Chapter 6 Black 8e Student Solutions Manual Ch06
User Manual:
Open the PDF directly: View PDF 
.
Page Count: 34
| Download | |
| Open PDF In Browser | View PDF |
Student’s Solutions Manual and Study Guide: Chapter 6 Page 1 Chapter 6 Continuous Distributions LEARNING OBJECTIVES The primary learning objective of Chapter 6 is to help you understand continuous distributions, thereby enabling you to: 1. 2. 3. 4. Solve for probabilities in a continuous uniform distribution Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution CHAPTER OUTLINE 6.1 The Uniform Distribution Determining Probabilities in a Uniform Distribution Using the Computer to Solve for Uniform Distribution Probabilities 6.2 Normal Distribution History of the Normal Distribution Probability Density Function of the Normal Distribution Standardized Normal Distribution Solving Normal Curve Problems Using the Computer to Solve for Normal Distribution Probabilities 6.3 Using the Normal Curve to Approximate Binomial Distribution Problems Correcting for Continuity 6.4 Exponential Distribution Probabilities of the Exponential Distribution Using the Computer to Determine Exponential Distribution Probabilities Student’s Solutions Manual and Study Guide: Chapter 6 Page 2 KEY TERMS Correction for Continuity Exponential Distribution Normal Distribution Rectangular Distribution Standardized Normal Distribution Uniform Distribution z Distribution z Score Student’s Solutions Manual and Study Guide: Chapter 6 Page 3 STUDY QUESTIONS 1. The uniform distribution is sometimes referred to as the _____________________ distribution. 2. Suppose a set of data are uniformly distributed from x = 5 to x = 13. The height of the distribution is ____________________. The mean of this distribution is ____________________. The standard deviation of this distribution is _____________________. 3. Suppose a set of data are uniformly distributed from x = 27 to x = 44. The height of this distribution is _____________________. The mean of this distribution is _____________________. The standard deviation of this distribution is ________________________. 4. A set of values is uniformly distributed from 84 to 98. The probability of a value occurring between 89 and 93 is _____________________. The probability of a value occurring between 80 and 90 is __________________. The probability of a value occurring that is greater than 75 is _____________________. 5. Probably the most widely known and used of all distributions is the _______________ distribution. 6. Many human characteristics can be described by the _______________ distribution. 7. The area under the curve of a normal distribution is ________. 8. In working normal curve problems using the raw values of x, the mean, and the standard deviation, a problem can be converted to ________ scores. 9. A z score value is the number of _______________ _______________ a value is from the mean. 10. Within a range of z scores of ± 1 from the mean, fall _________% of the values of a normal distribution. 11. Suppose a population of values is normally distributed with a mean of 155 and a standard deviation of 12. The z score for x = 170 is ________. Suppose a population of values is normally distributed with a mean of 76 and a standard deviation of 5.2. The z score for x = 73 is ________. 