Question

The table below shows the salaries of 280 persons : salary =5-10,10-15,15-20,20-25,25-30,30-35,35-40,40-45,45-50.no.of person 49,133,63,15,6,7,4,2,1 calculate the median salary of the data

Answer 1

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

$\sf\textsf{The table below shows the salaries of 280 persons.}\\ \sf\textsf{Calculate the median salary of the data}$

$\small{\begin{array}{|c|c|c|c|c}\cline{1-5}\bf Salary&5-10&10-15&15-20&20-25\\\cline{1-5}\bf No. of persons&49&133&63&15\\\cline{1-5}\end{array}}$
$\small{\begin{array}{c|c|c|c|c|c|}\cline{1-6}\bf Salary&25-30&30-35&35-40&40-45&45-50\\\cline{1-6}\bf No. of persons&6&7&4&2&1\\\cline{1-6}\end{array}}$



$\underline{\Huge{\textsf{Concept Behind:}}}$

$\boxed{\begin{minipage}{9 cm}\maltese \: \: \tt Median of Grouped data \approxeq \quad \bf l + \left(\dfrac{\frac{\bf N}{\bf 2}-\bf cf}{\bf f}\right) \times \bf h\\\\ where\\ l = Lower limit of median class \\\: N = Total no. of observations\\\: cf = Cumulative Frequency of the class before the median class \\\: f = Frequency of the median class\\\:h = Class width\\\\\maltese \: \: \tt Median Class is the class whose cumulative frequency is greater and nearest to \frac{N}{2} \end{minipage}}$



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

$\underline{\LARGE{\texttt{Step-1:}}}$

$\sf\textsf{Note the No. Of Observations and Class width}$

$\sf\textsf{No. Of observations , \textbf{N}= 280}$

$\textsf{Class Width is the difference between}\\\sf\textsf{Upper and Lower boundary of a class}$

$\therefore \textsf{Class width, \textbf{h}= 10-5 = 5}$

$\underline{\LARGE{\texttt{Step-2:}}}$

$\sf\textsf{Add the Frequencies cumulatively for each class}$

$\sf\textsf{Here, the Frequencies refer to the No. of persons}$

$\begin{array}{|c|c|cc|}\cline{1-4}\bf Salary &\bf Frequency & \bf Cumulative & \bf Frequency\\\cline{1-4}5-10&49&&=\: 49\\10-15&133&49+133&\: \; =\bf 182\\15-20&63&182+63&=245\\20-25&15&245+15&=260\\25-30&6&260+6&=266\\30-35&7&266+7&=273\\35-40&4&273+4&=277\\40-45&2&277+2&=279\\45-50&1&279+1&=280 \\\cline{1-4}\end{array}$

$\underline{\LARGE{\texttt{Step-3:}}}$

$\sf\textsf{Find the Median Class}$

$\sf\textsf{Here}\\\sf\textsf{The total Frequency, \textbf{N} = 280}$

$\dfrac{N}{2} = \dfrac{280}{2} = 140$

$\sf\textsf{The cumulative frequency greater than or}\\\sf\textsf{nearest to 140 is 182.}$

$\sf\textsf{The class having this Value (i.e., 182)}\\\sf\textsf{is the Median class for the given data.}$

$\therefore \textbf{Median Class = 10 - 15}$

$\underline{\LARGE{\texttt{Step-4:}}}$

$\sf\textsf{Note the lower limit and frequency of the Median Class}\\\sf\textsf{and cumulative frequency of the class before median class}$

$\sf\textsf{Lower limit of the median class, \textbf{l} = 10}$

$\sf\textsf{Frequency of the median class, \textbf{f} = 133}$

$\sf\textsf{Cumulative frequency of the class before the median class, \textbf{cf} = 49}$

$\underline{\LARGE{\texttt{Step-5:}}}$

$\sf\textsf{Find the median of the given data}$

$\mathbf{Median \approxeq l + \left(\dfrac{\frac{N}{2}-cf}{f}\right) \times h}$

$\sf\textsf{Substitute Values}$

$\implies \mathsf{median \approxeq 10 + \left(\dfrac{\frac{280}{2}-49}{133}\right) \times 5}$

$\implies \mathsf{median \approxeq 10 + \left(\dfrac{140-49}{133}\right)\times 5}$

$\implies \mathsf{median \approxeq 10 + \left(\dfrac{140-49}{133}\right)\times 5}$

$\implies \mathsf{median \approxeq 10 + \left(\dfrac{91\times 5}{133}\right)}$

$\implies \mathsf{median \approxeq 10 + \left(\dfrac{455}{133}\right)}$

$\small{\boxed{\begin{array}{rcl}133)&\: \: 455&(3.421\\&\underline{-399\; \; }&\\&\quad560&\\&\: \; \, \underline{-532\; \;}&\\&\qquad \: \, 280&\\&\qquad \: \underline{-266\; \; \: }\\&\qquad\quad\:\; 140&\\&\qquad\quad \, \underline{-133\; \; }&\\&\qquad\qquad7&\end{array}}}$

$\therefore \mathsf{\dfrac{455}{133} = 3.421}$

$\implies \mathsf{median \approxeq 10 + 3.421}$

$\therefore \mathbf{Median \approxeq 13.421}$


$\blacksquare \: \: \sf\textsf{The median Salary of the data =\Large{\underline{\LARGE{\textbf{13.421}}}}}$