InterviewSolution
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InterviewSolution
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Let `barx` be the mean of `x_(1), x_(2), ………, x_(n)` and `bary` be the mean of `y_(1), y_(2),……….,y_(n)`. If `barz` is the mean of `x_(1), x_(2), ……………..x_(n), y_(1), y_(2), …………,y_(n)`, then `barz` is equal toA. `bar(x)+bar(y)`B. `(bar(x)+bar(y))/(2)`C. `(bar(x)+bar(y))/(n)`D. `(bar(x)+bar(y))/(2n)` |
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Answer» We have `bar(x)` is the mean of `x_(1), x_(2),…., x_(n)` and `bar(y)` is the mean of `y_(1), y_(2),…. y_(n)` ` "So, " bar(x)=((x_(1)+ x_(2)+x_(3)+…+ x_(n)))/(n)` `rArr x_(1)+ x_(2)+x_(3)+…+ x_(n)=nbar(x)` `"and " bar(y)=((y_(1)+ y_(2)+y_(3)+…+ y_(n)))/(n)` `rArr y_(1)+ y_(2)+y_(3)+…+ y_(n)=nbar(y)` If `bar(z)` is the mean of `x_(1), x_(2),…., x_(n),y_(1), y_(2),…., y_(n),` then , `bar(z) = (nbar(x)+nbar(y))/(n+n) = (n(bar(x)+bar(y)))/(2n) = ((bar(x)+bar(y)))/(2)` Hence, (b) is the correct answer. |
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