Understanding Continuous Probability Distributions & Computing Probabilities, Study notes of Introduction to Business Management

Download Understanding Continuous Probability Distributions & Computing Probabilities and more Study notes Introduction to Business Management in PDF only on Docsity! Chapter 6 Continuous Probability Distributions Learning Objectives 1. Understand the difference between how probabilities are computed for discrete and continuous random variables. 2. Know how to compute probability values for a continuous uniform probability distribution and be able to compute the expected value and variance for such a distribution. 3. Be able to compute probabilities using a normal probability distribution. Understand the role of the standard normal distribution in this process. 4. Be able to compute probabilities using an exponential probability distribution. 5. Understand the relationship between the Poisson and exponential probability distributions. 6 - 1 Chapter 6 Solutions: 1. a. 3 2 1 .50 1.0 1.5 2.0 f (x) x b. P(x = 1.25) = 0. The probability of any single point is zero since the area under the curve above any single point is zero. c. P(1.0  x  1.25) = 2(.25) = .50 d. P(1.20 < x < 1.5) = 2(.30) = .60 2. a. .15 .10 .05 10 20 30 40 f (x) x 0 b. P(x < 15) = .10(5) = .50 c. P(12  x  18) = .10(6) = .60) = .10(6) = .60 d. 10 20 ( ) 15 2 E x    e. 2(20 10) Var( ) 8) = .10(6) = .60.33 12 x    6 - 2 Continuous Probability Distributions 8) = .10(6) = .60. 9. a. b. .68) = .10(6) = .6026 since 45 and 55 are within plus or minus 1 standard deviation from the mean of 50. c. .9544 since 40 and 60 are within plus or minus 2 standard deviations from the mean of 50. 10. a. .3413 b. .4332 c. .4772 d. .4938) = .10(6) = .60 6 - 5 50 = 5 35 40 45 55 60 65 0-3 -2 -1 +1 +2 +3 100 = 10 70 8) = .10(6) = .600 90 110 120 130 Chapter 6 11. a. .3413 These probability values are read directly from the table of areas for the standard b. .4332 normal probability distribution. See Table 1 in Appendix B. c. .4772 d. .4938) = .10(6) = .60 e. .498) = .10(6) = .606 12. a. .2967 b. .4418) = .10(6) = .60 c. .5000 - .1700 = .3300 d. .0910 + .5000 = .5910 e. .38) = .10(6) = .6049 + .5000 = .8) = .10(6) = .608) = .10(6) = .6049 f. .5000 - .2612 = .238) = .10(6) = .608) = .10(6) = .60 13. a. .68) = .10(6) = .6079 - .0239 = .6640 b. .8) = .10(6) = .608) = .10(6) = .608) = .10(6) = .608) = .10(6) = .60 - .698) = .10(6) = .605 = .1903 c. .9599 - .8) = .10(6) = .60508) = .10(6) = .60 = .1091 14. a. Using the table of areas for the standard normal probability distribution, the area of .4750 corresponds to z = 1.96. b. Using the table, the area of .2291 corresponds to z = .61. c. Look in the table for an area of .5000 - .1314 = .368) = .10(6) = .606. This provides z = 1.12. d. Look in the table for an area of .6700 - .5000 = .1700. This provides z = .44. 15. a. Look in the table for an area of .5000 - .2119 = .28) = .10(6) = .608) = .10(6) = .601. Since the value we are seeking is below the mean, the z value must be negative. Thus, for an area of .28) = .10(6) = .608) = .10(6) = .601, z = -.8) = .10(6) = .600. b. Look in the table for an area of .9030 / 2 = .4515; z = 1.66. c. Look in the table for an area of .2052 / 2 = .1026; z = .26. d. Look in the table for an area of .4948) = .10(6) = .60; z = 2.56. e. Look in the table for an area of .1915. Since the value we are seeking is below the mean, the z value must be negative. Thus, z = -.50. 