Question

★ A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained :-
$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c} \tt{Height\ (in\ cm)} & \tt{Number\ of\ girls} \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 140} &\dfrac{\qquad\qquad}{ \sf 4} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 145} &\dfrac{\qquad\qquad}{ \sf 11} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 150} &\dfrac{\qquad\qquad}{ \sf 29} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 155} &\dfrac{\qquad\qquad}{ \sf 40} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 160} &\dfrac{\qquad\qquad}{ \sf 46} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 165} &\dfrac{\qquad\qquad}{ \sf 51} &\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$
Find the median height.​

Answer 1

Explanation:

Answer:

Question :

⠀⠀⠀━━A survey regarding the heights (in cm) of 51 girls of Class X of a school was ⠀⠀⠀⠀⠀⠀⠀conducted and the following data was obtained :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c} \bf{Height\ (in\ cm)} & \bf{Number\ of\ girls} \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 140} &\dfrac{\qquad\qquad}{ \sf 4} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 145} &\dfrac{\qquad\qquad}{ \sf 11} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 150} &\dfrac{\qquad\qquad}{ \sf 29} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 155} &\dfrac{\qquad\qquad}{ \sf 40} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 160} &\dfrac{\qquad\qquad}{ \sf 46} & \\ \dfrac{\qquad\qquad}{ \sf Less\ than\ 165} &\dfrac{\qquad\qquad}{ \sf 51} &\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

⠀⠀⠀⠀⠀Find the median of Height .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀❍ Finding Median of Height :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small \boxed{\begin{array}{c |c} \bf{Height\ (in\ cm)} & \bf{Number\ of\ girls} \\ \dfrac{\qquad\qquad}{ \sf 0 \:- \:140 } &\dfrac{\qquad\qquad}{ \sf 4} & \\ \dfrac{\qquad\qquad}{ \sf 140\ -\ 145} &\dfrac{\qquad\qquad}{ \sf 7} & \\ \dfrac{\qquad\qquad}{ \sf 145 \ -\ 150} &\dfrac{\qquad\qquad}{ \sf 18} & \\ \dfrac{\qquad\qquad}{ \sf 150 \ - \ 155} &\dfrac{\qquad\qquad}{ \sf 11} & \\ \dfrac{\qquad\qquad}{ \sf 155 \ -\ 160} &\dfrac{\qquad\qquad}{ \sf 6} & \\ \dfrac{\qquad\qquad}{ \sf 160 \ - \ 165} &\dfrac{\qquad\qquad}{ \sf 5} &\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

⠀⠀⠀⠀⠀━━━ Now Finding Cumulative Frequency [ CF ] :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c|c} \bf{Height\ (in\ cm)} & \bf{Number\ of\ girls\:(f_{i} )\:} & \bf{Cumulative \ Frequency \ ( cf) } \\ \dfrac{\qquad\qquad}{ \sf 0 \:- \:140 } & \dfrac{\qquad\qquad}{ \sf 4} & \dfrac{\qquad\qquad}{ \sf 4} & \\ \dfrac{\qquad\qquad}{ \sf 140\ -\ 145} &\dfrac{\qquad\qquad}{ \sf 7} & \dfrac{\qquad\qquad}{ \sf 11} &\\ \dfrac{\qquad\qquad}{ \bf 145 \ -\ 150} &\dfrac{\qquad\qquad}{ \bf 18} & \dfrac{\qquad\qquad}{ \bf 29} & \\ \dfrac{\qquad\qquad}{ \sf 150 \ - \ 155} &\dfrac{\qquad\qquad}{ \sf 11} &\dfrac{\qquad\qquad}{ \sf 40} & \\ \dfrac{\qquad\qquad}{ \sf 155 \ -\ 160} &\dfrac{\qquad\qquad}{ \sf 6} &\dfrac{\qquad\qquad}{ \sf 46} & \\ \dfrac{\qquad\qquad}{ \sf 160 \ - \ 165} &\dfrac{\qquad\qquad}{ \sf 5} & \dfrac{\qquad\qquad}{ \sf 51} & \\ \dfrac{\qquad\qquad}{ \bf Total\::}& \dfrac{\qquad\qquad}{ \bf \Sigma \:f_{i}\: [ \ n \ ] \:\sf =\:51}&\dfrac{\qquad\qquad}{ \qquad\qquad}\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

[ Note : The letters in bold are median class . ]

$\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\ \\ \qquad\qquad \maltese \: \bf Formula\:for\:Median\:: \\\\ \end{gathered}$

$\begin{gathered}\qquad \dag\:\:\Bigg\lgroup \sf{Median \:: l +\Bigg( \dfrac{ \dfrac{n}{2} - cf }{f}\Bigg) \times h }\Bigg\rgroup \\\\\end{gathered}$

Here ,

n is the total of all Frequency = 51

l is the lower - limit of the Median class = 145

h is the class - interval = 145 - 140 = 5

cf is the Cumulative Frequency of the class before median class = 11

f is the Frequency of the median class = 18

⠀⠀⠀⠀⠀⠀$\begin{gathered}\underline$ ${\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Median \:= \: l +\Bigg( \dfrac{ \dfrac{n}{2} - cf }{f}\Bigg) \times h \:\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Median \:= \: 145 +\Bigg( \dfrac{ \dfrac{51}{2} - 11 }{18} \Bigg) \times 5 \:\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Median \:= \: 145 + \dfrac{ 25.5 - 11 }{18} \times 5 \:\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Median \:= \: 145 + \dfrac{ 14.5 }{18} \times 5 \:\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Median \:= \: 145 + 4.03 \:\\\end{gathered}$

$➠\begin{gathered}\qquad:\implies \sf Median \:= \: 149.03\:\\\end{gathered}$

$\begin{gathered}\qquad :\implies \frak{\underline{\purple{\: Median \:= \: 149.03 }} }\:\:\bigstar \\\end{gathered}$

$\begin{gathered}\therefore {\underline{ \mathrm {\:Median \:of\:Height \:is\:\bf{149.03}}.}}\\\end{gathered}$

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Answer 2

Answer:

hope it helps you

Refer to attachment