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Unformatted text preview: 84 Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition Chapter 6: Continuous Random Variables and Probability Distributions 6.1 a. 1 1.5 1.0 0.5 0.0 X f(x) Probability Density Function: f(x) b. 1 1.5 1.0 0.5 0.0 X Cumulative distribution function: F(x) F(X) c. P(X < .25) = .25 d. P(X >.75) = 1P(X < .75) = 1.75 = .25 85 Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition e. P(.2 < X < .8) = P(X <.8) – P(X <.2) = .8  .2 = .6 62 a. 4 3 2 1 0.75 0.50 0.25 0.00 X f(x) Probability density function: f(x) b. 1 2 3 4 0.00 0.25 0.50 0.75 X F(x) Cumulative density function: F(x) c. P(x < 1) = .25 d. P(X < .5) + P(X > 3.5)=P(X < .5) + 1 – P(X < 3.5) = .25 6.3 a. P(60,000 < X< 72,000) = P(X < 72,000) – P(X < 60,000) = .6  .5 = .1 Chapter 6: Continuous Random Variables and Probabilty Distributions 86 b. P(X < 60,000) < P(X < 65,000) < P(X < 72,000); .5 < P(X < 65,000) < .6 6.4 a. P(380 < X < 460) = P(X < 460) – P(X < 380) = .6  .4 = .2 b. P(X < 380) < (PX< 400) < P(X < 460); .4 < P(X < 400) < .6 6.5 μ Z = 10,000 + 1.5 μ X = 10,000 + 1.5 (30,000) = $55,000 σ Z = 1.5 σ X = 1.5 (8,000) = $12,000 6.6 μ Z = 20 + μ X = 20 + 4 = $24 million Bid = 1.1 z μ =1.1(24) = $26.4 million, σ π = $1 million 6.7 μ Z = 60 + .2 μ X = 60 + 140 = $200 σ Z = .2 σ X = .2 (130) = $26 6.8 μ Z = 6,000 + .08 μ X = 6,000 + 48,000 = $54,000 σ Z = .08 σ X = .08(180,000) = $14,400 6.9 a. P(Z < 1.20)  .8849 b. P(Z > 1.33) = 1 – F z (1.33) = 1  .9082 = .0918 c. P(Z < 1.70) = 1 – F z (1.7) = 1  .9554 = .0446 d. P(Z > 1.00) = F z (1) = .8413 e. P(1.2 < Z < 1.33) = F z (1.33) – F z (1.2) = .9082  .8849 = .0233 f. P(1.7 < Z < 1.2) = F z (1.2) – [1  F z (1.7)] = .8849 – .0446 = .8403 g. P(1.7 < Z < 1.0) = F z (1.7) – F z (1) = .9554  .8413 = .1141 6.10 a. Find Z such that P(Z < Z ) = .7, closest value of Z = .52 b. Find Z such that P(Z < Z ) = .25, closest value of Z = .67 c. Find Z such that P(Z > Z ) = .2, closest value of Z = .84 d. Find Z such that P(Z > Z ) = .6, closest value of Z = .25 6.11 a. P(Z < 400 380 50 ) = P(Z < .4) = .6554 b. P(Z > 360 380 50 ) = P(Z > .4) = F Z (.4) = .6554 c. The graph should show the property of symmetry – the area in the tails equidistant from the mean will be equal. d. P( 300 380 50 < Z < 400 380 50 ) = P(1.6 < Z < .4) = F Z (.4) – [1F Z (1.6)] = . 6554  .0548 = .6006 e. The area under the normal curve is equal to .8 for an infinite number of ranges – merely start at a point that is marginally higher. The shortest range will be the one that is centered on the z of zero. The z that corresponds to an area of .8 centered on the mean is a Z of 1.28. This yields an interval of the mean plus and minus 64: [316, 444] 87 Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition 6.12 a. P(Z > 1,000 1,200 100 ) = P(Z > 2) =F Z (2) = .9772 b. P( 1,100 1,200 100 < Z < 1,300 1,200 100 ) = P(1 < Z < 1) = 2F...
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 Spring '09
 StevenJordan
 Normal Distribution, Probability distribution, probability density function, fz

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