12. 13. Suppose a population of values is normally distributed with a mean of 250 and a variance of 225. The z score for x = 286 is ________. 14. Suppose a population of values is normally distributed with a mean of 9.8 and a standard deviation of 2.5. The probability that a value is greater than 11 in the distribution is ________. 15. A population is normally distributed with a mean of 80 and a variance of 400. The probability that x lies between 50 and 100 is ________. Student’s Solutions Manual and Study Guide: Chapter 6 Page 4 16. A population is normally distributed with a mean of 115 and a standard deviation of 13. The probability that a value is less than 85 is ________. 17. A population is normally distributed with a mean of 64. The probability that a value from this population is more than 70 is .0485. The standard deviation is __________. 18. A population is normally distributed with a mean of 90. 85.99% of the values in this population are greater than 75. The standard deviation of this population is ________. 19. A population is normally distributed with a standard deviation of 18.5. 69.85% of the values in this population are greater than 93. The mean of the population is ________. 20. A population is normally distributed with a variance of 50. 98.17% of the values of the population are less than 27. The mean of the population is ________. 21. A population is normally distributed with a mean of 340 and a standard deviation of 55. 10.93% of values in the population are less than ________. 22. In working a binomial distribution problem by using the normal distribution, the interval, ________, should lie between 0 and n. 23. A binomial distribution problem has an n of 10 and a p of .20. This problem _______________ be worked by the normal distribution because of the size of n and p. 24. A binomial distribution problem has an n of 15 and a p of .60. This problem _______________ be worked by the normal distribution because of the size of n and p. 25. A binomial distribution problem has an n of 30 and a p of .35. A researcher wants to determine the probability of x being greater than 13 and to use the normal distribution to work the problem. After correcting for continuity, the value of x that he/she will be solving for is ________. 26. A binomial distribution problem has an n of 48 and a p of .80. A researcher wants to determine the probability of x being less than or equal to 35 and wants to work the problem using the normal distribution. After correcting for continuity, the value of x that he/she will be solving for is _______________. Student’s Solutions Manual and Study Guide: Chapter 6 Page 5 27. A binomial distribution problem has an n of 60 and a p value of .72. A researcher wants to determine the probability of x being exactly 45 and use the normal distribution to work the problem. After correcting for continuity, he/she will be solving for the area between ________ and ________. 28. A binomial distribution problem has an n of 27 and a p of .53. If this problem were converted to a normal distribution problem, the mean of the distribution would be ________. The standard deviation of the distribution would be ________. 