ch04 - CHAPTER 4 Section 4-2 4-1. a) P (1 < X ) = e - x...

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CHAPTER 4 Section 4-2 4-1. a) 3679 . 0 ) ( ) 1 ( 1 1 1 = = = = < e e dx e X P x x b) 2858 . 0 ) ( ) 5 . 2 1 ( 5 . 2 1 5 . 2 1 5 . 2 1 = = = = < < e e e dx e X P x x c) 0 ) 3 ( 3 3 = = = dx e X P x d) 9817 . 0 1 ) ( ) 4 ( 4 4 0 4 0 = = = = < e e dx e X P x x e) 0498 . 0 ) ( ) 3 ( 3 3 3 = = = = e e dx e X P x x 4-2. a) 10 . 0 ) ( ) ( = = = = < x x x x x e e dx e X x P . T h e n , x = ln(0.10) = 2.3 b) 10 . 0 1 ) ( ) ( 0 0 = = = = x x x x x e e dx e x X P . Then, x = ln(0.9) = 0.1054 4-3 a) 4375 . 0 16 3 4 16 8 ) 4 ( 2 2 4 3 2 4 3 = = = = < x dx x X P , because 0 ) ( = x f X for x < 3. b) , 7969 . 0 16 5 . 3 5 16 8 ) 5 . 3 ( 2 2 5 5 . 3 2 5 5 . 3 = = = = > x dx x X P because for x > 5. 0 ) ( = x f X c) 5625 . 0 16 4 5 16 8 ) 5 4 ( 2 2 5 4 2 5 4 = = = = < < x dx x X P d) 7031 . 0 16 3 5 . 4 16 8 ) 5 . 4 ( 2 2 5 . 4 3 2 5 . 4 3 = = = = < x dx x X P e) 5 . 0 16 3 5 . 3 16 5 . 4 5 16 16 8 8 ) 5 . 3 ( ) 5 . 4 ( 2 2 2 2 5 . 3 3 2 5 5 . 4 2 5 . 3 3 5 5 . 4 = + = + = + = < + > x x dx x dx x X P X P . 4-1
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4-4 a) 1 ) 1 ( 4 ) 4 ( 4 ) 4 ( = = = < x x e dx e X P , because 0 ) ( = x f X for x < 4. This can also be obtained from the fact that is a probability density function for 4 < x. ) ( x f X b) 6321 . 0 1 ) 5 2 ( 1 5 4 ) 4 ( 5 4 ) 4 ( = = = = e e dx e X P x x c) ) 5 ( 1 ) 5 ( = < X P X P . From part b., 6321 . 0 ) 5 ( = X P . Therefore, . 3679 . 0 ) 5 ( = < X P d) 0180 . 0 ) 12 8 ( 8 4 12 8 ) 4 ( 12 8 ) 4 ( = = = = < < e e e dx e X P x x e) 90 . 0 1 ) ( ) 4 ( 4 ) 4 ( 4 ) 4 ( = = = = < x x x x x e e dx e x X P . Then, x = 4 ln(0.10) = 6.303 4-5 a) , by symmetry. 5 . 0 ) 0 ( = < X P b) 4375 . 0 0625 . 0 5 . 0 5 . 0 5 . 1 ) 5 . 0 ( 1 5 . 0 3 1 5 . 0 2 = = = = < x dx x X P c) 125 . 0 5 . 0 5 . 1 ) 5 . 0 5 . 0 ( 5 . 0 5 . 0 3 5 . 0 5 . 0 2 = = = x dx x X P d) P(X < 2) = 0 e) P(X < 0 or X > 0.5) = 1 f) 05 . 0 5 . 0 5 . 0 5 . 0 5 . 1 ) ( 3 1 3 1 2 = = = = < x x dx x X x P x x Then, x = 0.9655 4-6. a) 05 . 0 1000 ) 3000 ( 3 3000 3000 1000 1000 = = = = > e e dx e X P x x b) 233 . 0 1000 ) 2000 1000 ( 2 1 2000 1000 2000 1000 1000 1000 = = = = < < e e e dx e X P x x c) 6321 . 0 1 1000 ) 1000 ( 1 1000 0 1000 0 1000 1000 = = = = < e e dx e X P x x d) 10 . 0 1 1000 ) ( 1000 / 0 0 1000 1000 = = = = < x x x e e dx e x X P x x . Then, e , and x = 1000 ln 0.9 = 105.36. x = / . 1000 09 4-2
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4-7 a) 5 . 0 2 0 . 2 ) 50 ( 25 . 50 50 25 . 50 50 = = = > x dx X P b) x x dx x X P x x 2 5 . 100 2 0 . 2 90 . 0 ) ( 25 . 50 25 . 50 = = = = > Then, 2x = 99.6 and x = 49.8. 4-8. a) 25 . 0 25 . 1 25 . 1 ) 8 . 74 ( 8 . 74 6 . 74 8 . 74 6 . 74 = = = < x dx X P b) P(X < 74.8 or X > 75.2) = P(X < 74.8) + P(X > 75.2) because the two events are mutually exclusive. The result is 0.25 + 0.25 = 0.50. c) 750 . 0 ) 6 . 0 ( 25 . 1 25 . 1 25 . 1 ) 3 . 75 7 . 74 ( 3 . 75 7 . 74 3 . 75 7 . 74 = = = = < < x dx X P 4-9 a) P(X < 2.25 or X > 2.75) = P(X < 2.25) + P(X > 2.75) because the two events are mutually exclusive. Then, P(X < 2.25) = 0 and P(X > 2.75) = . 10 . 0 ) 05 . 0 ( 2 2 8 . 2 75 . 2 = = dx b) If the probability density function is centered at 2.5 meters, then for fx X () = 2 2.3 < x < 2.8 and all rods will meet specifications. 4-10. Because the integral is not changed whether or not any of the endpoints x fxdx x x 1 2 1 and x 2 are included in the integral, all the probabilities listed are equal. Section 4-3 4-11. a) P(X<2.8) = P(X 2.8) because X is a continuous random variable. Then, P(X<2.8) =F(2.8)=0.2(2.8) = 0.56. b) 7 . 0 ) 5 . 1 ( 2 . 0 1 ) 5 . 1 ( 1 ) 5 . 1 ( = = = > X P X P c) 0 ) 2 ( ) 2 ( = = < X F X P d) 0 ) 6 ( 1 ) 6 ( = = > X F X P 4-12. a) ) 8 . 1 ( ) 8 . 1 ( ) 8 . 1 ( X F X P X P = = < because X is a continuous random variable.
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ch04 - CHAPTER 4 Section 4-2 4-1. a) P (1 &lt; X ) = e - x...

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