\frac{\frac{3N}{4}− c.f}{f}\) × i
= 5 + \(\frac{30− 24}{7}\) ×5
= 5 + 4.28
= 9.28
Calculate the third quartile (upper quartile) from the following data.
Class  5 10  10 15  15  20  20  25  25  30  30  35  35  40 
Frequency  4  6  10  15  10  6  5 
Solution:
Class  frequency  c.f 
5  10  4  4 
10  15  6  10 
15  20  10  20 
20  25  15  35 
25  30  10  45 
30  35  6  51 
35  40  5  56 
N = 56 
Now,
or, Q_{3}= 3(\(\frac{N}{4}\))^{th} item
or, Q_{3}= 3(\(\frac{56}{4}\))^{th} item
or, Q_{3}= 3 \(\times\) 14^{th} item
or, Q_{3}= 42^{th} item
In c.f column, c.f is just greater than 42 is 45 and its corresponding value is 25  30.
Now,
l = 25, c.f = 35, f = 10, and i = 5
Then,
or, Q_{3}= L + \(\frac{\frac{3N}{4}c.f}{f}\)× i
or, Q_{3}= 25 + \(\frac{42  35}{10}\)× 5
or, Q_{3}= 25 + \(\frac{7}{2}\)
or, Q_{3}= 25 +3.5
or, Q_{3}= 28.5
\(\therefore\) The value of Q_{3} is 28.5.
Calculate first, second and third quartile from the following data:
Marks  3040  4050  5060  6070  7080  8090 
Students  5  12  25  15  8  3 
Solution:
Marks  Students (f)  c.f. 
3040  5  5 
4050  12  17 
5060  25  42 
6070  15  57 
7080  8  65 
8090  3  68 
N =68 
Now,
Q_{1} Class = value of (\(\frac{N}{4}\))^{th} item
= value of ( \(\frac{68}{4}\))^{th} item
= value of 17^{th} item
In c.f. column, c.f. just greater than 17 is 42 and its corresponding class is (5060).
\(\therefore\) Q_{1} lies in the class 5060
Here,
\(\frac{N}{4}\) = 17, l= 50, c.f. = 17, f = 25 and i =10
\(\therefore\)Q_{1}= l + \(\frac{\frac{N}{4}− c.f}{f}\) × i
= 50 + \(\frac{17− 17}{25}\) ×10
=50 + \(\frac{0}{25}\) ×10
= 50 +0
= 50
Again,
Q_{2}Class = value of (\(\frac{N}{2}\))^{th} item
= value of ( \(\frac{68}{2}\))^{th} item
= value of 34^{th} item
In c.f. column, c.f. just greater than 34 is 42 and its corresponding class is (5060).
\(\therefore\) Q_{2}lies in the class 5060
Here,
\(\frac{N}{2}\) = 34, l = 50, c.f. = 17, f = 25 and i =10
\(\therefore\)Q_{1}= l + \(\frac{\frac{N}{2}− c.f}{f}\) × i
= 50 + \(\frac{34− 17}{25}\) ×10
=50 + \(\frac{17}{25}\) ×10
= 50 +6.8
= 56.8
Q_{3} Class = value of(\(\frac{3N}{4}\))^{th} item
= value of(\(\frac{3 * 68}{4}\))^{th} item
=value of 51^{th} item
In c.f. column, c.f.just greater than 51 is 57 and its corresponding class is 6070.
Here,
\(\frac{3N}{4}\) = 51, l= 60, c.f. = 42, f = 15 and i = 10
\(\therefore\)Q_{3}= l + \(\frac{\frac{3N}{4}− c.f}{f}\) × i
= 60 + \(\frac{51− 42}{15}\) × 10
= 60 + 6
= 66
Find the Q_{1} and Q_{3} of the given data.
Weight (in kg)  50  60  70  80  90 
No. of teachers  8  16  12  4  5 
Solution:
x  f  c.f 
50  8  8 
60  16  24 
70  12  36 
80  4  40 
90  3  43 
N = 43 
Now,
or, Q_{1} =(\(\frac{N+1}{4}\))^{th} item
or, Q_{1} =(\(\frac{43+1}{4}\))^{th} item
or, Q_{1} =(\(\frac{44}{4}\))^{th} item
or, Q_{1} =11^{th} item
In c.f column, c.f is just greater than 11 is 24 and its corresponding value is 60.
So,Q_{1} = 60
Again,
or, Q_{3} =3(\(\frac{N+1}{4}\))^{th} item
or, Q_{3} =3(\(\frac{43+1}{4}\))^{th} item
or, Q_{3} =3(\(\frac{44}{4}\))^{th} item
or, Q_{3} =3 \(\times\) 11^{th} item
or, Q_{3} =33^{th} item
In c.f column, c.f is just greater than 33 is 36 and its corresponding value is 70.
i.e.Q_{3} = 70

In the data set below, what is the lower quartile?
1, 2, 4, 4, 5, 6, 6, 6, 7, 7, 8
6
7
2
4

In the data set below, what is the upper quartile?
1, 2. 4, 4, 5, 6, 6, 6 , 7, 7, 8
5
8
6
7

In the data set below, what is the lower quartile?
29, 34, 38, 41, 51, 57, 65, 65, 81, 83
41
34
29
38

In the data set below, what is the upper quartile?
17, 23, 32, 36, 52, 54, 67, 83, 98, 99
98
67
99
83

In the data set below, what is the lower quartile?
2, 2, 2, 2, 3 , 5, 5, 5, 7, 8, 9, 93
7
2
5

In the data set below, what is the upper quartile?
2, 2, 2, 2, 3 , 5, 5, 5, 7, 8, 9, 98
7
5
9

In the data set below, what is the upper quartile?
20, 38, 52, 54, 60, 87, 93
93
54
60
87

In the data set below, what is the lower quartile?
10, 27, 27, 29, 44, 672710
44
29
27

In the data set below, what is the lower quartile?
19, 37, 41, 43, 83, 85, 94
85
19
37
43

In the data set below, what is the upper quartile?
19, 37, 41, 43, 83, 85, 94
83
43
85
94

In the data set below, what is the lower quartile?
23, 26, 26, 30, 34, 42, 49, 60, 96, 96, 96
30
26
32
34

In the data set below, what is the upper quartile?
59, 86, 86, 88, 96, 9659
88
86
96

In the data set below, what is the lower quartile?
2, 4, 5, 6, 6, 7, 8, 8, 8, 9
4
6
5
2

In the data set below, what is the upper quartile?
2, 4, 5, 6, 6, 7, 8, 8, 8, 9
7
9
6
8

In the data set below, what is the upper quartile?
1, 1, 1, 2, 2, 4, 4, 5, 5, 6
6
4
1
5

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