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Question Number 50855 by peter frank last updated on 21/Dec/18
Fortherandomvariablexshowthata)Var(x)=E2(x)−[E(x)]2b)Var(ax+b)=a2Var(x)
Answered by peter frank last updated on 21/Dec/18
var(x)=Σf(x−x−)2Nvar(x)=Σf[x2−2xx−+(x−)2]Nvar(x)=Σfx2N−2x−ΣfxN+(x−)2var(x)=Σfx2N−2x−x−+(x−)2var(x)=Σfx2N−2(x−)2+(x−)2var(x)=Σfx2N−(x−)2x−=E(x)var(x)=E(x2)−[E(x)]2shownfromvar(x)=E(x2)−[E(x)]2var(ax+b)=E(ax+b)2−[E(ax+b)]2Σf(a2x2+2abx+b2)N−[a2[E(x)]2−2abE(x)+b2]simplifya2E(x2)−a2[E(x)]2a2var(x)shown
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