29. A binomial distribution problem has an n of 113 and a p of .29. If this problem were converted to a normal distribution problem, the mean of the distribution would be ________. The standard deviation of the distribution would be ________. 30. A binomial distribution problem is to determine the probability that x is less than 22 when the sample size is 40 and the value of p is .50. Using the normal distribution to work this problem produces a probability of ________. 31. A binomial distribution problem is to determine the probability that x is exactly 14 when the sample size is 20 and the value of p is .60. Using the normal distribution to work this problem produces a probability of ________. 32. A binomial distribution problem is to determine the probability that x is greater than or equal to 18 when the sample size is 30 and the value of p is .55. Using the normal distribution to work this problem produces a probability of ________. 33. A binomial distribution problem is to determine the probability that x is greater than 10 when the sample size is 20 and the value of p is .60. Using the normal distribution to work this problem produces a probability of ________. If this problem had been worked using the binomial tables, the obtained probability would have been ________. The difference in answers using these two techniques is ________. 34. The exponential distribution is a_______________ distribution. 35. The exponential distribution is closely related to the _______________ distribution. 36. The exponential distribution is skewed to the ________. 37. Suppose random arrivals occur at a rate of 5 per minute. Assuming that random arrivals are Poisson distributed, the probability of there being at least 30 seconds between arrivals is ________. 38. Suppose random arrivals occur at a rate of 1 per hour. Assuming that random arrivals are Poisson distributed, the probability of there being less than 2 hours between arrivals is ________. 39. Suppose random arrivals occur at a rate of 1.6 every five minutes. Assuming that random arrivals are Poisson distributed, the probability of there being between three minutes and six minutes between arrivals is ________. Student’s Solutions Manual and Study Guide: Chapter 6 Page 6 40. Suppose that the mean time between arrivals is 40 seconds and that random arrivals are Poisson distributed. The probability that at least one minute passes between two arrivals is ________. The probability that at least two minutes pass between two arrivals is ________. 41. Suppose that the mean time between arrivals is ten minutes and that random arrivals are Poisson distributed. The probability that no more than seven minutes pass between two arrivals is ________. 42. The mean of an exponential distribution equals ________. 43. Suppose that random arrivals are Poisson distributed with an average arrival of 2.4 per five minutes. The associated exponential distribution would have a mean of _______________ and a standard deviation of _______________. 44. An exponential distribution has an average interarrival time of 25 minutes. The standard deviation of this distribution is _______________. Student’s Solutions Manual and Study Guide: Chapter 6 Page 7 ANSWERS TO STUDY QUESTIONS 1. Rectangular 23. Cannot 2. 1/8, 9, 2.3094 24. Can 3. 1/17, 35.5, 4.9075 25. 13.5 4. .2857, .7143, 1.000 26. 35.5 5. Normal 27. 44.5, 45.5 6. Normal 28. 