# chapter9 - 1 Chapter 9 9–1 Since f x i = 1 σ √ 2 π...

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Unformatted text preview: 1 Chapter 9 9–1. Since f ( x i ) = 1 σ √ 2 π exp •- ( x i- μ ) 2 2 σ 2 ‚ , we have f ( x 1 ,x 2 ,...,x 5 ) = 5 Y i =1 f ( x i ) = 5 Y i =1 1 σ √ 2 π exp •- ( x i- μ ) 2 2 σ 2 ‚ = 1 σ 2 2 π ¶ 5 / 2 exp •- 1 2 σ 2 5 X i =1 ( x i- μ ) 2 ‚ 9–2. Since f ( x i ) = λe- λx i , we have f ( x 1 ,x 2 ,...,x n ) = n Y i =1 f ( x i ) = n Y i =1 λe- λx i = λ n exp •- λ n X i =1 x i ‚ 9–3. Since f ( x i ) = 1, we have f ( x 1 ,x 2 ,x 3 ,x 4 ) = 4 Y i =1 f ( x i ) = 1 2 9–4. The joint probability function for X 1 and X 2 is p X 1 ,X 2 (0 , 0) = N- M ¶ M 2 ¶ N 2 ¶ p X 1 ,X 2 (0 , 1) = N- M 1 ¶ M 1 ¶ 2 N 2 ¶ p X 1 ,X 2 (1 , 0) = N- M 1 ¶ M 1 ¶ 2 N 2 ¶ p X 1 ,X 2 (1 , 1) = N- M 2 ¶ M ¶ N 2 ¶ Of course, p X 1 ( x 1 ) = 1 X x 2 =0 p X 1 ,X 2 ( x 1 ,x 2 ) and p X 2 ( x 2 ) = 1 X x 1 =0 p X 1 ,X 2 ( x 1 ,x 2 ) So p X 1 (0) = M/N , p X 1 (1) = 1- ( M/N ), p X 2 (0) = M/N , p X 2 (1) = 1- ( M/N ). Thus, X 1 and X 2 are not independent since p X 1 ,X 2 (0 , 0) 6 = p X 1 (0) p X 2 (0) 9–5. N ( μ,σ 2 /n ) = N (5 , . 00125) 9–6. σ/ √ n = 0 . 1 / √ 8 = 0 . 0353 3 9–7. Use estimated standard error S/ √ n . 9–8. N (- 5 , . 22) 9–9. The standard error of ¯ X 1- ¯ X 2 is s σ 2 1 n 1 + σ 2 2 n 2 = r (1 . 5) 2 25 + (2 . 0) 2 30 = 0 . 473 9–10. Y = ¯ X 1- ¯ X 2 is a linear combination of the 55 variables...
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## This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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chapter9 - 1 Chapter 9 9–1 Since f x i = 1 σ √ 2 π...

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