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Dr. Hackney STA Solutions pg 81

# Dr. Hackney STA Solutions pg 81 - Second Edition 5-15 is in...

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Second Edition 5-15 is in the form n ( ( ¯ U - E U ) ) with n = 12, so X is approximately n(0 , 1) by the Central Limit Theorem. b. The approximation does not have the same range as Z n(0 , 1) where -∞ < Z < + , since - 6 < X < 6. c. E X = E 12 i =1 U i - 6 = 12 i =1 E U i - 6 = 12 i =1 1 2 - 6 = 6 - 6 = 0 . Var X = Var 12 i =1 U i - 6 = Var 12 i =1 U i = 12Var U 1 = 1 E X 3 = 0 since X is symmetric about 0. (In fact, all odd moments of X are 0.) Thus, the first three moments of X all agree with the first three moments of a n(0 , 1). The fourth moment is not easy to get, one way to do it is to get the mgf of X . Since E e tU = ( e t - 1) /t , E e t (∑ 12 i =1 U i - 6 ) = e - 6 t e t - 1 t 12 = e t/ 2 - e - t/ 2 t 12 . Computing the fourth derivative and evaluating it at t = 0 gives us E X 4 . This is a lengthy
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