14.31, 2.59 7. 1 29. 32.77, 4.82 8. z 30. .6808 9. Standard deviations 31. .1212 10. 68% 32. .3557 11. 1.25 33. .7517, .7550, .0033 12. -0.58 34. Continuous 13. 2.40 35. Poisson 14. .3156 36. Right 15. .7745 37. .0821 16. .0104 38. .8647 17. 3.614 39. .2363 18. 13.89 40. .2231, .0498 19. 102.62 41. .5034 20. 12.22 42. 1/ 21. 272.35 43. 2.08 Minutes, 2.08 Minutes 22. µ ± 3 44. 25 Minutes Student’s Solutions Manual and Study Guide: Chapter 6 Page 8 SOLUTIONS TO THE ODD-NUMBERED PROBLEMS IN CHAPTER 6 6.1 a = 200 b = 240 1 1 1 = .025 b a 240 200 40 a) f(x) = b) = = a b 200 240 = 220 2 2 b a 240 200 40 = 11.547 12 12 12 c) P(x > 230) = 240 230 10 = .250 240 200 40 d) P(205 < x < 220) = e) P(x < 225) = 6.3 a = 2.80 = = 220 205 15 = .375 240 200 40 225 200 25 = .625 240 200 40 b = 3.14 a b 2.80 314 . = 2.97 2 2 b a 3.14 2.80 = 0.098 12 12 P(3.00 < x < 3.10) = 3.10 3.00 = 0.2941 3.14 2.80 Student’s Solutions Manual and Study Guide: Chapter 6 6.5 µ = 639 a = 253 𝜎= Height = Page 9 b = 1025 𝑏−𝑎 √12 = 1025 − 253 √12 = 𝟐𝟐𝟐. 𝟖𝟓𝟕 1 1 = = . 𝟎𝟎𝟏𝟑 𝑏−𝑎 1025 − 253 P(x > 850) = 1025−850 1025−253 = . 𝟐𝟐𝟔𝟕 P(x > 1200) = .0000 since 1200 is above the upper limit of the data. P(350 < x < 480) = 6.7 µ = 22 480−350 1025−253 = . 𝟏𝟔𝟖𝟒 =4 a) P(x > 17): z = x 17 22 = -1.25 4 area between x = 17 and µ = 22 from table A.5 is .3944 P(x > 17) = .3944 + .5000 = .8944 b) P(x < 13): z = x 13 22 = -2.25 4 from table A.5, area = .4878 P(x < 13) = .5000 - .4878 = .0122 Student’s Solutions Manual and Study Guide: Chapter 6 Page 10 c) P(25 < x < 31): z = x 31 22 = 2.25 4 from table A.5, area = .4878 z = x 25 22 = 0.75 4 from table A.5, area = .2734 P(25 < x < 31) = .4878 - .2734 = .2144 6.9 µ = $1332 = $725 a) P(x > $2000): z = x 2000 1332 = 0.92 725 from Table A.5, the z = 0.92 yields: .3212 P(x > $2000) = .5000 - .3212 = .1788 b) P(owes money) = P(x < 0): z = x 0 1332 = -1.84 725 from Table A.5, the z = -1.84 yields: .4671 P(x < 0) = .5000 - .4671 = .0329 c) P($100 < x < $700): z = x 100 1332 = -1.70 725 from Table A.5, the z = -1.70 yields: z = x 700 1332 = -0.87 725 .4554 Student’s Solutions Manual and Study Guide: Chapter 6 Page 11 from Table A.5, the z = -0.87 yields: .3078 P($100 < x < $700) = .4554 - .3078 = .1476 6.11 µ = 200, = 47 Determine x a) 60% of the values are greater than x: Since 50% of the values are greater than the mean, µ = 200, 10% or .1000 lie between x and the mean. From Table A.5, the z value associated with an area of .1000 is z = -0.25. The z value is negative since x is below the mean. Substituting z = -0.25, µ = 200, and = 47 into the formula and solving for x: z = -0.25 = x x 200 47 x = 188.25 b) x is less than 17% of the values. Since x is only less than 17% of the values, 33% (.5000- .1700) or .3300 lie between x and the mean. Table A.5 yields a z value of 0.95 for an area of .3300. Using this z = 0.95, µ = 200, and = 47, x can be solved for: z = 0.95 = x x 200 47 x = 244.65 c) 22% of the values are less than x. Since 22% of the values lie below x, 28% lie between x and the mean (.5000 - .2200). Table A.5 yields a z of -0.77 for an area of .2800. Using the z value of -0.77, µ = 200, and = 47, x can be solved for: Student’s Solutions Manual and Study Guide: Chapter 6 Page 12 x z = -0.77 = x 200 47 x = 163.81 d) x is greater than 55% of the