Question

find mean median and mode of following data
class interval 0-10 10-20 20-30 30-40 40-50 frequency 5,7,8,6,4 I need full explanation pls​

Answer 1

☯ Find mean, median and mode of following data:

$\boxed{\begin{array}{c|c|c|c|c|c}\bf Class\: interval&\sf 0-10&\sf 10-20&\sf 20-30 &\sf 30-40&\sf 40-50\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}\\\bf Frequency&\sf 5&\sf 7&\sf 8&\sf 6&\sf 4\end{array}}$

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I. Finding Mean

⠀⠀⠀⠀

$\boxed{\begin{array}{ccccc}\sf Class\: interval&\sf F_i&\sf Mid\:Value&\sf u_i = \dfrac{x_i + 25}{10}&\sf f_i u_i\\\frac{\quad \qquad\qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \qquad\qquad}{}&\frac{ \qquad \qquad \qquad}{}&\frac{\qquad \qquad\qquad}{}\\\sf 0-10&\sf 5&\sf 5&\sf -2&\sf -10\\\\\sf 10-20 &\sf 7&\sf 15&\sf -1 &\sf -7 \\\\\sf 20-30 &\sf 8&\sf A = 25&\sf 0&\sf 0\\\\\sf 30-40&\sf 6&\sf 35&\sf 1 &\sf 6\\\\\sf 40-50 &\sf 4 &\sf 45&\sf 2&\sf 8\\\frac{\qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad}{}&\frac{\qquad\qquad\qquad}{}&\frac{\qquad \qquad\qquad}{}&\frac{\qquad \qquad\qquad}{}\\\sf Total& \sf \sum f = 30&&& \sf -3\end{array}}$

⠀⠀⠀⠀

$\sf We\:have \begin{cases} & \sf{A = \bf{25}} \\ & \sf{ \sum f_i = \bf{30}} \\ & \sf{\sum f_i u_i = \bf{ - 3}} \\ & \sf{h = \bf{10}} \end{cases}$

Now,

$\dag\;{\underline{\frak{Formula\:of\:mean\:is\:given\:by,}}}$

$\star\;{\boxed{\sf{\pink{ \bar{x} = A + \dfrac{ \sum f_i u_i}{\sum f_i} \times h}}}}$

$:\implies\sf 25 + \bigg(\cancel{\dfrac{-3}{30}} \bigg)\times 10\\ \\ \\ :\implies\sf 25 - 1 \times 10\\ \\ \\ :\implies\sf 25 - 10\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{\bar{x} = 24}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Mean\:of\:given\:data\:is\: {\textsf{\textbf{24}}}.}}}$

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II. Finding Median

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$\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency&\sf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0-10&\sf 5&\sf 5 \\\\\sf 10-20 &\sf 7&\sf 12 \\\\\sf 20-30 &\sf 8&\sf 20 \\\\\sf 30-40&\sf 6&\sf 26\\\\\sf 40-50 &\sf 4 &\sf 30\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\ &\sf \sum F_i = 30&\end{array}}$

⠀⠀⠀⠀

$\sf We\:have \begin{cases} & \sf{Lower\:limit,\:l = \bf{20}} \\ & \sf{Comulative\: frequency,\:C.F. = \bf{12}} \\ & \sf{N = \sum f_i = \bf{30}} \\ & \sf{Class\:interval,\:h = 30 - 20 = \bf{10}} \\ & \sf{Frequency,\:f = \bf{8}} \end{cases}$

Now,

$\dag\;{\underline{\frak{Formula\:of\:Median\:is\:given\:by,}}}$

$\star\;{\boxed{\sf{\pink{Median= l + \dfrac{ \frac{N}{2} - C.F.}{f} \times h}}}}$

