Question

please answer this question​

Answer 1

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

We know ,

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

Where

• l = lower limit of median class,
• n = number of observations,
• cf = cumulative frequency of class preceding the median class,
• f = frequency of median class,
• h = class size (assuming class size to be equal).

l= 125 , (n) = 67 , (cf) = 22 , (f) = 20 (h) = 20

• On substituting the values in Formula we get

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

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

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

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

Hance, median of electricity consumed is 136.5.

Answer 2

Answer:

The problems associated with this system, was that the amount of revenue was too high burdening the villagers. e. In reality, no periodic assessment was paid and company officials flouted the rules