values. Since x is greater than 55% of the values, 5% (.0500) lie between x and the mean. From Table A.5, a z value of 0.13 is associated with an area of .05. Using z = 0.13, µ = 200, and = 47, x can be solved for: x z = 0.13 = x 200 47 x = 206.11 6.13 µ = 750 = ?? Since 29.12% of the values are less than 500 and x = 500 is below the mean, then 20.88% lie between 500 and µ. From table A.5, z = - 0.55. −0.55 = 500 − 750 𝜎 -0.55 = -250 = −250 −0.55 = 454.55 Student’s Solutions Manual and Study Guide: Chapter 6 Page 13 6.15 = 6.2. Since 62.5% is greater than 21, x = 21 is in the lower half of the distribution and .1250 (.6250 - .5000) lie between x and the mean. Table A.5 yields a z = -0.32 for an area of .1255 (closest value to .1250)+. Solving for : z = x 21 − 𝜇 −0.32 = 6.2 -1.984 = 21 - = 22.984 6.17 a) P(x < 16 n = 30 and p = .70) µ = np = 30(.70) = 21 = n p q 30(.70)(.30) = 2.51 P(x < 16.5µ = 21 and = 2.51) b) P(10 < x < 20 n = 25 and p = .50) µ = np = 25(.50) = 12.5 = n p q 25(.50)(.50) = 2.5 P(10.5 < x < 20.5µ = 12.5 and = 2.5) c) P(x = 22 n = 40 and p = .60) µ = np = 40(.60) = 24 = n p q 40(.60)(.40) = 3.10 P(21.5 < x < 22.5µ = 24 and = 3.10) Student’s Solutions Manual and Study Guide: Chapter 6 Page 14 d) P(x > 14 n = 16 and p = .45) µ = np = 16(.45) = 7.2 = n p q 16(.45)(.55) = 1.99 P(x > 14.5µ = 7.2 and = 1.99) 6.19 a) P(x = 8n = 25 and p = .40) = µ = np = 25(.40) = 10 n p q 25(.40)(.60) = 2.449 µ ± 3 = 10 ± 3(2.449) = 10 ± 7.347 (2.653 to 17.347) lies between 0 and 25. Approximation by the normal curve is sufficient. P(7.5 < x < 8.5µ = 10 and = 2.449): z = 7.5 10 = -1.02 2.449 From Table A.5, area = .3461 z = 8.5 10 = -0.61 2.449 From Table A.5, area = .2291 P(7.5 < x < 8.5) = .3461 - .2291 = .1170 From Table A.2 (binomial tables) = .120 b) P(x > 13n = 20 and p = .60) = µ = np = 20(.60) = 12 n p q 20(.60)(.40) = 2.19 µ ± 3 = 12 ± 3(2.19) = 12 ± 6.57 (5.43 to 18.57) lies between 0 and 20. Approximation by the normal curve is sufficient. Student’s Solutions Manual and Study Guide: Chapter 6 Page 15 P(x > 12.5µ = 12 and = 2.19): z = x 12.5 12 = 0.23 2.19 From Table A.5, area = .0910 P(x > 12.5) = .5000 -.0910 = .4090 From Table A.2 (binomial tables) = .415 c) P(x = 7n = 15 and p = .50) = µ = np = 15(.50) = 7.5 n p q 15(.50)(.50) = 1.9365 µ ± 3 = 7.5 ± 3(1.9365) = 7.5 ± 5.81 (1.69 to 13.31) lies between 0 and 15. Approximation by the normal curve is sufficient. P(6.5 < x < 7.5µ = 7.5 and = 1.9365): z = x 6.5 7.5 = -0.52 1.9365 From Table A.5, area = .1985 From Table A.2 (binomial tables) = .196 d) P(x < 3n = 10 and p =.70): = µ = np = 10(.70) = 7 n p q 10(.70)(.30) µ ± 3 = 7 ± 3(1.449) = 7 ± 4.347 (2.653 to 11.347) does not lie between 0 and 10. The normal curve is not a good approximation to this problem. Student’s Solutions Manual and Study Guide: Chapter 6 6.21 n = 70, p = .59 Page 16 P(x < 35): Converting to the normal dist.: µ = n(p) = 70(.59) = 41.3 and = n p q 70(.59)(.41) = 4.115 Test for normalcy: 0 < µ + 3 < n, 0 < 41.3 + 3(4.115) < 70 0 < 28.955 to 53.645 < 70, passes the test correction for continuity, use x = 34.5 z = 34.5 41.3 = -1.65 4.115 from table A.5, area = .4505 P(x < 35) = .5000 - .4505 = .0495 Student’s Solutions Manual and Study Guide: Chapter 6 6.23 Page 17 p = .27 n = 130 Conversion to normal dist.: µ = n(p) = 130(.27) = 