$:\implies\sf 20 + \dfrac{ \frac{30}{2} - 12}{8} \times 10\\ \\ \\ :\implies\sf 20 - \dfrac{15 - 12}{8} \times 10\\ \\ \\ :\implies\sf 20 + \dfrac{3}{ \cancel{8}} \times \cancel{10}\\ \\ \\ :\implies\sf 20 + \dfrac{3}{4} \times 5\\ \\ \\ :\implies\sf 20 + \dfrac{15}{4}\\ \\ \\ :\implies\sf 20 + 3.75\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{23.75}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Median\:of\:given\:data\:is\: {\textsf{\textbf{23.75}}}.}}}$

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III. Finding Mode

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$\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}\\\sf 0-10&\sf 5\\\\\sf 10-20 &\sf 7\\\\\sf 20-30 &\sf 8\\\\\sf 30-40&\sf 6\\\\\sf 40-50 &\sf 4\end{array}}$

⠀⠀⠀⠀

Here,

• $\sf l = Lower\:limit\:of\;class,\:l = \bf{20}$

• $\sf Class\:interval,\:h = 30 - 20 = \bf{10}$

• $\sf Frequency\:of\:Median\;class\:,f_1 = \bf{8}$

• $\sf Frequency\:of\:preceding\;class\:,f_0 = \bf{7}$

• $\sf Frequency\:of\: succeeding\;class\:,f_2 = \bf{6}$

⠀⠀⠀⠀

$\dag\;{\underline{\frak{Formula\:of\:Mode\:is\:given\:by,}}}$

$\star\;{\boxed{\sf{\pink{Mode = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h}}}}$

$:\implies\sf 20 + \dfrac{8 - 7}{2 \times 8 - 7 - 6} \times 10\\ \\ \\:\implies\sf 20 + \dfrac{1}{16 - 7 - 6} \times 10\\ \\ \\:\implies\sf 20 + \dfrac{1}{16 - 7 - 6} \times 10\\ \\ \\:\implies\sf 20 + \dfrac{1}{9 - 6} \times 10\\ \\ \\:\implies\sf 20 + \dfrac{1}{3} \times 10\\ \\ \\ :\implies\sf 20 + \dfrac{10}{3}\\ \\ \\:\implies\sf 20 + 3.33\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{23.33}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Median\:of\:given\:data\:is\: {\textsf{\textbf{23.33}}}.}}}$

Answer 2

$\Large\bf{\color{cyan}GiVeN\:DaTa,}$

$\boxed{\begin{array}{c|c|c|c|c|c}\bf Class\: interval&\sf 0-10&\sf 10-20&\sf 20-30 &\sf 30-40&\sf 40-50\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}\\\bf Frequency&\sf 5&\sf 7&\sf 8&\sf 6&\sf 4\end{array}}$

$\Large\bf{\color{peru}CaLcUlAtIoN,}$

MEAN ;-

☆ To find the mean let us put the data in the table given below,

$\boxed{\begin{array}{cccc}\sf Class\: interval & \sf F_i & \sf Class\: Mark\: (x_i) & \sf f_i \:x_i\\ \frac{\quad \qquad \qquad}{}&\frac{\qquad \qquad\qquad}{}&\frac{ \qquad \qquad \qquad}{}&\frac{\qquad \qquad\qquad}{}\\\sf 0-10&\sf 5&\sf 5&\sf 25\\\\\sf 10-20 &\sf 7&\sf 15&\sf 105 \\\\\sf 20-30 &\sf 8&\sf25&\sf 200\\\\\sf 30-40&\sf 6&\sf 35&\sf 210\\\\\sf 40-50 &\sf 4 &\sf 45&\sf 180\\ \frac{\qquad \qquad \qquad}{}&\frac{\qquad\qquad\qquad}{}&\frac{\qquad \qquad\qquad}{}&\frac{\qquad \qquad\qquad}{}\\\sf Total& \sf \sum f_i = 30&& \sf 720\end{array}}$

$\bf\red{We\:know\:that}$

$\orange\bigstar\:\:\bf\green{Mean\:=\:\dfrac{\sum{f_i\:x_i}}{\sum{f_i}}\:}$

$:\implies\:\:\bf{Mean\:=\:\dfrac{720}{30}\:}$

$:\implies\:\:\bf\blue{Mean\:=\:24\:}$

$\Large\bold\therefore$ Mean of the given data is 24.

