Question

5. The median of the observations arranged in the ascending order as 36, 51, 52, x,
x + 4, 68, 82, 96 is 62.
Then, find the value of x.​

Answer 1

Given observation is arranged in ascending order.

》36, 51, 52, x, x + 4, 68, 82, 96.

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$\frak{We\:have} \begin{cases} \sf Median\:of\:data\: = \frak{62} & \\ \\ \sf Number\: of\: observations,\:(n)\: = \frak{8}& \end{cases}$

¤ Formula to find Median for an ( n = even number ) is,

⠀⠀⠀⠀

$\star\:{\underline{\boxed{\frak{\purple{Median = \dfrac{\bigg( \frac{n}{2} \bigg)^{th} + \bigg( \frac{n}{2} + 1 \bigg)^{th}\:observation}{2}}}}}}$

$\bf{\dag}\:{\underline{\frak{Now,\:Putting\: Given\: values\: in\: formula,}}}$

⠀⠀⠀⠀

$:\implies\sf 62 = \dfrac{\bigg( \cancel{\frac{8}{2}} \bigg)^{th} + \bigg( \cancel{\frac{8}{2}} + 1 \bigg)^{th}\:observation}{2}\\\\\\ :\implies\sf 62 = \dfrac{4^{th} + 5^{th}\:observation}{2}\\\\\\ :\implies\sf 62 = \dfrac{(x) + (x + 4)}{2}\\\\\\ :\implies\sf 62 = \dfrac{2x + 4}{2}\\\\\\ :\implies\sf 62 \times 2 = 2x + 4\\\\\\\ :\implies\sf 124 = 2x + 4\\\\\\\ :\implies\sf 124 - 4 = 2x\\\\\\\ :\implies\sf 120 = 2x\\\\\\ :\implies\sf x = \cancel{\dfrac{120}{2}}\\\\\\ :\implies{\underline{\boxed{\frak{\pink{x = 60}}}}}\:\bigstar$

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$\qquad\qquad\qquad\therefore\:{\underline{\sf{The\:value\:of\:x\:is\:{\textsf{\textbf{60}}}.}}}$

Answer 2

$\bf {Given} \begin {cases} \sf{ The \:observations\: arranged \:in \:the \:}\bf{ascending}\sf{\: order\: as\:\: :\:}\\\\ \sf { \qquad \quad\leadsto 36, \:51,\: 52,}\bf{ \:x\:,\: x\: +\:\: 4}\sf{\:,\: 68\:,\: 82\:, \:96\:. \:} \\\\ \sf{The\:Median \:of\:Given \:Data\:is\:}\bf{62} \end {cases}$

Exigency To Find : The Value of x from the given data .

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Given that ,

⠀⠀⠀⠀⠀36, 51, 52, x , x + 4, 68, 82, 96

Here ,

• Number of Observations = 8 [ Even Number ]
• Median of the given data is 62

$\dag\:\:\it{ As,\:We\:know\:that\::}\\\\ \sf \qquad \maltese \:\:Formula \:for\:Median\:for\:an\:\bf even\:number :$

$\qquad \dag\:\:\Bigg\lgroup \sf{ Median \:: \dfrac{\bigg( \dfrac{n}{2}\bigg)^{th}\:\: + \:\:\bigg( \dfrac{n}{2} + 1\bigg)^{th} \:\:Observation }{2}\:\:}\Bigg\rgroup$

⠀⠀Here n is the Number of Observations .

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⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \longmapsto \sf 62 \:= \dfrac{\bigg( \dfrac{8}{2}\bigg)^{th}\:\: + \:\:\bigg( \dfrac{8}{2} + 1\bigg)^{th} \:\:Observation }{2}\:\:$

$\qquad \longmapsto \sf 62 \:= \dfrac{\bigg( \cancel {\dfrac{8}{2}}\bigg)^{th}\:\: + \:\:\bigg( \cancel {\dfrac{8}{2}} + 1\bigg)^{th} \:\:Observation }{2}\:\:$

$\qquad \longmapsto \sf 62 \:= \dfrac{\bigg( 4\bigg)^{th}\:\: + \:\:\bigg( 4+ 1\bigg)^{th} \:\:Observation }{2}\:\:$

$\qquad \longmapsto \sf 62 \:= \dfrac{\bigg( 4\bigg)^{th}\:\: + \:\:\bigg( 5\bigg)^{th} \:\:Observation }{2}\:\:$

⠀⠀⠀⠀Here ,

$\qquad \leadsto \sf 4^{th} \:Observation \: =\:\: x\:$

$\qquad \leadsto \sf 5^{th} \:Observation \: =\:\: x+4\:$

$\qquad \longmapsto \sf 62 \:= \dfrac{ x \:\: + \:x + 4 }{2}\:\:$

$\qquad \longmapsto \sf 62 \:= \dfrac{ \:2x + 4 }{2}\:\:$

$\qquad \longmapsto \sf 62 \times 2 \:= \:2x + 4 \:\:$

$\qquad \longmapsto \sf 124 \:= \:2x + 4 \:\:$

$\qquad \longmapsto \sf 124 - 4 \:= \:2x \:\:$

$\qquad \longmapsto \sf 120 \: = \:2x \:\:$

$\qquad \longmapsto \sf \cancel {\dfrac{120}{2}} \:= \:x \:\:$

$\qquad \longmapsto \sf 60 \:= \:x \:\:$

$\qquad \longmapsto \frak{\underline{\purple{\:x = 60 }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:Value \:of\:x \:is\:\bf{60}}}}$

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