find the mode of the frequency distribution table given below class interval 0-10, 10-20, 20-30, 30-40, 40-50 frequency 7, 9, 15, 11, 8 About the author Adeline
Given data is [tex]\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ 0 – 10 & 7 \\10 – 20 & 9 \\20 – 30 & 15 \\30 – 40 & 11 \\40 – 50 & 8 \end{array}\end{gathered}[/tex] We know, Mode is given by [tex]\boxed{ \boxed{\sf{Mode = l + \bigg(\dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \bigg) \times h }}}[/tex] Where, [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: l \: is \: lower \: limit \: of \: modal \: class[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: \sf{f_0} \: is \: frequency \: of \: class \: preceding \: modal \: class[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: \sf{f_1} \: is \: frequency \: of \: modal \: class[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: \sf{f_2} \: is \: frequency \: of \: class \: succeeding \: modal \: class[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: h \: is \: height \: of \: modal \: class[/tex] Now, Here, Modal class = 20 – 30 [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: l = 20[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: h = 10[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: f_0 = 9[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: f_1 = 15[/tex] [tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: f_2 = 11[/tex] On substituting all these values in above formula, [tex]\rm :\longmapsto\:{{\bf{Mode = l + \bigg(\dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \bigg) \times h }}}[/tex] [tex]\rm :\longmapsto\:{{\bf{Mode = 20+ \bigg(\dfrac{15 – 9}{2 \times 15 – 9 – 11} \bigg) \times 10 }}}[/tex] [tex]\rm :\longmapsto\:{{\bf{Mode = 20+ \bigg(\dfrac{6}{30 -20} \bigg) \times 10 }}}[/tex] [tex]\rm :\longmapsto\:{{\bf{Mode = 20+ \bigg(\dfrac{6}{10} \bigg) \times 10 }}}[/tex] [tex]\rm :\longmapsto\:{{\bf{Mode = 20+ 6 }}}[/tex] [tex]\rm :\longmapsto\:{{\bf{Mode = 26 }}}[/tex] Additional Information :- 1. Mean using Direct Method :- [tex]\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}[/tex] 2. Mean using Short Cut Method :- [tex]\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i}[/tex] 3. Mean using Step Deviation Method [tex]\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h[/tex] 4. Median :- [tex]\dashrightarrow\sf Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} – cf \bigg)}{f} \Bigg \}[/tex] Reply
Given data is
[tex]\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ 0 – 10 & 7 \\10 – 20 & 9 \\20 – 30 & 15 \\30 – 40 & 11 \\40 – 50 & 8 \end{array}\end{gathered}[/tex]
We know,
Mode is given by
[tex]\boxed{ \boxed{\sf{Mode = l + \bigg(\dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \bigg) \times h }}}[/tex]
Where,
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: l \: is \: lower \: limit \: of \: modal \: class[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: \sf{f_0} \: is \: frequency \: of \: class \: preceding \: modal \: class[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: \sf{f_1} \: is \: frequency \: of \: modal \: class[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: \sf{f_2} \: is \: frequency \: of \: class \: succeeding \: modal \: class[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: h \: is \: height \: of \: modal \: class[/tex]
Now,
Here,
Modal class = 20 – 30
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: l = 20[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: h = 10[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: f_0 = 9[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: f_1 = 15[/tex]
[tex] \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \sf \: f_2 = 11[/tex]
On substituting all these values in above formula,
[tex]\rm :\longmapsto\:{{\bf{Mode = l + \bigg(\dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \bigg) \times h }}}[/tex]
[tex]\rm :\longmapsto\:{{\bf{Mode = 20+ \bigg(\dfrac{15 – 9}{2 \times 15 – 9 – 11} \bigg) \times 10 }}}[/tex]
[tex]\rm :\longmapsto\:{{\bf{Mode = 20+ \bigg(\dfrac{6}{30 -20} \bigg) \times 10 }}}[/tex]
[tex]\rm :\longmapsto\:{{\bf{Mode = 20+ \bigg(\dfrac{6}{10} \bigg) \times 10 }}}[/tex]
[tex]\rm :\longmapsto\:{{\bf{Mode = 20+ 6 }}}[/tex]
[tex]\rm :\longmapsto\:{{\bf{Mode = 26 }}}[/tex]
Additional Information :-
1. Mean using Direct Method :-
[tex]\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}[/tex]
2. Mean using Short Cut Method :-
[tex]\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i}[/tex]
3. Mean using Step Deviation Method
[tex]\dashrightarrow\sf Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h[/tex]
4. Median :-
[tex]\dashrightarrow\sf Median= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} – cf \bigg)}{f} \Bigg \}[/tex]