Question

Class interval:100-200,200-300,300-400,400-500,500-600
Frequency: 10,18,12,20,40
Find mean by step deviation method

Answer 1

Hope this is helpful!

Answer 2

$\underline{\underline{\Huge{\textbf{Question:}}}}$

Find the mean of following frequency distribution using step deviation method

$\begin{array}{|l|c|c|c|c|c|}\cline{1-6}Class\: Interval:& 100-200&200-300&300-400&400-500&500-600\\\cline{1-6}Frequency:& 10&18&12&20&40\\\cline{1-6}\end{array}$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

Using Step-deviation method involves using the formula
$\bigstar\: \: \Large{\mathbf{\overline{x} = A+h\left[\dfrac{1}{N}\sum\limits_{i}\: f_{i}\: x_{i}\right]}}$

Where,
N is Class Width
A is Assumed Mean


$\underline{\Large{\textsf{Step-1:}}}$
Find the Assumed Mean ($\mathbf{A}$)

To find Assumed Mean($\mathbf{A}$), Check whether No. of frequencies are Odd or Even

$\left \{ {{\textsf{Odd;}\quad\quad A=x_{\left(\frac{n+1}{2}\right)}} \\ \atop {\quad\quad\textsf{Even;}\quad\quad A = x_{\left(\frac{n}{2}\right)}\: or\: A = x_{\left(\frac{n}{2}+1\right)}}} \right.$


Here,
$n = 5\textsf{(odd)}$
$\implies A = x_{\left(\frac{5+1}{2}\right)}$
$\implies A = x_{\left(\frac{6}{2}\right)}$
$\implies A =x_{3} = 350$



$\underline{\Large{\textsf{Step-2:}}}$
Note the class width ($\mathbf{h}$)

$\sf{\textsf{Class width is the difference between the lower limits of consecutive classes}}$

(or)

$\bigstar\: \: \textbf{Class\: Width = Upper\: Class\: Limit \: - \: Lower\: Class\: Limit}$

Here,
Class width $h= 200-100= 100$



$\underline{\Large{\textsf{Step-3:}}}$
Calculate Class Marks for each Class Interval

$\bigstar\: \: \textbf{Class\: Mark = \dfrac{Lower\: Class\: Limit+Upper\: Class\: Limit}{2} }$



$\underline{\Large{\textsf{Step-4:}}}$
Calculate $d_{i}$ for each Class Interval

$d_{i}= x_{i}-A$



$\underline{\Large{\textsf{Step-5:}}}$
Calculate $u_{i}$ for each Class Interval

$u_{i}= \dfrac{d_{i}}{h}$



$\underline{\Large{\textsf{Step-6:}}}$
Calculate $f_{i}u_{i}$ for each Class Interval



$\underline{\Large{\textsf{Step-7:}}}$
Find the Sum of Frequencies ($\mathbf{N}$)

$N =\sum\limits_{i}\: f_{i}$



$\underline{\Large{\textsf{Step-8:}}}$
Find sum of all $f_{i}u_{i}$



$\begin{array}{|c|c|c|c|c|c|}\cline{1-6}Class\: Interval&Class\: Marks(x_{i})& Frequency(f_{i})&d_{i}&u_{i}=\dfrac{d_{i}}{h}&f_{i}u_{i}\\\cline{1-6}100-200&150&10&-200&-2&-20\\200-300&250&18&-100&-1&-18\\300-400&350\: \: \: (=\: A)&12&0&0&0\\400-500&450&20&100&1&20\\500-600&550&40&200&2&80\\\cline{1-6}&&N= \sum\limits_{i}\: f_{i}= 100&&&\sum\limits_{i}\: f_{i}u_{i}=62\\\cline{1-6}\end{array}$



$\displaystyle\overline{x} = A+h\left[\dfrac{1}{N}\sum\limits_{i}\: f_{i}\: x_{i}\right]$

Substitute Values

$\overline{x} = 350+100\left[\dfrac{1}{100}(-20-18+0+20+80)\right]$

$\overline{x} = 350+\cancel{100}\left[\dfrac{1}{\cancel{100}}(62)\right]$

$\overline{x} = 350 + 62$

$\overline{x} = 412$

$\therefore$
$\blacksquare \: \: \textsf{The Required Mean ( \overline{x} ) = \underline{\Large{\textbf{412}}}}$