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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 ($$\frac{N+1}{4}$$)th item.

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 ($$\frac{N+1}{4}$$)th item

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.

.

#### Click on the questions below to reveal the answers

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

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