Answer.We wantPSnn-3.5> .05.We rewrite this asP(|Sn-nEX1|>(.05)(3600)) =PSn-nEX1√n√VarX1>180(60)3512≈P(|Z|>1.756)≈.08.Example.Suppose the lifetime of a human has expectation 72 and variance 36. What is the probability thatthe average of the lifetimes of 100 people exceeds 73?Answer.We wantPSnn>73=P(Sn>7300)=PSn-nEX1√n√VarX1>7300-(100)(72)√100√36≈P(Z >1.667)≈.047.The idea behind proving the central limit theorem is the following.It turns out that ifmYn(t)→mZ(t) for everyt, thenP(a≤Yn≤b)→P(a≤Z≤b).(We won’t prove this.)We are going to letYn= (Sn-nμ)/σ√n. LetWi= (Xi-μ)/σ. ThenEWi= 0, VarWi=VarXiσ2= 1, theWiare independent,andSn-nμσ√n=∑ni=1Wi√n.So there is no loss of generality in assuming thatμ= 0 andσ= 1. ThenmYn(t) =EetYn=Ee(t/√n)(Sn)=mSn(t/√
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