MAU_HMWK2SOLUTIONS

# MAU_HMWK2SOLUTIONS - Chapter 5 1 a P(X = 1 Y = 1 = p(1,1...

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Unformatted text preview: Chapter 5 1. a. P(X = 1, Y = 1) = p(1,1) = .20 b. P(X ≤ 1 and Y ≤ 1) = p(0,0) + p(0,1) + p(1,0) + p(1,1) = .42 c. At least one hose is in use at both islands. P(X ≠ 0 and Y ≠ 0) = p(1,1) + p(1,2) + p(2,1) + p(2,2) = .70 d. By summing row probabilities, p x (x) = .16, .34, .50 for x = 0, 1, 2, and by summing column probabilities, p y (y) = .24, .38, .38 for y = 0, 1, 2. P(X ≤ 1) = p x (0) + p x (1) = .50 e. P(0,0) = .10, but p x (0) ⋅ p y (0) = (.16)(.24) = .0384 ≠ .10, so X and Y are not independent. 26. Revenue = 3X + 10Y, so E (revenue) = E (3X + 10Y) 4 . 15 ) 2 , 5 ( 35 ... ) , ( ) , ( ) 10 3 ( 5 2 = ⋅ + + ⋅ = ⋅ + = ∑∑ = = p p y x p y x x y 37. P(x 1 ) .20 .50 .30 P(x 2 ) x 2 | x 1 25 40 65 .20 25 .04 .10 .06 .50 40 .10 .25 .15 .30 65 .06 .15 .09 a. x 25 32.5 40 45 52.5 65 ( 29 x p .04 .20 .25 .12 .30 .09 ( 29 μ = = + + + = 5 . 44 ) 09 (. 65 ... ) 20 (. 5 . 32 ) 04 )(. 25 ( x E b. s 2 112.5 312.5 800 P(s 2 ) .38 .20 .30 .12 E(s 2 ) = 212.25 = σ 2 49. a. 11 P.M. – 6:50 P.M. = 250 minutes. With T = X 1 + … + X 40 = total grading time, 240 ) 6 )( 40 ( = = = μ μ n T and , 95 . 37 = = n T σ σ so P( T ≤ 250) ≈ ( 29 6026 . 26 . 95 . 37 240 250 = ≤ = - ≤ Z P Z P b. ( 29 ( 29 2981 . 53 . 95 . 37 240 260 260 = = - = Z P Z P T P 51. X ~ N(10,4). For day 1, n = 5 P( X ≤ 11)= 8686 . ) 12 . 1 ( 5 / 2 10 11 = ≤ = - ≤ Z P Z P For day 2, n = 6 P( X ≤ 11)= 8888 . ) 22 . 1 ( 6 / 2 10 11 = ≤ = - ≤ Z P Z P For both days, P( X ≤ 11)= (.8686)(.8888) = .7720 Chapter 7 3....
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MAU_HMWK2SOLUTIONS - Chapter 5 1 a P(X = 1 Y = 1 = p(1,1...

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