Question

The following distribution gives the monthly consumption of electricity of 68 thousands of a locality. Find the median. 65-85 --4, 85-105--5, 105-125--13, 125-145--20, 145-165--14, 165-185--8, 185-205--4.​

Answer 1

Step-by-step explanation:

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c} \qquad&\qquad&\qquad \\ \sf Class( Marks)&\sf Frequency(f)&\sf ~ \sf Cumulative ~ \\ && \sf{frequency(cf)}\\ \hline \sf 65-85& \sf4& \sf4\\\hline \sf85-105&\sf5&\sf9 \\\hline \sf105-125&\sf 13&\sf 22 \\ \hline \sf125-145 &\sf 20&\sf 42\\\hline \sf145-165&\sf 14&\sf 56 \\\hline \sf165-185&\sf 7&\sf 63\\\hline \sf185-205&\sf 4&\sf 67\end{array}} \\ { \sf N = \sum}f = 67\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\textsf{We know that,}$

${\bf{Median = l+ \Bigg\{ h × \dfrac{\left(\dfrac{N}{2} - cf\right)}{f} } \Bigg \}}$

$\textsf{Where}$

$\textsf{l = lower limit of median class,}$

$\textsf{n = number of observations,}$

$\textsf{cf = cumulative frequency of class preceding the median class, }$

$\textsf{f = frequency of median class,}$

$\textsf{h = class size (assuming class size to be equal).}$

$\textsf{l= 125 , (n) = 67 , (cf) = 22 , (f) = 20 (h) = 20}$

$\textsf{On substituting the values in Formula we get}$

$\sf{~~~~~:~~\implies Median = 125+ \Bigg\{ 20 × \dfrac{\left(\dfrac{67}{2} - 20\right)}{20} } \Bigg \}$

$\sf{~~~~~:~~\implies Median = 125+ \Bigg\{ \xcancel{20} × \dfrac{33.5 - 22}{ \xcancel{20}} } \Bigg \}$

$\sf{~~~~~:~~\implies Median = 125+ 11.5}$

$\sf{~~~~~:~~\implies Median = 136.5}$

$\bf \underline{Hence, the\:median \:of \:electricity\: consumed \:is\: 136.5.}$