Question

Please answer this question.

Answer 1

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

The frequency distribution table is as follow :

$\begin{gathered}\begin{gathered}\begin{gathered} \boxed{\begin{array}{|c|c|c|c|c|}\bf Class&\bf Frequency (f_i)&\bf class \: mark (x_i) & \bf d_i = (x_i - A)& \bf f_id_i\\\sf 0-50&\sf 4&\sf25 & \sf -100 & \sf -400 \\ \sf 50-100&\sf 8&\sf 75& \sf -50 & \sf -400\\\sf 100-150&\sf 16 &\sf 125 & \sf 0 & \sf 0\\\sf 150-200&\sf 13& \sf 175 & \sf 50 & \sf 650\\\sf 200 - 250&\sf 6&\sf 225 & \sf 100 & \sf 60\\\sf 250 - 300&\sf 3&\sf 275 & \sf 150 & \sf 450 \\\sf &\sf \sum f_i= 50& & & \sf \sum f_id_i = 900\\\end{array}}\end{gathered} \end{gathered}\end{gathered}$

So, we have following data now

$\rm :\longmapsto\:A = 125$

$\rm :\longmapsto\: \sum \: f_i \: = \: 50$

$\rm :\longmapsto\: \sum \: f_id_i \: = \: 900$

$\rm :\longmapsto\:h \: = \: 50$

We know,

Mean using Short Cut Method is given by

$\red{\rm :\longmapsto\: \boxed{ \sf{ \:Mean \: = \: A \: + \: \frac{ \sum \: f_i \: d_i}{ \sum \: f_i}}}}$

On substituting the values, we get

$\rm :\longmapsto\: \:Mean \: = \: 125 \: + \: \dfrac{ 900}{ 50}$

$\rm :\longmapsto\: \:Mean \: = \: 125 \: + \: \dfrac{ 90}{ 5}$

$\rm :\longmapsto\: \:Mean \: = \: 125 \: + \: 18$

$\red{\rm :\longmapsto\: \boxed{ \tt{ \: \:Mean \: = \: 143\: }}}$

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Additional Information :-

1. Mean using Direct Method :-

$\red{ \boxed{ \sf{ \:Mean \: = \: \: \frac{ \sum \: f_i \: x_i}{ \sum \: f_i}}}}$

2. Mean using Step Deviation Method :-

$\red{ \boxed{ \sf{ \:Mean \: = \: A \: + \: \frac{ \sum \: f_i \: u_i}{ \sum \: f_i} \: \times \: h \: }}}$