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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.
 Spring '09
 wen
 Probability, Variance

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