35.1 = = 5.062 a) P(x > 39): z Correct for continuity: x = 39.5 39.5 35.1 .87 5.062 from table A.5, area = .3078 P(x > 39) = .5000 - .3078 = .1922 b) P(28 < x < 38): z Correct for continuity: 27.5 to 38.5 27.5 35.1 1.50 5.062 z 38.5 35.1 0.67 5.062 from table A.5, area for z = -1.50 is .4322 area for z = 0.67 is .2486 P(28 < x < 38) = .4322 + .2486 = .6808 c) P(x < 23): z correct for continuity: x = 22.5 22.5 35.1 2.49 5.062 from table A.5, area for z = -2.49 is .4936 P(x < 23) = .5000 - .4936 = .0064 d) P(x = 33): z correct for continuity: 32.5 to 33.5 32.5 35.1 33.5 35.1 0.51 z 0.32 5.062 5.062 from table A.5, area for -0.51 = .1950 area for -0.32 = .1255 P(x = 33) = .1950 - .1255 = .0695 Student’s Solutions Manual and Study Guide: Chapter 6 6.25 Page 18 a) = 0.1 x0 0 1 2 3 4 5 6 7 8 9 10 y .1000 .0905 .0819 .0741 .0670 .0607 .0549 .0497 .0449 .0407 .0368 b) = 0.3 x0 0 1 2 3 4 5 6 7 8 9 y .3000 .2222 .1646 .1220 .0904 .0669 .0496 .0367 .0272 .0202 Student’s Solutions Manual and Study Guide: Chapter 6 Page 19 c) = 0.8 x0 0 1 2 3 4 5 6 7 8 9 y .8000 .3595 .1615 .0726 .0326 .0147 .0066 .0030 .0013 .0006 Student’s Solutions Manual and Study Guide: Chapter 6 Page 20 d) = 3.0 x0 0 1 2 3 4 5 y 3.0000 .1494 .0074 .0004 .0000 .0000 Student’s Solutions Manual and Study Guide: Chapter 6 Page 21 6.27 a) P(x > 5 = 1.35) = for x0 = 5: P(x) = e-x = e-1.35(5) = e-6.75 = .0012 b) P(x < 3 = 0.68) = 1 - P(x < 3 = .68) = for x0 = 3: 1 – e-x = 1 – e-0.68(3) = 1 – e –2.04 = 1 - .1300 = .8700 c) P(x > 4 = 1.7) = for x0 = 4: P(x) = e-x = e-1.7(4) = e-6.8 = .0011 d) P(x < 6 = 0.80) = 1 - P(x > 6 = 0.80) = for x0 = 6: P(x) = 1 – e-x = 1 – e-0.80(6) = 1 – e-4.8 = 1 - .0082 = .9918 6.29 = 2.44/min. a) P(x > 10 min = 2.44/min) = Let x0 = 10, e-x = e-2.44(10) = e-24.4 = .0000 b) P(x > 5 min = 2.44/min) = Let x0 = 5, e-x = e-2.44(5) = e-12.20 = .0000 c) P(x > 1 min = 2.44/min) = Let x0 = 1, e-x = e-2.44(1) = e-2.44 = .0872 d) Expected time = µ = 1 1 min. = .41 min = 24.6 sec. 2.44 Student’s Solutions Manual and Study Guide: Chapter 6 Page 22 6.31 = 1.31/ 1000 passengers 𝜇= (0.7634)(1,000) = 763.4 1 = 1 = .7634 1.31 a) P(x > 500): Let x0 = 500/1,000 passengers = .5 e-x = e-1.31(.5) = e-.655 = .5194 b) P(x < 200): Let x0 = 200/1,000 passengers = .2 e-x = e-1.31(.2) = e-.262 = .7695 P(x < 200) = 1 - .7695 = .2305 6.33 = 2/month Average number of time between rain = µ = = µ = 15 days 1 1 month = 15 days 2 P(x < 2 days = 2/month): 2 = .067/day 30 P(x < 2 days = .067/day) = Change to days: = 1 – P(x > 2 days = .067/day) let x0 = 2, 1 – e-x = 1 – e-.067(2) = 1 – .8746 = .1254 Student’s Solutions Manual and Study Guide: Chapter 6 6.35 a) P(x < 21 µ = 25 and = 4): z = x 21 25 = -1.00 4 From Table A.5, area = .3413 P(x < 21) = .5000 -.3413 = .1587 b) P(x > 77 µ = 50 and = 9): z = x 77 50 = 3.00 9 From Table A.5, area = .4987 P(x > 77) = .5000 -.4987 = .0013 c) P(x > 47 µ = 50 and = 6): z = x 47 50 = -0.50 6 From Table A.5, area = .1915 P(x > 47) = .5000 + .1915 = .6915 d) P(13 < x < 29 µ = 23 and = 4): z = x 13 23 = -2.50 4 From Table A.5, area = .4938 z = x 29 23 = 1.50 4 From Table A.5, area = .4332 P(13 < x < 29) = .4938 + 4332 = .9270 e) P(x > 105 µ = 90 and = 2.86): z = x 105 90 = 5.24 2.86 Page 23 Student’s Solutions Manual and Study Guide: Chapter 6 From Table A.5, area = .5000 P(x > 105) = .5000 - .5000 = .0000 6.37 a) P(x > 3 = 1.3): let x0 = 3 P(x > 3 = 1.3) = e-x = e-1.3(3) = e-3.9 = .0202 b) P(x < 2 = 2.0): Let x0 = 2 P(x < 2 = 2.0) = 1 - P(x > 2 = 2.0) = 1 – e-x = 1 – e-2(2) = 1 – e-4 = 1 - .0183 = .9817 c) P(1 < x < 3 = 1.65): P(x > 1 = 1.65): Let x0 = 1 e-x = e-1.65(1) = e-1.65 = .1920 P(x > 3 = 1.65): Let x0 = 3 e-x = e-1.65(3) = e-4.95 = .0071 P(1 < x < 3) = P(x > 1) - P(x > 3) = .1920 - .0071 = .1849 d) P(x > 2 = 0.405): Let x0 = 2 e-x = e-(.405)(2) = e-.81 = .4449 Page 24 Student’s Solutions Manual and Study Guide: Chapter 6 6.39 p = 1/5 = .20 Page 25 n = 150 P(x > 50): µ = 150(.20) = 30 = z = 150(.20)(.80) = 4.899 50.5 30 = 4.18 4.899 Area associated with z = 4.18 is .5000 P(x > 50) = .5000 - .5000 = .0000 6.41 µ = 237 = 54 a) P(x < 150): 𝑧= 150 − 237 = −1.61 54 from Table A.5, area for z = -1.61 is .4463 P(x < 80) = .5000 - .4463 = .0537 b) P(x > 400): 𝑧= 400 − 237 = 3.02 54 from Table A.5, area for z = 3.02 is .4987 P(x > 400) = .5000 - .4987 = .0013 Student’s Solutions Manual and Study Guide: Chapter 6 Page 26 c) P(120 < x < 185): 𝑧= 120 − 237 = −2.17 54 𝑧= 185 − 237 = −0.96 54 from Table A.5, area for z = -2.17 is .4850 area for z = -0.96 is .3315 P(120 < x < 185) = .4850 - .3315 = .1535 6.43 a = 18 b = 65 P(25 < x < 50) = = 50 25 25 = .5319 65 18 47 a b 65 18 = 41.5 2 2 1 1 1 = .0213 b a 65 18 47 f(x) = 6.45 µ = 951 = 96 a) P(x > 1000): z = x 1000 951 = 0.51 96 from Table A.5, the area for z = 0.51 is .1950 P(x > 1000) = .5000 - .1950 = .3050 Student’s Solutions Manual and Study Guide: Chapter 6 b) P(900 < x < 1100): z = z = x x 900 951 = -0.53 96 1100 951 = 1.55 96 from Table A.5, the area for z = -0.53 is .2019 the area for z = 1.55 is .4394 P(900 < x < 1100) = .2019 + .4394 = .6413 c) P(825 < x < 925): z = z = x x 825 951 = -1.31 96 925 951 = -0.27 96 from Table A.5, the area for z = -1.31 is .4049 the area for z = -0.27 is .1064 P(825 < x < 925) = .4049 - .1064 = .2985 d) P(x < 700): z = x 700 951 = -2.61 96 from Table A.5, the area for z = -2.61 is .4955 P(x < 700) = .5000 - .4955 = .0045 Page 27 Student’s Solutions Manual and Study Guide: Chapter 6 6.47 µ = 50,542 = 4,246 a) P(x > 60,000): from Table A.5, the area for z = 2.23 is .4871 P(x > 50,000) = .5000 - .4871 = .0129 b) P(x < 45,000): from Table A.5, the area for z = -1.31 is .4049 P(x < 40,000) = .5000 - .4049 = .0951 c) P(x > 40,000): from Table A.5, the area for z = -2.48 is .4934 P(x > 35,000) = .5000 + .4934 = .9934 d) P(44,000 < x < 52,000): from Table A.5, the area for z = -1.54 is .4382 the area for z = 0.34 is .1331 P(44,000 < x < 52,000) = .4382 + .1331 = .5713 Page 28 Student’s Solutions Manual and Study Guide: Chapter 6 6.49 = 88 = 6.4 a) P(x < 70): z = x 70 88 = -2.81 6.4 From Table A.5, area = .4975 P(x < 70) = .5000 - .4975 = .0025 b) P(x > 80): z = x 80 88 = -1.25 6.4 From Table A.5, area = .3944 P(x > 80) = .5000 + .3944 = .8944 c) P(90 < x < 100): z = x 100 88 = 1.88 6.4 From Table A.5, area = .4699 z = x 90 88 = 0.31 6.4 From Table A.5, area = .1217 P(90 < x < 100) = .4699 - .1217 = .3482 Page 29 Student’s Solutions Manual and Study Guide: Chapter 6 6.51 n = 150 p = .75 µ = np = 150(.75) = 112.5 = n p q 150(.75)(.25) = 5.3033 a) P(x < 105): correcting for continuity: x = 104.5 z = x 104.5 112.5 = -1.51 5.3033 