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# ch03a - Chapter 3 Section 3-2 3-1 Continuous 3-2 Discrete...

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Chapter 3 Section 3-2 3-1. Continuous 3-2. Discrete 3-3. Continuous 3-4. Discrete 3-5. Discrete 3-6. Continuous 3-7. Discrete Section 3-3 3-8. a) Engineers with at least 36 months of full-time employment. b) Samples of cement blocks with compressive strength of at least 6000 kg per square centimeter. c) Measurements of the diameter of forged pistons that conform to engineering specifications. d) Cholesterol levels at most 180 or at least 220. 3-9. The intersection of A and B is empty, therefore P(X A B) = 0. a) Yes, since P(X A B) = 0. b) P(X A ) = 1 - P(X A) = 1 – 0.4 = 0.6 c) P(X B ) = 1 - P(X B) = 1 – 0.6 = 0.4 d) P(X A B) = P(X A) + P(X B) - P(X A B) = 0.4 + 0.6 – 0 = 1 3-10. a) P(X A ) = 1 - P(X A) = 1 – 0.3 = 0.7 b) P(X B ) = 1 - P(X B) = 1 – 0.25 = 0.75 c) P(X C ) = 1 - P(X C) = 1 – 0.6 = 0.4 d) A and B are mutually exclusive if P(X A B) = 0. To determine if A and B are mutually exclusive, solve the following for P(X A B): P(X A B) = P(X A) + P(X B) - P(X A B) 0.55 = 0.3 + 0.25 - P(X A B) 0.55 = 0.55 - P(X A B) and P(X A B) = 0. Therefore, A and B are mutually exclusive. e) B and C are mutually exclusive if P(X B C) = 0. To determine if B and C are mutually exclusive, solve the following for P(X B C): P(X B C) = P(X B) + P(X C) - P(X B C) 0.70 = 0.25 + 0.60 - P(X B C) 0.70 = 0.85 - P(X B C) and P(X B C) = 0.15. Therefore, B and C are not mutually exclusive. 3-11. a) P(X > 15) = 1 – P(X 15) = 1 – 0.3 = 0.7 b) P(X 24) = P(X 15) + P(15 < X 24) = 0.3 + 0.6 = 0.9 c) P(15 < X 20) = P(X 20) – P(X 15) = 0.5 – 0.3 = 0.2 d) P(X 18) = P(15 < X 18) + P(X 15) where P(15 < X 18) = P(15 < X 24) – P(18 < X 24) = 0.6 – 0.4 = 0.2 Therefore, P(X 18) = P(15 < X 18) + P(X 15) = 0.2 + 0.3 = 0.5 Alternatively, P(X 18) = P(X 24) - P(18 < X 24) = 0.9 – 0.4 = 0.5. 3-12. A - Overfilled, B - Medium filled, C - Underfilled a) P(X C’) = 1 - P(X C) = 1 – 0.15 = 0.85 b) P(X A C) = P(X A) + P(X C) – P(X A C) = 0.40 + 0.15 – 0 = 0.55 (P(X A C) = 0 since A and C are mutually exclusive) 3-13. a) P(X 7000) = 1 – P(X > 7000) = 1 – 0.45 = 0.55 b) P(X > 5000) = 1 – P(X 5000) = 1 – 0.05 = 0.95 c) P(5000 < X 7000) = P(X 7000) – P(X 5000) = 0.55 – 0.05 = 0.50 3-14. a) Probability that a component does not fail: P ) E ( 1 = 1 – P(E 1 ) = 1 – 0.15 = 0.85 b) P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) = 0.15 + 0.30 = 0.45

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c) P(E 1 or E 2 ) = 1 – 0.45 = 0.55 Section 3-4 3-15. a) kx . Therefore, k = 3/64. dx k k x 2 0 4 3 0 4 3 64 3 = = E X x dx x ( ) = = 3 64 3 64 4 3 3 4 0 4 0 4 = ( ) V X x x dx x x x dx x x x ( ) ( ) ( ) . = = + = + = 3 64 3 3 64 6 9 3 64 0 6 2 2 0 4 4 3 2 0 4 5 6 4 9 3 0 4 5 4 3 b) k x dx k x x k ( ) ( ) 1 2 6 0 2 2 0 2 + = + = . Therefore, k = 1/6. E X x x dx x x dx x x ( ) ( ) ( ) ( ) / = + = + = + = 1 6 1 2 1 6 2 1 6 2 2 3 11 9 2 0 2 0 2 2 3 0 2 V X x x dx x x x x x dx x x x x x x ( ) ( )( ) ( ) . . = + = + + + = + + + + = = 1 6 11 9 2 0 2 1 6 2 22 9 121 81 3 44 9 2 242 81 0 2 1 6 3 22 9 2 121 81 4 44 9 3 242 81 2 0 2 1 2 2 2 1704 6 0 284 3 2 4 3 2 c) ke dx k e k x x = = 0 0 ( ) . Therefore, k = 1.
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