ch04 - CHAPTER 4 Section 4-2 4-1 a 3679 1 1 1 1 = = = =<...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 4 Section 4-2 4-1. a) 3679 . ) ( ) 1 ( 1 1 1 = =- = = <- ∞- ∞- ∫ e e dx e X P x x b) 2858 . ) ( ) 5 . 2 1 ( 5 . 2 1 5 . 2 1 5 . 2 1 =- =- = = < <---- ∫ e e e dx e X P x x c) ) 3 ( 3 3 = = = ∫- dx e X P x d) 9817 . 1 ) ( ) 4 ( 4 4 4 =- =- = = <--- ∫ e e dx e X P x x e) 0498 . ) ( ) 3 ( 3 3 3 = =- = = ≤- ∞- ∞- ∫ e e dx e X P x x 4-7 a) 5 . 2 . 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 . 2 90 . ) ( 25 . 50 25 . 50- = = = = ∫ Then, 2x = 99.6 and x = 49.8. 4-15. Now, x e x f- = ) ( for 0 < x and x x x x x X e e dx e x F---- =- = = ∫ 1 ) ( for 0 < x. Then, - ≤ =- , 1 , ) ( x e x x F x X 4-21. < ≤ < < = 9 4 , 04 . 4 , 2 . ) ( x x x f Section 4-4 4-25. 4 5 . 1 5 . 1 ) ( 1 1 1 1 4 3 = = =-- ∫ x dx x X E 6 . 5 5 . 1 5 . 1 ) ( 5 . 1 ) ( 1 1 5 1 1 1 1 4 2 3 = = =- =--- ∫ ∫ x dx x dx x x X V 4-29. a) 39 . 109 ln 600 600 ) ( 120 100 120 100 2 = = = ∫ x dx x x X E 19 . 33 ) 39 . 109 ln 78 . 218 ( 600 1 600 600 ) 39 . 109 ( ) ( 120 100 1 2 ) 39 . 109 ( 120 100 ) 39 . 109 ( 2 2 120 100 2 2 2 =-- = +- =- =- ∫ ∫ x x x dx dx x x X V x x b.) Average cost per part = $0.50*109.39 = $54.70 Section 4-5 4-33. a) E(X) = (-1+1)/2 = 0, 577 . , 3 / 1 12 )) 1 ( 1 ( ) ( 2 = =-- = x and X V σ b) x x t dt x X x P x x x x = = = = < <--- ∫ ) 2 ( 5 . 5 . ) ( 2 1 Therefore, x should equal 0.90. c) 0, 1 ( ) 0.5 0.5, 1 1 1, 1 x F x x x x < - = +- < 4-37. a) The distribution of X is f ( x ) = 100 for 0.2050 < x < 0.2150. Therefore, ≤ < ≤- < = x x x x x F 2150 . , 1 2150 . 2050 . , 50 . 20 100 2050 . , ) ( b) 25 . ] 50 . 20 ) 2125 . ( 100 [ 1 ) 2125 . ( 1 ) 2125 . ( =-- =- = F X P c) If P(X > x) = 0.10, then 1 - F(X) = 0.10 and F(X) = 0.90. Therefore, 100x - 20.50 = 0.90 and x = 0.2140. d) E(X) = (0.2050 + 0.2150)/2 = 0.2100 μ m and V(X) = 2 6 2 m 10 33 . 8 12 ) 2050 . 2150 . ( μ- × =- 2 (253 247 1) 1 12 4- +- = Section 4-6 4-45. a) P(X < 13) = P(Z < (13- 10)/2) = P(Z < 1.5) = 0.93319 b) P(X > 9) = 1 - P(X < 9) = 1 - P(Z < (9- 10)/2) = 1 - P(Z < - 0.5) = 0.69146. c) P(6 < X < 14) = P Z 6 10 2 14 10 2- < <- = P(- 2 < Z < 2) = P(Z < 2) - P(Z < - 2)] = 0.9545. d) P(2 < X < 4) = P Z 2 10 2 4 10 2- < <- = P(- 4 < Z < - 3) = P(Z < - 3) - P(Z < - 4) = 0.00132 e) P(- 2 < X < 8) = P(X < 8) - P(X < - 2) = P Z P Z <- - <-- 8 10 2 2 10 2 = P(Z < - 1) - P(Z < - 6) = 0.15866....
View Full Document

This note was uploaded on 11/05/2008 for the course EGN 3443 taught by Professor Unknown during the Fall '08 term at University of South Florida.

Page1 / 9

ch04 - CHAPTER 4 Section 4-2 4-1 a 3679 1 1 1 1 = = = =<...

This preview shows document pages 1 - 4. Sign up to view the full document.

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