# HW1 - N ( , 2 ) . By theorem 7.3, ( n-1) S 2 2 2 n-1 ....

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4.129 Find y 0 satisfying P ( Y < y 0 ) = 0 . 9 . P ( Y < y 0 ) = 0 . 9 P ( Y y 0 ) = 0 . 1 . Since Y N (70 , 12 2 ) , Z = Y - 70 12 N (0 , 1) . 0 . 1 = P ( Z > 1 . 28) = P ( Y > 85 . 36) by Normal table. 4.151 a Since 0 a 1 , 1. f ( y ) = af 1 ( y ) + 1 - af 2 ( y ) 0 2. R f ( y ) dy = R ( af 1 ( y ) + (1 - a ) f 2 ( y )) dy = R af 1 ( y ) dy + R 1 - af 2 ( y ) dy = a + 1 - a = 1 b Let f ( y ) = af 1 ( y ) + 1 - af 2 ( y ) 1. E ( Y ) = R yf ( y ) = R ayf 1 ( y ) dy + R 1 - ayf 2 ( y ) dy = 1 + (1 - a ) μ 2 2. V ar ( Y ) = E ( Y 2 ) - E ( Y ) 2 = a ( μ 2 1 + σ 2 1 ) + (1 - a )( μ 2 2 + σ 2 2 ) - [ 1 - (1 - a ) μ 2 ] 2 = 2 1 + (1 - a ) σ 2 2 + a (1 - a )[ μ 1 + μ 2 ] 2 E ( Y 2 ) = R y 2 f ( y ) dy = R ay 2 f 1 ( y ) dy + R 1 - ay 2 f 2 ( y ) dy = a ( μ 2 1 + σ 2 1 ) + (1 - a )( μ 2 2 + σ 2 2 ) V ( Y 1 ) = E ( Y 2 1 ) - E ( Y 1 ) 5.133 E ( ¯ Y - ¯ X ) = μ 1 - μ 2 V ( ¯ Y - ¯ X ) = V ( ¯ Y ) + V ( ¯ X ) = 2 1 n 2 + 2 2 n 2 = σ 2 1 n + σ 2 2 n since Y,X are independent. 7.10 a Let U χ 2 . By Deﬁnition 4.9 (p 178), χ ν = gamma ( ν/ 2 , 2) . E ( U ) = ν/ 2 × 2 = ν, Var ( U ) = ν/ 2 × 2 2 = 2 ν by theorem 4.8 b Let Y 1 ,...Y n
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Unformatted text preview: N ( , 2 ) . By theorem 7.3, ( n-1) S 2 2 2 n-1 . Therefore E ( n-1) S 2 2 = n-1 E ( S 2 ) = 2 V ar ( n-1) S 2 2 = 2( n-1) , V ar ( S 2 ) = 2 2 n-1 by a. 7.42 X i : the time to process the ith persons order P ( X i > 240) = P ( X > 240 100 ) = P Z > 2 . 4-2 . 5 2 / 100 = 0 . 6915 1...
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## This note was uploaded on 05/17/2008 for the course MATH 415 taught by Professor Yuzhang during the Spring '08 term at Pennsylvania State University, University Park.

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