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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 == EX xdx x () 3 64 3 64 4 3 3 4 0 4 0 4 = ( ) VX x x dx x x x dx xxx ( ) ( ) . =− = + += ∫∫ 3 64 3 3 64 69 3 64 06 22 0 4 432 0 4 5 6 4 9 3 0 4 543 b) kx d x k x x k () ( ) 12 6 0 2 2 0 2 . Therefore, k = 1/6. E X x x dx x x dx xx ( ) ( ) ( ) / =+=+ =+ = 1 6 1 6 2 1 62 2 3 11 9 2 0 2 0 2 23 0 2 xx x d 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 63 22 92 121 81 4 44 93 242 81 2 0 2 2 2 1704 6 0 284 32 43 2 c) ke dx k e k = 0 0 ( ) . Therefore, k = 1. , using integration by parts E X xe dx x = 0 xe e dx e + = 0 0 0 1 . Now, using integration by parts V X x e dx x e xe e dx x ( ) ( ) = + −− 2 0 2 0 x x = . Therefore, , because is a probability density x e dx xe x 2 00 2 = V X e dx x = 0 1 e x function.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 35

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online