from Table A.5, the area for z = -1.51 is .4345 P(x < 105) = .5000 - .4345 = .0655 b) P(110 < x < 120): correcting for continuity: x = 109.5, x = 120.5 z = 109.5 112.5 = -0.57 5.3033 z = 120.5 112.5 = 1.51 5.3033 from Table A.5, the area for z = -0.57 is .2157 the area for z = 1.51 is .4345 P(110 < x < 120) = .2157 + .4345 = .6502 c) P(x > 95): correcting for continuity: x = 95.5 z = 95.5 112.5 = -3.21 5.3033 from Table A.5, the area for -3.21 is .4993 P(x > 95) = .5000 + .4993 = .9993 Page 30 Student’s Solutions Manual and Study Guide: Chapter 6 6.53 Page 31 µ = 85,200 60% are between 75,600 and 94,800 94,800 –85,200 = 9,600 75,600 – 85,200 = 9,600 The 60% can be split into 30% and 30% because the two x values are equal distance from the mean. The z value associated with .3000 area is 0.84 z = .84 = x 94,800 85,200 = 11,428.57 6.55 = 3 hurricanes5 months a) P(x > 1 month = 3 hurricanes per 5 months): x = 1 month is 1/5 of the 5 month interval associated with . Thus, x0 = 1/5 P(x0 > 1) = e x0 = e-3(1/5) = e-0.6 = .5488 b) P(x < 2 weeks): 2 weeks = 0.5 month = 1/10 of the 5 month interval P(x < 2 weeks = 0.6 per month) = 1 - P(x > 2 weeks = 0.6 per month) But P(x > 2 weeks = 0.6 per month): Let x0 = 0.1 P(x > 2 weeks) = e x0 = e-3(1/10) = e-0.3 = .7408 P(x < 2 weeks) = 1 - P(x > 2 weeks) = 1 - .7408 = .2592 c) Average time = Expected time = µ = 1/ = 1.67 months Student’s Solutions Manual and Study Guide: Chapter 6 = 175 6.57 µ = 2087 If 20% are less, then 30% lie between x and µ. z.30 = -.84 z = x x 2087 175 -.84 = x = 1940 If 65% are more, then 15% lie between x and µ z.15 = -0.39 z = -.39 = x x 2087 175 x = 2018.75 If x is more than 85%, then 35% lie between x and µ. z.35 = 1.04 z = 1.04 = x x 2087 175 x = 2269 Page 32 Student’s Solutions Manual and Study Guide: Chapter 6 6.59 µ = 1,717,000 Page 33 = 50,940 P(x > 1,800,000): from table A.5 the area for z = 1.63 is .4484 P(x > 1,800,000) = .5000 - .4484 = .0516 P(x < 1,600,000): from table A.5 the area for z = -2.30 is .4893 P(x < 1,600,000) = .5000 - .4893 = .0107 1.07% of the time. 6.61 This is a uniform distribution with a = 11 and b = 32. The mean is (11 + 32)/2 = 21.5 and the standard deviation is (32 - 11)/ 12 = 6.06. Almost 81% of the time there are less than or equal to 28 sales associates working. One hundred percent of the time there are less than or equal to 34 sales associates working and never more than 34. About 23.8% of the time there are 16 or fewer sales associates working. There are 21 or fewer sales associates working about 48% of the time. Student’s Solutions Manual and Study Guide: Chapter 6 Page 34 6.63 The lengths of cell phone calls are normally distributed with a mean of 2.35 minutes and a standard deviation of .11 minutes. Almost 99% of the calls are less than or equal to 2.60 minutes, almost 82% are less than or equal to 2.45 minutes, over 32% are less than 2.3 minutes, and almost none are less than 2 minutes.
Source Exif Data:
File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.5 Linearized : No Page Count : 34 Language : en-US Tagged PDF : Yes Title : Chapter 6 Author : Ken Black Creator : Microsoft® Word 2010 Create Date : 2014:12:08 19:26:54-07:00 Modify Date : 2014:12:08 19:26:54-07:00 Producer : Microsoft® Word 2010EXIF Metadata provided by EXIF.tools