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If b_i=1-a_i na = Sigma_(i=1)^(n)a_i, nb = Sigma_(i=1)^(n) b_i " then " Sigma_(i=1)^(n) a_b_i+Sigma_(i=1)^(n)(a_i-a)^2= |
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Answer» ab `=NA-Sigmaa_(i)^(2)` `=na-Sigma(a_(i)-a+a)^(2)` `=na-Sigma[(a_(i)-a)^(2)+a^(2)+2A(a_(i)-a)]` `=na-Sigma(a_(i)-a)^(2)-Sigmaa^(2)-2aSigma(a_(i)-a)` `rArrSigmaa_(i)b_(i)+Sigma(a_(i)-a)^(2)=na-na^(2)-2a(na-na)` `=na(1-a)=nab` `[{:(becauseSigmab_(i)=Sigma1-Sigmaa_(i)),(thereforenb=n-na),(or a+b=1):}]` |
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