__________________________

MEDIAN ;-

☆ To find the median let us put the data in the table given below,

$\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency&\sf Cumulative\:Frequency\:(C.F)\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0-10&\sf 5&\sf 5 \\\\\sf 10-20 &\sf 7&\sf 12 \\\\\sf 20-30 &\sf 8&\sf 20 \\\\\sf 30-40&\sf 6&\sf 26\\\\\sf 40-50 &\sf 4 &\sf 30\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\ &\sf \sum F_i = 30&\end{array}}$

$\bf\pink{Now,}$

$\longmapsto\:\:\bf{N\:=\:30}$

$\longmapsto\:\:\bf{\dfrac{N}{2}\:=\:\dfrac{30}{2}\:=\:15\:}$

☆ The cumulative frequency just greater than 15, i.e. 20 and the corresponding class is 20-30.

♨ Thus the median class is 20-30.

$\bf\red{We\:know\:that}$

$\green\bigstar\:\:\bf\orange{Median\:=\:I\:+\:\dfrac{\frac{N}{2}\:-\:C.F}{f}\times{h}\:}$

$\bf\blue{Where,}$

• I = Lower limit = 20

• N = 30 & N/2 = 15

• C.F = 12

• f = Frequency = 8

• h = Class interval = 30 - 20 = 10

$:\implies\:\:\bf{Median\:=\:20\:+\:\dfrac{15\:-\:12}{8}\times{10}\:}$

$:\implies\:\:\bf{Median\:=\:20\:+\:\dfrac{3}{4}\times{5}\:}$

$:\implies\:\:\bf{Median\:=\:20\:+\:\dfrac{15}{4}\:}$

$:\implies\:\:\bf{Median\:=\:20\:+\:3.75\:}$

$:\implies\:\:\bf\purple{Median\:=\:23.75\:}$

$\Large\bold\therefore$ Median of the given data is 23.75.

__________________________

MODE ;-

$\bf\red{We\:know\:that}$

$\pink\bigstar\:\:\bf{\color{indigo}Mode\:=\:I\:+\:\dfrac{f_1\:-\:f_o}{2f_1\:-\:f_o\:-\:f_2}\times{h}\:}$

$\bf\purple{Where,}$

• I = 20

• $\bf{f_1\:=\:Max^m\:frequency\:=\:8}$

• $\bf{f_o\:=\:7}$

• $\bf{f_2\:=\:6}$

• h = 10

$:\implies\:\:\bf{Mode\:=\:20\:+\:\dfrac{8\:-\:7}{2\times{8}\:-\:7\:-\:6}\times{10}\:}$

$:\implies\:\:\bf{Mode\:=\:20\:+\:\dfrac{1}{16\:-\:13}\times{10}\:}$

$:\implies\:\:\bf{Mode\:=\:20\:+\:\dfrac{10}{3}\:}$

$:\implies\:\:\bf{Mode\:=\:20\:+\:3.32\:}$

$:\implies\:\:\bf{\color{lime}Mode\:=\:23.32\:}$

Or

$\bf\pink{We\:know\:that}$

$\purple\bigstar\:\:\bf{\color{olive}Mode\:=\:(3\times{Median})\:-\:(2\times{Mean})\:}$

$:\implies\:\:\bf{Mode\:=\:3\times{23.75}\:-\:2\times{24}\:}$

$:\implies\:\:\bf{Mode\:=\:71.25\:-\:48\:}$

$:\implies\:\:\bf{\color{coral}Mode\:=\:23.25\:}$

$\Large\bold\therefore$ Mode of the given data is 23.32.