Question

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Answer 1

Given:-

$\begin{lgathered} \boxed{\boxed{\begin{array}{c|c|c|c|c|c|c} 0-20 & 20-40&40-60&60-80&80-100&100-120&120-140\\4&15&23&12&16&8&2{}\end{array}}} \end{lgathered}$

Find:-

• Mean, median and mode of the given data

Solution:-

$\huge{\underline{\bf{Mean:}}}$

$\begin{gathered}\boxed{\begin{array}{ccccc}\sf Class\: interval&\sf Frequency \: f_{i}&\sf x_{i} & \sf f_i x_i\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{} &\frac{\qquad \qquad \qquad \qquad\qquad}{} \\\sf 0-20&\sf 4&\sf 10 & \sf40 \\\\\sf 20-40 &\sf 15&\sf 30 & \sf450 \\\\\sf 40-60 &\sf 23 &\sf 50 & \sf1150 \\\\\sf 60-80&\sf 12&\sf 70 & \sf840 \\\\\sf 80-100 &\sf 16 &\sf 90 & \sf1440 \\\\\sf 100-120 &\sf 8 &\sf 110 & \sf880 \\\\\sf 120-140 &\sf 2 &\sf 130 & \sf260 \\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{\bf{\sum\limits\:f_{i}=80}}&\frac{\qquad \qquad \qquad \qquad\qquad}{} & \frac{\qquad \qquad \qquad \qquad}{\bf{\sum\limits\:f_{i}x_{i}=5060}} \end{array}} \end{gathered}$

we, know that

$\underline{\boxed{\sf Mean = \dfrac{\Sigma f_i x_i}{\Sigma f_i}}}$

$\sf where \small{ \begin{cases} \sf\Sigma f_i x_i = 5060\\\sf \Sigma f_i = 80 \end{cases}}$

$\bigstar$ Substituting these values:

$\implies\sf Mean = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$

$\implies\sf Mean = \dfrac{5060}{80}$

$\implies\sf Mean = \dfrac{506}{8}$

$\implies\sf Mean =63.5$

$\underline{\boxed{\sf \therefore Mean \:of\:the\: following\:data\:is\:63.5}}$

$\qquad$ ____________________

$\huge{\underline{\bf{Mode:}}}$

we, know that the mode is that the no. which appears the most or the no. which hav higher frequency.

Here, we have

4,5,23,12,16,8,2

Here, all no. are in equal frequency

So, There is No Mode.

$\underline{\boxed{\sf \therefore Mode \:of\:the\: following\:data\:is\:0}}$

$\qquad$ ____________________

$\huge{\underline{\bf{Median:}}}$

we, have

4,5,23,12,16,8,2

Arranging it in ascending order:

2,4,5,8,12,16,23

Here, we have 7 no. which is odd in no.

Now, finding median

$\underline{\boxed{\sf Median = {\bigg \lgroup\dfrac{n + 1}{2} \bigg \rgroup}^{th} }}$

$\sf where \small{ \begin{cases} \sf n = 7 \end{cases}}$

$\bigstar$ Substituting thus value:

$\implies\sf Median = {\bigg \lgroup\dfrac{n + 1}{2} \bigg \rgroup}^{th}$

$\implies\sf Median = {\bigg \lgroup\dfrac{7 + 1}{2} \bigg \rgroup}^{th}$

$\implies\sf Median = {\bigg \lgroup\dfrac{8}{2} \bigg \rgroup}^{th}$

$\implies\sf Median = 4^{th} \: term$

$:\to\sf Median = 4^{th} \: term = 8$

$:\to\sf Median = 8$

$\underline{\boxed{\sf \therefore Median \:of\:the\: following\:data\:is\:8}}$

Answer 2

Yeah I'll follow back.....