ismch06 - 84 Instructors Solutions Manual for Statistics...

Info iconThis preview shows pages 1–5. 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: 84 Instructors 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) = 1-P(X < .75) = 1-.75 = .25 85 Instructors Solutions Manual for Statistics for Business & Economics, 5 th Edition e. P(.2 < X < .8) = P(X <.8) P(X <.2) = .8 - .2 = .6 6-2 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 Z = 10,000 + 1.5 X = 10,000 + 1.5 (30,000) = $55,000 Z = |1.5| X = 1.5 (8,000) = $12,000 6.6 Z = 20 + X = 20 + 4 = $24 million Bid = 1.1 z =1.1(24) = $26.4 million, = $1 million 6.7 Z = 60 + .2 X = 60 + 140 = $200 Z = |.2| X = .2 (130) = $26 6.8 Z = 6,000 + .08 X = 6,000 + 48,000 = $54,000 Z = |.08| 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) [1-F 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 Instructors 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...
View Full Document

This note was uploaded on 03/25/2009 for the course FIN FIN534 taught by Professor Stevenjordan during the Spring '09 term at Korea Advanced Institute of Science and Technology.

Page1 / 15

ismch06 - 84 Instructors Solutions Manual for Statistics...

This preview shows document pages 1 - 5. Sign up to view the full document.

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