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: -

  1. Lower Quartile (Q1)
  2. Middle Quartile (Q2)
  3. Upper Quartile (Q3)

.

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 (Q1) = value of (\(\frac{N+1}{4}\))thitem.

Upper Quartile (Q3) =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 (Q1) = value of (\(\frac{N+1}{4}\))thitem

Upper quartile (Q3) = 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 (Q1) = 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 (Q3) = 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, Q3 = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)thitem
or, Q3 = 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)thitem
or, Q3 = 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)thitem
or, Q3 = 3 \(\times\) 2thitem
or, Q3 = 6thitem
\(\therefore\) The value of Q3is 90.







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\(\begin{pmatrix}\frac{n+1}{4} \end{pmatrix}\)thterm

= value of\(\begin{pmatrix}\frac{7+1}{4} \end{pmatrix}\)thterm

= value of\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)thterm

= value of 2nditem

\(\therefore\) Lower Quartile = 20

Upper Quartile = value of\(\begin{pmatrix}\frac{3(n+1)}{4} \end{pmatrix}\)thterm

= value of\(\begin{pmatrix}\frac{3(7+1)}{4} \end{pmatrix}\)thterm

= value of\(\begin{pmatrix}\frac{3 \times 8}{4} \end{pmatrix}\)thterm

= value of\(\begin{pmatrix}\frac{24}{4} \end{pmatrix}\)thterm

= value of 6thterm

\(\therefore\) Upper Quartile = 27

Solution:

Here,
6, x+5, 12, 14, 17, 20, 21
Q1= 8
numbers = 7
We know that,
or, Q1 = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)th item
or, Q1 = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)th item
or, Q1 = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)th item
or, Q1 = 2th item
or, Q1 =x + 5
Now,
or,Q1= x + 5
or, 8= x + 5
or, x = 8 - 5
\(\therefore\) x = 5




Solution:

Here,
numbers = 7
Q3= 60
Now,
or, Q3= 3(\(\frac{N + 1}{4}\))th item
or, Q3= 3(\(\frac{7 + 1}{4}\))th item
or, Q3= 3(\(\frac{8}{4}\))th item
or, Q3= 3 \(\times\) 2th item
or, Q3= 6th item
i.e.Q3= x + 25
Then,
or,Q3= 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, Q1 = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)thitem
or, Q1 = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)thitem
or, Q1 = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)thitem
or, Q1 = 2thitem
or,Q1 = 52
Again,
or, Q2= \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)thitem
or, Q2= \(\begin{pmatrix}\frac{7 + 1}{2} \end{pmatrix}\)thitem
or, Q2= \(\begin{pmatrix}\frac{8}{2} \end{pmatrix}\)thitem
or, Q2= 4thitem
or, Q2= 61
Similarly,
or, Q3= 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)thitem
or, Q3= 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)thitem
or, Q3= 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)thitem
or, Q3= 3 \(\times\)2thitem
or, Q3= 6thitem
or, Q3 = 68

\(\therefore\) The value ofQ1, Q2 and Q3is 52, 61 and 68 respectively.

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 10th term

In c.f. column just greater than 10 is 12 and its coressponding value is 30.

\(\therefore\) Q1 = 30

Q2 =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 20th term

In c.f. column just greater than 20 is 24and its coressponding value is 40.

\(\therefore\) Q2=40

Q3 =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 30th term

In c.f. column just greater than 30 is 32 and its coressponding value is 50.

\(\therefore\) Q3= 50

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,Q1= \(\begin{pmatrix}\frac{N + 1}{4} \end{pmatrix}\)th item
or,Q1= \(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)th item
or,Q1= \(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)th item
or,Q1= 11th item

In c.f column, c.f is just greater than 11 is 15 and its corresponding value is 25. So, Q1 = 25
Again,
or, Q3 = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)
or, Q3 = 3\(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)
or, Q3 = 3\(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)
or, Q3 = 3 \(\times\) 11thitem
or, Q3 = 33th item

In c.f column, c.f is just greater than 33 is 38 and its corresponding value is 45.
So, Q3 is 45.

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, Q1 = \(\begin{pmatrix}\frac{N}{4} \end{pmatrix}\)th item
or, Q1 = \(\begin{pmatrix}\frac{36}{4} \end{pmatrix}\)th item
or, Q1 = 9th item
In c.f column, c.f is just greater than 9 is 14 and its corresponding value is 10 - 20.
So, Q1 lies in class 10 - 20
Then,
l = 10, c.f = 6, f = 8, and i = 10

or, Q1 =L + \(\frac{\frac{N}{4}-C.F}{f}\) × i
or,Q1 = 10 + \(\frac{\frac{36}{4}-6}{8}\) × 10
or,Q1 = 10 + \(\frac{9 - 6}{8}\) × 10
or,Q1 = 10 + \(\frac{3}{8}\) × 10
or,Q1 = 10 + \(\frac{3}{4}\) × 5
or,Q1 = 10 + 0.75× 5
or,Q1 = 10 +3.75
or,Q1 = 13.75
\(\therefore\) The value of Q1is 13.75.

