Quartiles
If three numbers divide the series into four equal parts then, these three numbers are called quartiles.
Here 9, 13 and 27 divide the series into 4 equal parts. Therefore, 9, 13 and 27 are called quartiles. 9 is called the first quartile. (Q_{1}). 25% of the items are less than Q_{1 }and 75% of the items are more thanQ_{1}. 13 is called the second quartile. 50% of the items are below and above it. Similarly, 27 is called the third quartile. 75% of the items are less than Q_{3}and 25% are more than Q3.
Therefore, quartiles are also the positional value of the items.
Calculation of quartiles in continuous series
Steps:
a. Construct the cumulative frequency table.
b. Use the formula to find the quartiles class.
The first quartile, Q_{1} = (\(\frac{N}{4}\))^{th} class
The third quartile, Q_{3} = (\(\frac{3N}{4}\))^{th} class
c. Use the formula to find the exact value of quartiles.
\( Q_1 = L + \frac {\frac{N}{4}  C.f }{f} \times h \)
\( Q_3 = L + \frac {\frac{3N}{4}  C.f }{f} \times h \)
Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of classinterval
Ogive (Cumulative frequency curve )
Ogive is a graphical representation of the distribution of a continuous data. While drawing ogive, points are plotted with cumulative frequency along with yaxis and its corresponding variables along with xaxis.There are two types of the ogive. They are less than ogive and more than ogive.
\( Q_1 = L + \frac {\frac{N}{4}  C.f }{f} \times h \)
\( Q_3 = L + \frac {\frac{3N}{4}  C.f }{f} \times h \)
Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of classinterval
Solution:
The given data write in ascending order: 40, 49, 50, 55, 60, 61, 70 where N = 7
\begin{align*} Position \: of\: Q_3 &= \frac{3(N + 1)^{th}}{4}term \\ &= \frac{3(7+1)^{th}}{4}\\ &= 6^{th} \: term \\ 6^{th} \: term \: &represent \: 61 \\ \therefore Upper \: quartile \: &(Q_3) = 61 \end{align*}
Solution:
The given data write in ascending order
30, 31, 37,40, 42, 43, 45, 48, where N = 8
\begin{align*} Position \: of \: Q_3 &= \frac{3(N + 1)^{th}}{4}term \\ &= \frac{3(8 + 1)^{th}}{4} term \\ &= 3 \times 2.25^{th} \: term \\ &= 6.75^{th} \: term \\ \: \\ Q_3 = 6^{th}\:term + &(7^{th}term  6^{th} term) \times 0.75\\ Q_3 =43 + (45  &43) \times 0.75 \\ Q_3 = 43 + 1.5 \\ \therefore \: \: Q_3 = 44.5 \: \: _{Ans} \end{align*}
Solution:
N = 7
\begin{align*} Position \: of \: Q_1 &= \frac{N + 1^{th}}{4}term \\ &= \frac{7 + 1^{th}}{4}term \\ &= 2^{nd} \: term \\ \: \\ 2^{th}\: term \: repr&esent \: 3x 1 \\ \: \\ Q_1 &= 3x 1 \\ 20 &= 3x 1 \\ or, 3x&= 20 +1 \\ or, x &= \frac{21}{3}\\ \therefore x &= 7 \: _{Ans} \end{align*}
Solution:
The given data arranging in ascending order.
11, 12, 14, 17, 18, 19, 22, 25 where N = 8.
\begin{align*} Position \: of \: first \: quartile \: (Q_1) &= \frac{N+1^{th}}{4}term \\ &= \frac{8+1}{4}term \\ &= 2.25^{th} \: term \\ \: \\ Q_1 = 2^{nd} term + (3^{rd}  &2^{nd}) term \times 0.25 \\ Q_1 = 12 + (1412) \times &0.25 \\ Q_1 = 12 + 2 \times 0.25\\ \therefore Q_1 = 12.5 \: \: _{Ans} \end{align*}
Find the class of lower quartile (Q_{1}) from the following data.
Marks  50  60  70  80  90  100 
No. of students  3  4  7  5  2  9 
Solution:
Calculating lower quartile (Q_{1})
Marks (X)  Frequency (f)  Cumulative frequency (cf) 
50  3  3 
60  4  7 
70  7  14 
80  5  19 
90  2  21 
100  9  30 
N = 30 
\begin{align*}Position \: of \: lower \: quartile (Q_1) &= \frac{N + 1^{th}}{4}term \\ &= \frac{30 + 1^{th}}{4}term \\ &= 7.75^{th} term \end{align*}
7.75^{th} term represent cf value 14.
\(\therefore\) lower quartile (Q_{1}) = 70 \(_{Ans}\)
Solution:
The given data write in ascending order
\(1,5,7,2x4,x+7,2x+1,3x+2\) where N = 7
\begin{align*} Position \: of \: Q_3 &= \frac{3(N + 1)^{th}}{4}term \\ &= \frac{3(7+1)^{th}}{4}term \\ &= 3 \times 2^{th} \: term \\ &= 6^{th} term \\ 6^{th} term \: repr&esent \: 2x + 1 \end{align*}
\begin{align*} Q_1 &= 2x+1 \\ 2x + 1 &= 15 \\ or, 2x &= 15 1 \\ \therefore x &= \frac{14}{2} =7 \: _{Ans} \end{align*}
Calculate the class of third quartile from the given data.
Marks  0  10  10  20  20  30  30  40 
No. of students  4  8  12  4 
Solution:
Calculating the Q_{3} class
Marks (x)  No. of students(f)  Cumulative frequenc (cf) 
0  10  4  4 
10  20  8  12 
20  30  12  24 
30  40  4  28 
N = 28 
\begin{align*} Position \: of \: Q_3 \: class &= \frac{3N^{th}}{4}term \\ &= \frac{3 \times 28^{th}}{4} term \\ &= 21^{th} \: term \\ Class \: of \: Q_3 &= 20  30 \: _{Ans}\end{align*}
Find the upper quartile from the following data.
Marks  10  20  30  40  50 
No. of students  5  4  5  6  7 
solution:
Calculating the Q_{3} class
Marks (x)  No. of students(f)  Cumulative frequenc (cf) 
10  5  5 
20  4  9 
30  5  14 
40  6  20 
50  7  27 
N = 27 
\begin{align*} Position \: of \: upper\:quartile (Q_3) &= \frac{3(N+1)^{th}}{4}term \\ &= \frac{3(27 + 1)^{th}}{4}term \\ &= 3 \times 7^{th} \: term \\ &=21 ^{th} \: term\\ \: \\ 21^{th} \: term \: represent \: c.f \: 27 \\ \therefore Upper \: quartile \: (Q_3) &= 50 \: \: _{Ans} \end{align*}
Solution:
From graph,
Total number of boys (N) = 60
\begin{align*} Position \: of \: first \: quartile &= \frac{N^{th}}{4}term \\ &= \frac{60^{th}}{4}term \\ &= 15^{th} term \\ Class \: of \: first \: quartile &= 5  10 \: _{Ans} \end{align*}

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q3if the third quartile is 45 
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how to find frequency from graph 
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