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 class-interval
Ogive is a graphical representation of the distribution of a continuous data. While drawing ogive, points are plotted with cumulative frequency along with y-axis and its corresponding variables along with x-axis.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 class-interval
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 + (14-12) \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
\(1,5,7,2x-4,x+7,2x+1,3x+2\) 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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if the third quartile is 45
Mar 19, 2017
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how to find frequency from graph
Jan 01, 2017
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