InterviewSolution
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InterviewSolution
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The mean and variance of 7 observations are 8 and 16,respectively. If five of the observations are 2, 4, 10, 12, 14. Find theremaining two observations. |
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Answer» Let remaining tw observations are `x_(1)` and `x_(2)`. Here `n=7` Give that `barx=8implies(sumx_(i))/n=8` `implies x_(1)+x_(2)+2+4+10+12+14=8xx7` `implies x_(1)+x_(2)=14`…………1 and variance `=16` `implies (sumx_(i)^(2))/n-((sumx)/n)^(2)=16implies(sumx_(i)^(2))/7=16+(8)^(2)=80` `implies sumx_(i)^(2)=560` `implies x_(1)^(2)+x_(2)^(2)+2^(2)+4^(2)+10^(2)+12^(2)+14^(2)=560` `=x_(1)^(2)+14-x_(1)^(2)+4+16+100144+196=560` `implies x_(1)^(2)+196-28x_(1)+x_(1)^(2)+460-560=0` `implies 2x_(1)^(2)-28x_(1)+96=0` `=x_(1)^(2)-14x_(1)+48=0` `implies(x_(1)-6)(x_(1)-8)=0` `implies x_(1)-6=0` or `x_(1)-8=0` `implies x_(1)=6` or `x_(1)=8` If `x_(1)=6` then `x_(2)=14-6=8` `x_(1)=8` then `x_(2)=14-8=6` `:.` Remaining two observations `=6,8`. |
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