hw3sol_fall03 - (f 9 6;9,0.7 k b k = = 1 B(5 9 0.7 =...

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AMS 310.01 Fall 2003 Homework 3 Solution 3.74 P( from agency F | bad tires ) = P (from agency F and bad tires) / P( bad tires ) = (0.60) * (0.04) / 0.068 = 0.353 << in 3.73 , P(car had bad tires) = (0.20)(0.10)+(0.20)(0.12) +(0.60)(0.04) = 0.068 >> 4.4 (a) Yes. Since , 1 ) ( 0 i f and = = 5 0 1 ) ( i i f (b) No. f (3) = -- 4 / 6 < 0 (c) Yes. Since , 1 ) ( 0 i f and = = 6 3 1 ) ( i i f (d) No. = < = 5 1 1 25 / 20 ) ( i i f 4.6 = + + + = - = - = - - = = = x o i x x x i x i k k i f x F ) ) 2 1 ( 1 ( 31 32 2 1 ) ) 2 1 ( 1 ( 31 16 1 2 1 1 ) 2 1 ( 2 ) ( ) ( 1 1 1 0 << in 4.5 Using the identity 1 0 ( 1) 1 n i n i x x x + = - = - or 1 0 1 , 1 n n i i x x x + = - = - we have 4 1 4 0 1 ( ) 1 31 2 1 2 16 1 2 x x k k k + = - = = - This must equal 1 , so k = 16/31 >> 4.10. From Table 1: (a) B(8; 16, 0.4) = 0.8577. (b) b(8 ; 16, 0.4) = (0.8577)-(0.7161) = 0.1416 (c) B(9 ; 12, 0.6) = 0.9166 (d) b(9; 12, 0.6) = (0.9166) –(0.7747) = 0.1419 (e) 20 6 ( ;20,0.15) k b k = = 1 – B(5; 20, 0.15) = 1-(0.9327) = 0.0673.
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Unformatted text preview: (f) 9 6 ( ;9,0.7) k b k = = 1 - B(5; 9, 0.7) = 1-(0.2703) = 0.7297 4.16 (a) b(1;12, 0.05) = B(1;12, 0.05) – B(0;12,0.05) = (0.8816)-(0.5404) = 0.3412 (b) B(2; 12, 0.05) = 0.9804 (c) 1- B(1;12, 0.05) = 1-(0.8816)=0.1184 4.18 (a) The probability that 1 or more components in a sample of 15 is defective when the true probability of being good is 0.95 , is 1 - b(0;15, 0.05) = 1- 15 (0.95) = 0.5367 (b) The probability that 0 are defective when the true probability is 0.90 is b(0; 15, 0.10) = 0.2059 (c) When the true probability is 0.90, we have b(0; 15, 0.20)= 0.0352...
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