Question

solve using ASSUME MEAN METHOD​

Answer 1

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

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c|c|c}\sf Class\: interval&\sf Frequency\: (f_i)&\sf \: midvalue \: (x_i)&\sf \: u_i&\sf \: f_iu_i\\\frac{\qquad  \qquad}{}&\frac{\qquad  \qquad}{}\\\sf 11 - 13&\sf 3&\sf12&\sf - 3&\sf - 9\\\\\sf 13 - 15 &\sf 6&\sf14&\sf - 2&\sf - 12\\\\\sf 15-17 &\sf 9 &\sf16&\sf - 1&\sf - 9\\\\\sf 17 - 19&\sf 13&\sf18 - A&\sf0&\sf0\\\\\sf 19-21&\sf f&\sf20&\sf1&\sf \: f\\\\\sf 21-23&\sf 5&\sf22&\sf2&\sf 10\\\\\sf 23-25&\sf 4&\sf24&\sf3&\sf12\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Now,

↝ We have,

$\: \: \: \: \: \: \: \: \bull \sf \: A \: = \: 18$

$\: \: \: \: \: \: \: \: \bull \sf \: h \: = \: 2$

$\: \: \: \: \: \: \: \: \bull \sf \: \sum \: f_i \: = \: 40 + f$

$\: \: \: \: \: \: \: \: \bull \sf \: \sum \: f_i \: u_i \: = \: f \: - \: 8$

$\: \: \: \: \: \: \: \: \bull \sf \: \overline{x} \: = \: 18$

Now,

↝ Mean using Step Deviation Method is given by

$\dashrightarrow\sf \: \overline{x} \: = \: A \: + \: \dfrac{ \sum f_i u_i}{ \sum f_i} \: \times \: h$

On substituting all the above values, we get

$\rm :\longmapsto\:\cancel{18} = \cancel{18} + \dfrac{(f - 8)}{(40 + f)} \times 2$

$\rm :\longmapsto\:\dfrac{f - 8}{f + 40} \times 2 = 0$

$\rm :\longmapsto\:f - 8 = 0$

$\bf\implies \:f \: = \: 8$

$\overbrace{ \underline { \boxed { \bf \therefore \: The \: value \: of \:f \: = \: 8}}}$

Additional Information :-

Mean using Direct Method :-

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

Mean using Short Cut Method :-

$\dashrightarrow\sf Mean \: = \: A \: + \: \dfrac{ \sum f_i d_i}{ \sum f_i}$