Solution:

Class Frequency(f) c.f.
0-5 6 6
5-10 8 14
10-15 10 24
15-20 7 31
20-25 5 36
25-30 4 40
N =40

Now,

Q1 Class = value of (\(\frac{N}{4}\))th item

= value of ( \(\frac{40}{4}\))th item

= value of 10th item

In c.f. column, c.f. just greater than 10 is 14 and its corresponding class is (5-10).

\(\therefore\) Q1 lies in the class 5-10

Here,

\(\frac{N}{4}\) = 10,l= 5, c.f. = 6, f = 8 and i = 5

\(\therefore\)Q1= 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,

Q3 Class = value of(\(\frac{3N}{4}\))th item

= value of(\(\frac{3 * 40}{4}\))th item

=value of30th item

In c.f. column, c.f.just greater than 30 is 31 and its corresponding class is 15-20.

Here,

\(\frac{3N}{4}\) = 30,l= 5, c.f. = 24, f = 7and i = 5

\(\therefore\)Q3= l + \(\frac{\frac{3N}{4}− c.f}{f}\) × i

= 5 + \(\frac{30− 24}{7}\) ×5

= 5 + 4.28

= 9.28

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, Q3= 3(\(\frac{N}{4}\))th item
or, Q3= 3(\(\frac{56}{4}\))th item
or, Q3= 3 \(\times\) 14th item
or, Q3= 42th 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, Q3= L + \(\frac{\frac{3N}{4}-c.f}{f}\)× i
or, Q3= 25 + \(\frac{42 - 35}{10}\)× 5
or, Q3= 25 + \(\frac{7}{2}\)
or, Q3= 25 +3.5
or, Q3= 28.5
\(\therefore\) The value of Q3 is 28.5.

Solution:

Marks Students (f) c.f.
30-40 5 5
40-50 12 17
50-60 25 42
60-70 15 57
70-80 8 65
80-90 3 68
N =68

Now,

Q1 Class = value of (\(\frac{N}{4}\))th item

= value of ( \(\frac{68}{4}\))th item

= value of 17th item

In c.f. column, c.f. just greater than 17 is 42 and its corresponding class is (50-60).

\(\therefore\) Q1 lies in the class 50-60

Here,

\(\frac{N}{4}\) = 17, l= 50, c.f. = 17, f = 25 and i =10

\(\therefore\)Q1= l + \(\frac{\frac{N}{4}− c.f}{f}\) × i

= 50 + \(\frac{17− 17}{25}\) ×10

=50 + \(\frac{0}{25}\) ×10

= 50 +0

= 50

Again,

Q2Class = value of (\(\frac{N}{2}\))th item

= value of ( \(\frac{68}{2}\))th item

= value of 34th item

In c.f. column, c.f. just greater than 34 is 42 and its corresponding class is (50-60).

\(\therefore\) Q2lies in the class 50-60

Here,

\(\frac{N}{2}\) = 34, l = 50, c.f. = 17, f = 25 and i =10

\(\therefore\)Q1= 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

Q3 Class = value of(\(\frac{3N}{4}\))th item

= value of(\(\frac{3 * 68}{4}\))th item

=value of 51th item

In c.f. column, c.f.just greater than 51 is 57 and its corresponding class is 60-70.

Here,

\(\frac{3N}{4}\) = 51, l= 60, c.f. = 42, f = 15 and i = 10

\(\therefore\)Q3= l + \(\frac{\frac{3N}{4}− c.f}{f}\) × i

= 60 + \(\frac{51− 42}{15}\) × 10

= 60 + 6

= 66

Solution:

x f c.f
50 8 8
60 16 24
70 12 36
80 4 40
90 3 43
N = 43

Now,
or, Q1 =(\(\frac{N+1}{4}\))th item
or, Q1 =(\(\frac{43+1}{4}\))th item
or, Q1 =(\(\frac{44}{4}\))th item
or, Q1 =11th item
In c.f column, c.f is just greater than 11 is 24 and its corresponding value is 60.
So,Q1 = 60
Again,
or, Q3 =3(\(\frac{N+1}{4}\))th item
or, Q3 =3(\(\frac{43+1}{4}\))th item
or, Q3 =3(\(\frac{44}{4}\))th item
or, Q3 =3 \(\times\) 11th item
or, Q3 =33th item
In c.f column, c.f is just greater than 33 is 36 and its corresponding value is 70.
i.e.Q3 = 70







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    2
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