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| 1. |
If the median of the distribution given below is 28.5find the values of x and y |
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| Answer» | Monthly Consumption | \t\t\t0-105510-20x5 + x20-302025 + x30-401540 + x40-50y40 + x + y50-60545 + x + yTotal\xa0{tex}\\sum f _ { i } = n = 60{/tex}\xa0\tHere, {tex}\\sum f _ { i } = n = 60{/tex}, then {tex}\\frac { n } { 2 } = \\frac { 60 } { 2 } = 30{/tex}, also, median of the distribution is 28.5, which lies in interval 20 – 30.{tex}\\therefore{/tex}\xa0Median class = 20 – 30So, l = 20, n = 60, f = 20, cf = 5 + x\xa0and h = 10{tex}\\because 45 + x + y = 60{/tex}{tex}\\Rightarrow x + y = 15{/tex}\xa0………...........(i)Now, Median = {tex}l + \\left[ \\frac { \\frac { n } { 2 } - c f } { f } \\right] \\times h{/tex}{tex}\\Rightarrow { 28.5 = 20 + \\left[ \\frac { 30 - ( 5 + x ) } { 20 } \\right] \\times 10 }{/tex}{tex}\\Rightarrow 28.5 = 20 + \\frac { 30 - 5 - x } { 2 }{/tex}{tex}\\Rightarrow { 28.5 } = \\frac { 40 + 25 - x } { 2 }{/tex}{tex}\\Rightarrow 2 ( 28.5 ) = 65 - x{/tex}{tex}\\Rightarrow 57.0 = 65 - x{/tex}{tex}\\Rightarrow x = 65 - 57 = 8{/tex}{tex}\\Rightarrow{/tex}\xa0x = 8Putting the value of x\xa0in eq. (i), we get,8 + y = 15{tex}\\Rightarrow{/tex}\xa0y\xa0= 7Hence the value of x\xa0and y\xa0are\xa08 and 7 respectively. | |
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