Question

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Answer 1

$\underline{ \mathfrak{ \: Solution :- \: }}$

$\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency&\sf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0-15&\sf 5&\sf 5 \\\\\sf 15-30 &\sf 20&\sf 25 \\\\\sf 30-45 &\sf 40 &\sf 65 \\\\\sf 45-60&\sf 50&\sf 115 \\\\\sf 60-75&\sf 25 &\sf 140 \\\\\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{\bf{\sum\limits\:f=140}}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\end{array}}$

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$\dashrightarrow \sf N=\sum\limits f$

$\dashrightarrow \sf \dfrac{N}{2} = \dfrac{\sum\limits f}{2}$

$\dashrightarrow \sf \dfrac{N}{2} = \dfrac{140}{2}$

$\dashrightarrow \sf\dfrac{N}{2} = 70$

$\underline{\textsf{Hence, 45 -60 is the median class.}}$

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$\bigstar\:\:\sf Median = l + \sf\dfrac{\frac{n}{2}-C.f.}{f}\times h$

$\sf Here:$

$\sf 1) \:l=Lower\:limit\:of\:median\:class=45$

$\sf 2)\:C.f.=Cumulative\:frequency\:of\:class \: preceeding\:the\:median\:class=65$

$\sf 3)\:f= frequency\:of\:median\:class=50$

$\sf 4)\:h= Class\:interval = 15$

$\underline{ \mathfrak{ \: Putting\:\:the\:\:values:- \: }}$

$\dashrightarrow\sf\:\:Median = l +\dfrac{\frac{n}{2}-C.f.}{f}\times h$

$\dashrightarrow\sf\:\:Median = 45 +\bigg\lgroup\dfrac{70- 65}{50}\times 15\bigg\rgroup$

$\dashrightarrow\sf\:\:Median = 45 +\bigg\lgroup\dfrac{5}{50}\times 15\bigg\rgroup$

$\dashrightarrow\sf\:\:Median = 45 + 1.5$

$\dashrightarrow {\boxed{\frak{\purple{Median = 46.5}}}}$

⠀

$\therefore\:\underline{\textsf{Median Height of the distribution is \textbf{46.5}}}.$

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