st422%20hw02%20solution

st422%20hw02%20solution - Homework 2: Solutions 7.13: Y F...

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Homework 2: Solutions 7.13: 2 1 2 2 1 22 12 21 (1 , 2 ) /1 /2 1/ ( 2, 1) v v vv YF v v v Y v UY F v v χ χχ = ⎛⎞ ⇒= = = ⎜⎟ ⎝⎠ 7.15: () 2 2 2 (1, ) / / v v v Z UT Fv v v 2 1 2 == = 5 2 1 2 5 2 2 i5 1 1 6 61 2 64 1 5 7.19: (0,1) 1,2. ..5 a) (thm 7.2) 4 b) U= Y (5 1) (thm 7.3) c) Given Y (0,1) Y Thus, S=U+Y i i i i Given Y N i WY YS N 2 5 = = + ∀= = −= = =
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64 6 16 5 6 26 4 22 6 66 7.20: From 7.19, Given Y (0,1), and U a) T = 5 / 5 b) T =2 / 4 c) (0,1/ 5) 5 and Y 2(5 +Y ) (5 +Y ) / 2 (2,4) /4 NW Y YW t W Y YU t U YN Y YY F UU χ χχ 2 2 5 11 = = ⇒= ∼∼ 7.22: 44 . 5 4 a) P( 4.5) 2/ 100 ( 2.5) 0.0062 b) Given =14 and =2 P(| | ) 0.95 | | 0.95 // || 0 . 9 5 / 1.96 / (1.96 2) / 100 0.392 Thus interval that in X XP PZ Xa P nn a n a n a μσ μ σσ σ ⎛⎞ −− >= > ⎜⎟ ⎝⎠ ⇒>= −< = ⇒< = = × = cludes ( ) (14 0.392) (13.608,14.392) =±= ± =
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() 7.30: Given =10 and n=100 (1 1 ) ( | | 1 ) 11 || | | / / 10/ 100 | | 1 1 2 ( 1) 1 (2 0.1587) 0.6826 PY P Y Y PP Z nn PZ σ μμ μ σσ −<<+= −< ⎛⎞ =< = < ⎜⎟ ⎝⎠ = −> = × = 0.01/2 2 7.31: Given =10 P(| | 1) 0.99 (| | 1/( / )) (| | /10) /10 2.576 (25.76) 663.57 664 X n n nZ n −<= ⇒< ⇒= = = 7.34: Given =1.4, =0.05 and n=25 1.3 1.4 P( 1.3) 0.05/ 25 ( 10) 1 ( 10) 0 YP Z μσ ≤≈ =≤ = 100 1 7.42: Given =2.5, =2 and n=100 (2.5,4/100) (2.5,0.04) 2.4 2.5
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st422%20hw02%20solution - Homework 2: Solutions 7.13: Y F...

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