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Unformatted text preview: Second Edition 511 5.29 Let Xi = weight of ith booklet in package. The Xi s are iid with EXi = 1 and VarXi = .052 . 100 100 We want to approximate P i=1 Xi > 100.4 = P i=1 Xi /100 > 1.004 = P (X > 1.004). By the CLT, P (X > 1.004) P (Z > (1.004  1)/(.05/10)) = P (Z > .8) = .2119. 5.30 From the CLT we have, approximately, X1 n(, 2 /n), X2 n(, 2 /n). Since X1 and X2 1  X2 n(0, 2 2 /n). Thus, we want are independent, X .99 P = P P X1 X2 < /5 /5 /  1 5 n/2 < X1 X2 / n/2 < /5 / n/2 n 1 n <Z< , 2 5 2 where Z n(0, 1). Thus we need P (Z n/5( 2)) .005. From Table 1, n/5 2 = 2.576, which implies n = 50(2.576)2 332. 2 5.31 We know that X = 9/100. Use Chebyshev's Inequality to get P 3k/10 < X < 3k/10 1  1/k 2 . We need 1  1/k 2 .9 which implies k 10 = 3.16 and 3k/10 = .9487. Thus P (.9487 < X  < .9487) .9
2 by Chebychev's Inequality. Using the CLT, X is approximately n , X with X =  )/.3 n(0, 1). Thus and (X .09 = .3 .9 = P 1.645 < X < 1.645 .3 = P (.4935 < X  < .4935). Thus, we again see the conservativeness of Chebychev's Inequality, yielding bounds on X  that are almost twice as big as the normal approximation. Moreover, with a sample of size 100, X is probably very close to normally distributed, even if the underlying X distribution is not close to normal. 5.32 a. For any > 0, P Xn  a > = P Xn  a Xn + a > Xn + a = P Xn  a > Xn + a P Xn  a > a 0, as n , since Xn a in probability. Thus Xn a in probability. b. For any > 0, P a 1 Xn = P = P P = P a a Xn 1+ 1 a a a Xn a + 1+ 1 a a a Xn a + 1+ 1+ a Xn  a 1, 1+ a+ a a <a+ 1+ 1 as n , since Xn a in probability. Thus a/Xn 1 in probability. ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

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