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If mu the mean of distribution (y_(i),f_(i)), then sum f_(i)(y_(i)-mu)= |
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Answer» M.D. `sumf_(i)(y_(i)-MU)=SUM f_(i)y_(i)-mu sum f_(i)=mu sum f_(i)-mu sum f_(i)=0 "" [because mu=(sum f_(i)y_(i))/(sum f_(i))]` |
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