53. Find median of given data.Class (less then)102030405060Cumulative

1.	53. Find median of given data.Class (less then)102030405060Cumulative frequency 21017222630
Answer» $\large\underline{\sf{Given \:Question - }}$ FIND the median of the following data :- $\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c\|cc}\sf Class\: (less \: than)&\sf Cumulative \: Frequency\: (c.f.)\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}\\\sf 10&\sf2\\\\\sf 20 &\sf 10\\\\\sf 30 &\sf 17\\\\\sf 40&\sf 22\\\\\sf 50&\sf 26 \\ \\\sf 60 &\sf 30&\sf \\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$ $\large\underline{\sf{Solution-}}$ First we have to convert the Cumulative frequency series to Exclusive Series. $\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c\|cc}\sf Class\: interval&\sf Frequency\: (f)\\\frac{\qquad }{}&\frac{\qquad \qquad \qquad}{}\\\sf 0 - 10&\sf 2\\\\\sf 10 - 20 &\sf 8\\\\\sf 20-30 &\sf 7\\\\\sf 30 - 40&\sf 5\\\\\sf 40-50&\sf 4\\ \\ \sf 50 - 60&\sf 4&\sf\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf \sum f = 30& \end{array}}\end{gathered}\end{gathered}\end{gathered}$ So, $\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c\|c\|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: Cumulative \: frequency\\\frac{\qquad }{}&\frac{\qquad \qquad \qquad}{}\\\sf 0 - 10&\sf 2&\sf2\\\\\sf 10 - 20 &\sf 8&\sf10\\\\\sf 20-30 &\sf 7&\sf17\\\\\sf 30 - 40&\sf 5&\sf22\\\\\sf 40-50&\sf 4&\sf26\\ \\ \sf 50 - 60&\sf 4&\sf30\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf \sum f = 30& \end{array}}\end{gathered}\end{gathered}\end{gathered}$ We know, Formula of Median :- $\: \: \: \: \: \: \boxed{ \sf M= l + \Bigg \{<klux>H</klux> \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$ Here, l denotes lower limit of median class h denotes width of median class f denotes frequency of median class cf denotes cumulative frequency of the class preceding the median class N denotes sum of frequency ACCORDING to the question, N = 30, this implies N/2 = 15 So, Median class is 20-30 l = 20, h = 10, f = 7, cf = cf of preceding class = 10 N/2 = 15 By substituting all the given values in the formula, $\dashrightarrow\sf M= 20 + \Bigg \{10 \times \dfrac{ ( 15 - 10)}{7} \Bigg \}$ $\dashrightarrow\sf M= 20 + \Bigg \{10 \times \dfrac{ ( 5)}{7} \Bigg \}$ $\sf \dashrightarrow \: M= 20 + \dfrac{50}{7} = 20 + 7.14 = 27.14 \: (approx)$ Additional Information :- 1. Mode :- $\begin{gathered} \dashrightarrow\sf Mode = x_{k} + \bigg \{h \times \dfrac{(f_{k} - f_{k - 1})}{(2f_{k} - _{k - 1} - f_{k + 1}} \bigg \} \\ \end{gathered}$ 2. Mean using direct Formula :- $\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$ 3. Mean using Short - Cut Method :- $\dashrightarrow\sf Mean = \: A \: + \dfrac{ \sum f_i d_i}{ \sum f_i}$ 4. Mean using STEP - deviation method :- $\dashrightarrow\sf Mean = \: A \: + \: \dfrac{ \sum f_i u_i}{ \sum f_i} \: \times \: h$

53. Find median of given data.Class (less then)102030405060Cumulative frequency 21017222630

Answer»

$\large\underline{\sf{Given \:Question - }}$

FIND the median of the following data :-

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|cc}\sf Class\: (less \: than)&\sf Cumulative \: Frequency\: (c.f.)\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}\\\sf 10&\sf2\\\\\sf 20 &\sf 10\\\\\sf 30 &\sf 17\\\\\sf 40&\sf 22\\\\\sf 50&\sf 26 \\ \\\sf 60 &\sf 30&\sf \\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

First we have to convert the Cumulative frequency series to Exclusive Series.

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|cc}\sf Class\: interval&\sf Frequency\: (f)\\\frac{\qquad }{}&\frac{\qquad \qquad \qquad}{}\\\sf 0 - 10&\sf 2\\\\\sf 10 - 20 &\sf 8\\\\\sf 20-30 &\sf 7\\\\\sf 30 - 40&\sf 5\\\\\sf 40-50&\sf 4\\ \\ \sf 50 - 60&\sf 4&\sf\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf \sum f = 30& \end{array}}\end{gathered}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: Cumulative \: frequency\\\frac{\qquad }{}&\frac{\qquad \qquad \qquad}{}\\\sf 0 - 10&\sf 2&\sf2\\\\\sf 10 - 20 &\sf 8&\sf10\\\\\sf 20-30 &\sf 7&\sf17\\\\\sf 30 - 40&\sf 5&\sf22\\\\\sf 40-50&\sf 4&\sf26\\ \\ \sf 50 - 60&\sf 4&\sf30\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf \sum f = 30& \end{array}}\end{gathered}\end{gathered}\end{gathered}$

We know,

Formula of Median :-

$\: \: \: \: \: \: \boxed{ \sf M= l + \Bigg \{<klux>H</klux> \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$

Here,

l denotes lower limit of median class

h denotes width of median class

f denotes frequency of median class

cf denotes cumulative frequency of the class preceding the median class

N denotes sum of frequency

ACCORDING to the question,

N = 30, this implies N/2 = 15

So,

Median class is 20-30

l = 20,

h = 10,

f = 7,

cf = cf of preceding class = 10

N/2 = 15

By substituting all the given values in the formula,

$\dashrightarrow\sf M= 20 + \Bigg \{10 \times \dfrac{ ( 15 - 10)}{7} \Bigg \}$

$\dashrightarrow\sf M= 20 + \Bigg \{10 \times \dfrac{ ( 5)}{7} \Bigg \}$

$\sf \dashrightarrow \: M= 20 + \dfrac{50}{7} = 20 + 7.14 = 27.14 \: (approx)$

Additional Information :-

1. Mode :-

$\begin{gathered} \dashrightarrow\sf Mode = x_{k} + \bigg \{h \times \dfrac{(f_{k} - f_{k - 1})}{(2f_{k} - _{k - 1} - f_{k + 1}} \bigg \} \\ \end{gathered}$

2. Mean using direct Formula :-

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

3. Mean using Short - Cut Method :-

$\dashrightarrow\sf Mean = \: A \: + \dfrac{ \sum f_i d_i}{ \sum f_i}$

4. Mean using STEP - deviation method :-

$\dashrightarrow\sf Mean = \: A \: + \: \dfrac{ \sum f_i u_i}{ \sum f_i} \: \times \: h$