Question

Q26. Find the mode of the following frequency distribution.
Class: 0- 10, 10-20,20-30,30-40,40-50,50-60,60-70
Frequency: 8,10,10,16,12,6,7
​

Answer 1

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

$\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ 0 - 10 & 8  \\10 - 20 & 10 \\20 - 30 & 10 \\30 - 40 & 16 \\40 - 50 & 12\\50 - 60 & 6\\60 - 70 & 7 \end{array}\end{gathered}$

We know,

$\boxed{ \boxed{\sf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}$

Where,

$\: \: \: \: \: \: \sf \: \bull \: l \: is \: lower \: limit \: of \: modal \: class$

$\: \: \: \: \: \: \sf \: \bull \:\sf{f_1} \:  is \: frequency \: of \: modal \: class</p><p>$

$\: \: \: \: \: \: \sf \: \bull \:\sf{f_0} \: is \: frequency \: of \: class \: preceding modal \: class$

$\: \: \: \: \: \: \sf \: \bull \:\sf{f_2} \:  is \: frequency \: of \: class \: succeeding \: modal \: class$

$\: \: \: \: \: \: \sf \: \bull \:h \: is \: height \: of \: modal \: class$

Here, Modal class = 30 - 40,

So,

$\rm :\longmapsto\:l \: = \: 30$

$\rm :\longmapsto\:f_0 = 10$

$\rm :\longmapsto\:f_1 = 16$

$\rm :\longmapsto\:f_2 = 12$

$\rm :\longmapsto\:h = 10$

Now,

On substituting all these values in formula of mode,

$\rm :\longmapsto\:{{\bf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}$

$\rm :\longmapsto\:{{\bf{Mode = 30 + \bigg(\dfrac{16 - 10}{2 \times 16 - 10 - 12} \bigg) \times 10 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 30 + \bigg(\dfrac{6}{32 - 22} \bigg) \times 10 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 30 + \bigg(\dfrac{6}{10} \bigg) \times 10 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 30 + 6 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 36 }}}$

Additional Information :-

$\dashrightarrow\sf Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}$

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

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

$\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$