sec_3_3_sol

# sec_3_3_sol - x 2 = 1 n Â n n 1(2 n 1 6 = n 1(2 n 1 6...

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APPM 4/5570 Solutions to Problems from Section 3.3 29. a. E [ X ] = (0)(0 . 08) + (1)(0 . 15) + (2)(0 . 45) + (3)(0 . 27) + (4)(0 . 05) = 2 . 06 b. σ 2 = V ( X ) = E [( X - μ ) 2 ] = E [( X - 2 . 06) 2 ] = (0 - 2 . 06) 2 (0 . 08) + (1 - 2 . 06) 2 (0 . 15) + (2 - 2 . 06) 2 (0 . 45) +(3 - 2 . 06) 2 (0 . 27) + (4 - 2 . 06) 2 (0 . 05) = 0 . 9364 c. σ = σ 2 = 0 . 9364 0 . 9677 d. V ( X ) = E [ X 2 ] - ( E [ X ]) 2 E [ X 2 ] = (0) 2 (0 . 08) + (1) 2 (0 . 15) + (2) 2 (0 . 45) + (3) 2 (0 . 27) + (4) 2 (0 . 05) = 5 . 18 So, V ( X ) = E [ X 2 ] - ( E [ X ]) 2 = 5 . 18 - (2 . 06) 2 = 0 . 9364 30. a. E [ Y ] = (0)(0 . 60) + (1)(0 . 25) + (2)(0 . 10) + (3)(0 . 05) = 0 . 60 b. E [100 Y 2 ] = 100 E [ Y 2 ] = 100 [(0) 2 (0 . 60) + (1) 2 (0 . 25) + (2) 2 (0 . 10) + (3) 2 (0 . 05)] = 110 31. a. E [ X ] = (13 . 5)(0 . 2) + (15 . 9)(0 . 5) + (19 . 1)(0 . 3) = 16 . 38 E [ X 2 ] = (13 . 5) 2 (0 . 2) + (15 . 9) 2 (0 . 5) + (19 . 1) 2 (0 . 3) = 272 . 298 V ( X ) = E [ X 2 ] - ( E [ X ]) 2 = 272 . 298 - (16 . 38) 2 = 3 . 9936

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b. E [25 X - 8 . 5] = 25 E [ X ] - 8 . 5 = 25(16 . 38) - 8 . 5 = 401 c. V [25 X - 8 . 5] = V [25 X ] = 25 2 · V [ X ] = (625)(3 . 9936) = 2496 d. E [ h ( X )] = E [ X - 0 . 01 X 2 ] = E [ X ] - 0 . 01 E [ X 2 ] = 16 . 38 - 0 . 01(271 . 298) 13 . 668 33. The pdf is x 0 1 P ( X = x ) 1 - p p a. E [ X 2 ] = (0) 2 (1 - p ) + (1) 2 ( p ) = p b. E [ X ] = (0)(1 - p ) + (1)( p ) = p , so V ar ( X ) = E [ X 2 ] - ( E [ X ]) 2 = p - p 2 = p (1 - p ) . c. E [ X 79 ] = (0) 79 (1 - p ) + (1) 79 ( p ) = p 37. E [ X ] = n X x =1 x · P ( X = x ) = n X x =1 x · 1 n = 1 n · n X x =1 x = 1 n · n ( n + 1) 2 = n + 1 2 E [ X 2 ] = n X x =1 x 2 · P ( X = x ) = n X x =1 x 2 = 1 n · n X x =1
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Unformatted text preview: x 2 = 1 n Â· n ( n + 1)(2 n + 1) 6 = ( n + 1)(2 n + 1) 6 Therefore, V ar ( X ) = E [ X 2 ]-( E [ X ]) 2 = ( n + 1)(2 n + 1) 6-Â± n + 1 2 Â² 2 = 1 12 ( n 2-1) 39. E [ X ] = (1)(0 . 2) + (2)(0 . 4) + (3)(0 . 3) + (4)(0 . 1) = 2 . 3 E [ X 2 ] = (1) 2 (0 . 2) + (2) 2 (0 . 4) + (3) 2 (0 . 3) + (4) 2 (0 . 1) = 6 . 1 So, V ar ( X ) = E [ X 2 ]-( E [ X ]) 2 = 6 . 1-2 . 3 2 = 0 . 81 The number of pounds left after the next customerâ€™s order is shipped is 100-5 X . So, E [100-5 X ] = 100-5 E [ X ] = 100-5(2 . 3) = 88 . 5 and V ar (100-5 X ) = V ar (-5 X ) = (-5) 2 V ar ( X ) = 25 V ar ( X ) = 25(0 . 81) = 20 . 25 ....
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## This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.

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sec_3_3_sol - x 2 = 1 n Â n n 1(2 n 1 6 = n 1(2 n 1 6...

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