Videos Related with Quartiles

Note on Quartiles

W3Schools
  • Note
  • Things to remember
  • Videos
  • Exercise

If three numbers divide the series into four equal parts then, these three numbers are called quartiles.

sdf

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. (Q1). 25% of the items are less than Q1 and 75% of the items are more thanQ1. 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 Q3and 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, Q1 = (\(\frac{N}{4}\))th class

The third quartile, Q3 = (\(\frac{3N}{4}\))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 (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 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

.

Very Short Questions

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*}

Solution:

Calculating lower quartile (Q1)

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*}

7.75th term represent cf value 14.

\(\therefore\) lower quartile (Q1) = 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*}

Solution:

Calculating the Q3 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*}

solution:

Calculating the Q3 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*}

0%

DISCUSSIONS ABOUT THIS NOTE

You must login to reply

Forum Time Replies Report
q3

if the third quartile is 45


You must login to reply

how to find frequency from graph


You must login to reply