Question

The mode for the table is Class: 0-6 6-12 12-18 18-24
24-30 Frequency: 8 10 15 5 6​

Answer 1

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

$\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ 0 - 6 & 8  \\6 - 12 & 10 \\12 - 18 & 15 \\18 - 24 & 5 \\24 - 30 & 6 \end{array}\end{gathered}$

We know,

Formula of Mode,

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

where,

$\: \: \: \: \: \: \sf \: \bull \:l \: is \: lower \: limit$

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

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

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

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

Here,

• Modal class is 12 - 18

Thus,

$\rm :\longmapsto\:l \: = \: 12$

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

$\rm :\longmapsto\:f_1 = 15$

$\rm :\longmapsto\:f_2 = 5$

$\rm :\longmapsto\:h \: = \: 6$

Now,

On substituting the values in formula of mode,

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

$\rm :\longmapsto\:{{\bf{Mode = 12 + \bigg(\dfrac{15 - 10}{2 \times 15 - 10 - 5} \bigg) \times 6 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 12 + \bigg(\dfrac{5}{30 - 15} \bigg) \times 6 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 12 + \bigg(\dfrac{5}{15} \bigg) \times 6 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 12 + 2 }}}$

$\rm :\longmapsto\:{{\bf{Mode = 14}}}$

Additional Information :-

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

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

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

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