5 the area for z 112 is 3686 the area for z 301 is

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from A.5, the area for z = -1.12 is .3686 the area for z = -3.01 is .4987 Prob(155 < x < 170) = .4987 - .3686 = .1301 d) Prob(x < 200 personnel): correcting for continuity: x = 199.5 z = 485 . 8 180 5 . 199 = 2.30 from A.5, the area for z = 2.30 is .4893 Prob(x < 200) = .5000 + .4893 = .9893
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Chapter 6: Continuous Distributions 6 6.24 n = 95 a) Prob(44 < x < 52) agree with direct investments, p = .52 By the normal distribution: µ = n( p ) = 95(.52) = 49.4 σ = ) 48 )(. 52 (. 95 = q p n = 4.87 test: µ + 3 σ = 49.4 + 3(4.87) = 49.4 + 14.61 0 < 34.79 to 64.01 < 95 test passed z = 87 . 4 4 . 49 5 . 43 = -1.21 from table A.5, area = .3869 z = 87 . 4 4 . 49 5 . 52 = 0.64 from table A.5, area = .2389 Prob(44 < x < 52) = .3869 + .2389 = .6258 b) Prob( x > 56): correcting for continuity, x = 56.5 z = 87 . 4 4 . 49 5 . 56 = 1.46 from table A.5, area = .4279 Prob( x > 56) = .5000 - .4279 = .0721 c) Joint Venture: p = .70, n = 95 By the normal dist.: µ = n( p ) = 95(.70) = 66.5 σ = ) 30 )(. 70 (. 95 = q p n test for normalcy: 66.5 + 3(4.47) = 66.5 + 13.41 0 < 53.09 to 79.91 < 95 test passed Prob( x < 60): correcting for continuity: x = 59.5 z = 47 . 4 5 . 66 5 . 59 = -1.57 from table A.5, area = .4418 Prob( x < 60) = .5000 - .4418 = .0582 d) Prob(55 < x < 62): correcting for continuity: 54.5 to 62.5 z = 47 . 4 5 . 66 5 . 54 = -2.68 from table A.5, area = .4963 z = 47 . 4 5 . 66 5 . 62 = -0.89 from table A.5, area = .3133 Prob(55 < x < 62) = .4963 - .3133 = .1830 6.25 a) λ = 0.1 x 0 0 1 2 3 4 5 6 7 8 9 10 y .1000 .0905 .0819 .0741 .0670 .0607 .0549 .0497 .0449 .0407 .0368 b) λ = 0.3 x 0 y 0 .3000 1 .2222 2 .1646 3 .1220 4 .0904 5 .0669 6 .0496 7 .0367 8 .0272 9 .0202
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Chapter 6: Continuous Distributions 7 c) λ = 0.8 x 0 y 0 .8000 1 .3595 2 .1615 3 .0726 4 .0326 5 .0147 6 .0066 7 .0030 8 .0013 9 .0006 d) λ = 3.0 x 0 y 0 3.0000 1 .1494 2 .0074 3 .0004 4 .0000 5 .0000 6.26 a) λ = 3.25 µ = 25 . 3 1 1 = λ = 0.31 σ = 25 . 3 1 1 = λ = 0.31 b) λ = 0.7 µ = 007 . 1 1 = λ = 1.43 σ = 007 . 1 1 = λ = 1.43 c) λ = 1.1 µ = 1 . 1 1 1 = λ = 0.91 σ = 1 . 1 1 1 = λ = 0.91 d) λ = 6.0 μ = 6 1 1 = λ = 0.17 σ = 6 1 1 = λ = 0.17 6.27 a) Prob(x > 5 λ = 1.35) = for x 0 = 5: Prob(x) = e - λ x = e -1.35(5) = e -6.75 = .0012 b) Prob(x < 3 λ = 0.68) = 1 - Prob(x < 3 λ = .68) = for x 0 = 3: 1 – e - λ x = 1 – e -0.68(3) = 1 – e –2.04 = 1 - .1300 = .8700 c) Prob(x > 4 λ = 1.7) = for x 0 = 4: Prob(x) = e - λ x = e -1.7(4) = e -6.8 = .0011 d) Prob(x < 6 λ = 0.80) = 1 - Prob(x > 6 λ = 0.80) = for x 0 = 6: Prob(x) = 1 – e - λ x = 1 – e -0.80(6) = 1 – e -4.8 = 1 - .0082 = .9918
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Chapter 6: Continuous Distributions 8 6.28 µ = 23 sec. λ = μ 1 = .0435 per second a) Prob(x > 1 min λ = .0435/sec.) Change λ to minutes: λ = .0435(60) = 2.61 min Prob(x > 1 min λ = 2.61/min) = for x 0 = 1: Prob(x) = e - λ x = e -2.61(1) = .0735 b) λ = .0435/sec Prob(x > 5 min λ = .0435/sec.) Change λ to minutes: λ = (.0435)(60) = 2.61 min P(x > 5 λ = 2.61/min) = for x 0 = 5: Prob(x) = e - λ x = e -2.61(5) = e -13.05 = .0000 6.29 λ = 2.44/min. a) Prob(x > 10 min λ = 2.44/min) = Let x 0 = 10, e - λ x = e -2.44(10) = e -24.4 = .0000 b) Prob(x > 5 min λ = 2.44/min) = Let x 0 = 5, e - λ x = e -2.44(5) = e -12.20 = .0000 c) Prob(x > 1 min λ = 2.44/min) = Let x 0 = 1, e - λ x = e -2.44(1) = e -2.44 = .0872 d) Epected time = µ = 44 . 2 1 1 = λ min. = .41 min = 24.6 sec. 6.30 λ = 1.12 planes/hr. a) µ = 12 . 1 1 1 = λ = .89 hr. = 53.4 min . b) Prob(x > 2 hrs λ = 1.12 planes/hr.) = Let x 0 = 2, e - λ x = e -1.12(2) = e -2.24 = .1065 c) Prob(x < 10 min λ = 1.12/hr.) = 1 - P(x > 10 min λ = 1.12/hr.) Change λ to 1.12/60 min. = .01867/min. 1 - Prob(x > 10 min λ = .01867/min) = Let x 0 = 10, 1 – e - λ x =1 – e -.01867(10) =1 – e -.1861 =1 - .8297 = .1703 6.31 λ = 3.39/ 1000 passengers µ = 39 . 3 1 1 = λ = 0.295 (0.295)(1,000) = 295 Prob(x > 500): Let x 0 = 500/1,000 passengers = .5 e - λ x = e -3.39(.5) = e -1.695 = .1836 Prob(x < 200): Let x 0 = 200/1,000 passengers = .2 e - λ x = e -3.39(.2) = e -.678 = .5076 Prob(x < 200) = 1 - .5076 = .4924 6.32 µ = 20 years λ = 20 1 = .05/year x 0 Prob(x>x 0 )=e - λ x 1 .9512 2 .9048 3 .8607 If the foundation is guaranteed for 2 years, based on past history, 90.48% of the foundations will last at least 2 years without major repair and only 9.52% will require a major repair before 2 years.
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