elemprob-page27 - Answer. We want P Sn − 3.5 > .05 . n...

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Unformatted text preview: Answer. We want P Sn − 3.5 > .05 . n We rewrite this as P(|Sn − nE X1 | > (.05)(3600)) = P Sn − n E X 1 180 > √√ n Var X1 (60) 35 12 ≈ P(|Z | > 1.756) ≈ .08. Example. Suppose the lifetime of a human has expectation 72 and variance 36. What is the probability that the average of the lifetimes of 100 people exceeds 73? Answer. We want P Sn > 73 = P(Sn > 7300) n 7300 − (100)(72) Sn − n E X 1 √ √ > =P √ √ n Var X1 100 36 ≈ P(Z > 1.667) ≈ .047. The idea behind proving the central limit theorem is the following. It turns out that if mYn (t) → mZ (t) for every t, then P(a ≤ Yn ≤ b) → P(a ≤ Z ≤ b). (We won’t prove this.) We are going to let √ Yn = (Sn − nµ)/σ n. Let Wi = (Xi − µ)/σ . Then E Wi = 0, Var Wi = Var2Xi = 1, the Wi are independent, σ and n Sn − nµ =1 Wi √ = i√ . σn n So there is no loss of generality in assuming that µ = 0 and σ = 1. Then √ mYn (t) = E etYn = E e(t/ n)(Sn ) √ = mSn (t/ n). Since the Xi are i.i.d., all the Xi have the same moment generating function. Since Sn = X1 + · + Xn , then mSn (t) = mX1 (t) · · · mXn (t) = [mX1 (t)]n . If we expand etX1 as a power series, mX1 (t) = E etX1 = 1 + tE X1 + t3 t2 E (X1 )2 + E (X1 )3 + · · · . 2! 3! We put the above together and obtain √ mYn (t) = mSn (t/ n) √ = [mX1 (t/ n)]n √ (t/ n)2 = 1+t·0+ + Rn 2! t2 = 1+ + Rn ]n , 2n 2 where |Rn |/n → 0 as n → ∞. This converges to et /2 n = mZ (t) as n → ∞. 27 ...
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This note was uploaded on 12/29/2011 for the course MATH 317 taught by Professor Wen during the Spring '09 term at SUNY Stony Brook.

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