hw3 - STAT 350 Spring 2009 Homework #3 Solution covers...

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STAT 350 – Spring 2009 Homework #3 Solution covers through Lecture F Text Exercises Chapter 2, Problems #10, 14, 24 2.10 () 1 b a fx EX x d x ba ⎛⎞ = ⎜⎟ ⎝⎠ ±²³²´ = 2 ab + 2.14 a. = 2 () () 2 22 0 0.5 x xd x = ±³´ b. for a continuous RV: ( )() () () () () () () 1 x x a b xf xd x a f x b x f x d x a f xd x b x x a f xd x b x x μ +∞ +∞ −∞ −∞ +∞ +∞ +∞ +∞ −∞ −∞ −∞ −∞ == += + ⎡⎤ ⎣⎦ =+ = + ∫∫ ±²³ ²´ ± ²³ ²´ OR for a discrete RV: ( ) ( ) ( ) 1 x x xpx a px b x px a b x a px b x + = + ∑∑ ±²³ ² ´± ²³ ²´ If the mean repair time is μ x = 0.5, the mean repair revenue is μ 25+40 x = 25 + 40 μ x = 45 2.24 a. () ( ) ( ) ( ) N ( ) N ( ) N ( ) N ( N 2 22222 0.4 0.1 0.1 0.1 0.3 Var 1.8 01 . 8 0 11 . 8 1 21 . 8 2 31 . 8 3 41 . 8 4 Xx p x ppppp =− +− ) = 2.96 2.96 SD X = =1.72 b. P(1.8 - 1.72 < X < 1.8 + 1.72) = P(0.8 < X < 3.52) = p(1) + p(2) + p(3) = 0.1 + 0.1 + 0.1 = 0.3 P(X < 1.8 - 3 × 1.72 X > 1.8 + 3 × 1.72) = P(X < -3.36) + P(X > 6.96) = 0 Homework #3 - Solution Page 1 of 7
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Then answer the following problems: 1. X is a discrete random variable with the following PMF: () 2 0.04 0.24 0.40 if 0,1,2,. ..,5 0o t h e r w X xx x px −+ = = i s e a. Find P( X = 2) = p (2) = 0.04(2) 2 – 0.24(2) + 0.40 = 0.08 b. Find P( X < 2) = P(X = 0) + P(X = 1) = p (0) + p (1) = [0.04(0) 2 – 0.24(0) + 0.40] + [0.04(1) 2 – 0.24(1) + 0.40] = 0.60 c. Find P( X 2) Complementation Rule: P( X 2) = 1 – P( X < 2) = 1 – 0.60 = 0.40 d. Find P( X = 6) 0 (6 is not in the sample space) e. Find E( X ) 5 0 x x = = = (0)[0.04(0) 2 – 0.24(0) + 0.40] + (1)[0.04(1) 2 – 0.24(1) + 0.40] +(2)[0.04(2) 2 – 0.24(2) + 0.40] + (3)[0.04(3) 2 – 0.24(3) + 0.40] +(4)[0.04(4) 2 – 0.24(4) + 0.40] + (5)[0.04(5) 2 – 0.24(5) + 0.40] = 1.80 f. Find Var( X ) ( ) ( ) ( 55 22 00 1.8 ) x x μ == =− ∑∑ = (0-1.8) 2 [0.04(0) 2 – 0.24(0) + 0.40] + (1-1.8) 2 [0.04(1) 2 – 0.24(1) + 0.40] +(2-1.8) 2 [0.04(2) 2 – 0.24(2) + 0.40] + (3-1.8) 2 [0.04(3) 2 – 0.24(3) + 0.40] +(4-1.8) 2 [0.04(4) 2 – 0.24(4) + 0.40] + (5-1.8) 2 [0.04(5) 2 – 0.24(5) + 0.40] = 3.920 g. Find SD( X ) 3.920 Var X =1.979899 Homework #3 - Solution Page 2 of 7
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2. X is a continuous random variable with the following PDF: () 0.04 0.26 if 1 4 0o t h e r w i X xx fx s e +− < = < a. Find P(2 < X < 3) () () () () 3 3 2 2 2 22 0.04 0.26 0.02 0.26 0.02 3 0.26 3 0.02 2 0.26 2 x x x dx x x = = =− + = + ⎡⎤ + −− + ⎣⎦ = 0.16 b. Find P(0.2 < X < 0.3) 0.3 0.3 2 0.2 0.2 0.04 0.26 0.02 0.26 0.02 0.3 0.26 0.3 0.02 0.2 0.26 0.2 x x xd x = = + =
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hw3 - STAT 350 Spring 2009 Homework #3 Solution covers...

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