Math, asked by giciydutdtudigchvhoh, 5 months ago

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​

Answers

Answered by RoyalChori
0

\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.

Answered by Anonymous
0

Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device trades off input forces against movement to obtain a desired amplification in the output force. The model for this is the law of the lever.

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