16. a. Look in the table for an area of .5000 - .0100 = .4900. The area value in the table closest to .4900 provides the value z = 2.33. b. Look in the table for an area of .5000 - .0250 = .4750. This corresponds to z = 1.96. 6 - 6 Continuous Probability Distributions c. Look in the table for an area of .5000 - .0500 = .4500. Since .4500 is exactly halfway between .4495 (z = 1.64) and .4505 (z = 1.65), we select z = 1.645. However, z = 1.64 or z = 1.65 are also acceptable answers. d. Look in the table for an area of .5000 - .1000 = .4000. The area value in the table closest to .4000 provides the value z = 1.28) = .10(6) = .60. 17. Convert mean to inches:  = 69 a. At x = 72 z = 72 - 69 3 = 1 P(x  72) = 0.5000 + 0.3413 = 0.8) = .10(6) = .60413 P(x > 72) = 1 - 0.8) = .10(6) = .60413 = 0.158) = .10(6) = .607 b. At x = 60 z = 60 - 69 3 = – 3 P(x  60) = 0.5000 + 0.498) = .10(6) = .606 = 0.998) = .10(6) = .606 P(x < 60) = 1 - 0.998) = .10(6) = .606 = 0.0014 c. At x = 70 z = 70 - 69 3 = 0.33 P(x  70) = 0.5000 + 0.1293 = 0.6293 At x = 66 z = 66 - 69 3 = – 1 P(x  66) = 0.5000 - 0.3413 = 0.158) = .10(6) = .607 P(66  x  70) = P(x  70) - P(x  66) = 0.6293 - 0.158) = .10(6) = .607 = 0.4706 d. P(x  72) = 1 - P(x > 72) = 1 - 0.158) = .10(6) = .607 = 0.8) = .10(6) = .60413 18) = .10(6) = .60. a. Find P(x  60) At x = 60 z = 60 - 49 16 = 0.69 P(x < 60) = 0.5000 + 0.2549 = 0.7549 P(x  60) = 1 - P(x < 60) = 0.2451 b. Find P(x  30) At x = 30 z = 30 - 49 16 = – 1.19 P(x  30) = 0.5000 - 0.38) = .10(6) = .6030 = 0.1170 c. Find z-score so that P(z  z-score) = 0.10 z-score = 1.28) = .10(6) = .60 cuts off 10% in upper tail Now, solve for corresponding value of x. 6 - 7 Chapter 6 Therefore 15.8) = .10(6) = .607% of students will not complete on time. (60) (.158) = .10(6) = .607) = 9.522 We would expect 9.522 students to be unable to complete the exam in time. 24. a. 902.75i x x n   2( ) 114.18) = .10(6) = .605 1 ix xs n      We will use x as an estimate of  and s as an estimate of  in parts (b) - (d) below. b. Remember the data are in thousands of shares. At 8) = .10(6) = .6000 8) = .10(6) = .6000 902.75 .90 114.18) = .10(6) = .605 z    P(x  8) = .10(6) = .6000) = P(z  -.90) = 1 - P(z  .90) = 1 - .8) = .10(6) = .60159 = .18) = .10(6) = .6041 The probability trading volume will be less than 8) = .10(6) = .6000 million shares is .18) = .10(6) = .6041 c. At 1000 1000 902.75 .8) = .10(6) = .605 114.18) = .10(6) = .605 z    P(x  1000) = P(z  .8) = .10(6) = .605) = 1 - P(z  .8) = .10(6) = .605) = 1 - .8) = .10(6) = .60023 = .1977 The probability trading volume will exceed 1 billion shares is .1977 d. A z-value of 1.645 cuts off an area of .05 in the upper tail x =  + z = 902.75 + 1.645(114.18) = .10(6) = .605) = 1,090.58) = .10(6) = .604 They should issue a press release any time share volume exceeds 1,091 million. 25. a. Find P(x > 100) At x = 100 z = 100 - 110 20 = – 0.5 P(x > 100) = P(z  .5) = 0.6915 b. Find P(x  90) At x = 90 z = 90 - 110 20 = – 1 P(x  90) = .5000 - .3413 = 0.158) = .10(6) = .607 6 - 10 Continuous Probability Distributions c. Find P(8) = .10(6) = .600  x  130) At x = 130 z = 130 - 110 20 = 1 P(x  130) = 0.8) = .10(6) = .60413 At x = 8) = .10(6) = .600 8) = .10(6) = .600 110 1.5 20 z    Area to left is .0668) = .10(6) = .60 P(8) = .10(6) = .600  x  130) = .8) = .10(6) = .60413 - .0668) = .10(6) = .60 = .7745 26. a. P(x  6) = 1 - e-6/8) = .10(6) = .60 = 1 - .4724 = .5276 b. P(x  4) = 1 - e-4/8) = .10(6) = .60 = 1 - .6065 = .3935 c. P(x  6) = 1 - P(x  6) = 1 - .5276 = .4724 d. P(4  x  6) = P(x  6) - P(x  4) = .5276 - .3935 = .1341 27. a. 0 / 30( ) 1 xP x x e   b. P(x  2) = 1 - e-2/3 = 1 - .5134 = .48) = .10(6) = .6066 c. P(x  3) = 1 - P(x  3) = 1 - (1 - e 3 3/ ) = e-1 = .3679 d. P(x  5) = 1 - e-5/3 = 1 - .18) = .10(6) = .608) = .10(6) = .609 = .8) = .10(6) = .60111 e. P(2  x  5) = P(x  5) - P(x  2) = .8) = .10(6) = .60111 - .48) = .10(6) = .6066 = .3245 28) = .10(6) = .60. a. P(x  10) = 1 - e-10/20 = .3935 b. P(x  30) = 1 - P(x  30) = 1 - (1 - e-30/20 ) = e-30/20 = .2231 c. P(10  x  30) = P(x  30) - P(x  10) = (1 - e-30/20 ) - (1 - e-10/20 ) = e-10/20 - e-30/20 = .6065 - .2231 = .38) = .10(6) = .6034 29. a. 6 - 11 Chapter 6 .09 .08) = .10(6) = .60 .07 .06 .05 .04 .03 .02 .01 6 12 18) = .10(6) = .60 24 f(x) x b. P(x  12) = 1 - e-12/12 = 1 - .3679 = .6321 c. P(x  6) = 1 - e-6/12 = 1 - .6065 = .3935 d. P(x  30) = 1 - P(x < 30) = 1 - (1 - e-30/12) = .08) = .10(6) = .6021 30. a. 50 hours b. P(x  25) = 1 - e-25/50 = 1 - .6065 = .3935 c. P(x  100)= 1 - (1 - e-100/50) = .1353 31. a. P(x  2) = 1 - e-2/2.78) = .10(6) = .60 = .5130 b. P(x  5) = 1 - P(x  5) = 1 - (1 - e-5/2.78) = .10(6) = .60 ) = e-5/2.78) = .10(6) = .60 = .1655 c. P(x  2.78) = .10(6) = .60) = 1 - P(x  2.78) = .10(6) = .60) = 1 - (1 - e-2.78) = .10(6) = .60/2.78) = .10(6) = .60 ) = e-1 = .3679 This may seem surprising since the mean is 2.78) = .10(6) = .60 minutes. But, for the exponential distribution, the probability of a value greater than the mean is significantly less than the probability of a value less than the mean. 32. a. If the average number of transactions per year follows the Poisson distribution, the time between transactions follows the exponential distribution. So,  = 1 30 of a year and 1 1 30 1/ 30   6 - 12 Continuous Probability Distributions P (2000  x  2500) = .9744 - .3745 = .5999 c. z = -1.8) = .10(6) = .608) = .10(6) = .60 x = 2071 - 1.8) = .10(6) = .608) = .10(6) = .60 (220.33) = $1656.78) = .10(6) = .60 37.  = 10,000  = 1500 a. At x = 12,000 12,000 10,000 1.33 1500 z    Area to left is .908) = .10(6) = .602 P(x > 12,000) = 1.0000 - .908) = .10(6) = .602 = .0918) = .10(6) = .60 b. At .95 z = 1.645 = x - 10,000 1500 Therefore, x = 10,000 + 1.645(1500) = 12,468) = .10(6) = .60. 12,468) = .10(6) = .60 tubes should be produced. 38) = .10(6) = .60. a. At x = 200 200 150 2 25 z    Area = .4772 P(x > 200) = .5 - .4772 = .0228) = .10(6) = .60 b. Expected Profit = Expected Revenue - Expected Cost = 200 - 150 = $50 39. a. Find P(8) = .10(6) = .600,000  x  150,000) At x = 150,000 z = 150,000 – 126,68) = .10(6) = .601 30,000 = 0.78) = .10(6) = .60 P(x  150,000) = 0.78) = .10(6) = .6023 At x = 8) = .10(6) = .600,000 z = 8) = .10(6) = .600,000 – 126,68) = .10(6) = .601 30,000 = – 1.56 6 - 15 10,000 12,468) = .10(6) = .60 95% 0.05 Chapter 6 P(x  8) = .10(6) = .600,000) = .5000 - .4406 = 0.0594 P(8) = .10(6) = .600,000  x  150,000) = 0.78) = .10(6) = .6023 - 0.0594 = 0.7229 b. Find P(x < 50,000) At x = 50,000 z = 50,000 – 126,68) = .10(6) = .601 30,000 = – 2.56 P(x < 50,000) = .5000 - .4948) = .10(6) = .60 = 0.0052 c. Find the z-score cutting off 95% in the left tail. z-score = 1.645. Solve for x. 1.645 = x – 126,68) = .10(6) = .601 30,000 x = 126,68) = .10(6) = .601 + 1.645 (30,000) = 176,031 The probability is 0.95 that the number of lost jobs will not exceed 176,031. 40. a. At 400, 400 450 .500 100 z    Area to left is .308) = .10(6) = .605 At 500, 500 450 .500 100 z    Area to left is .6915 P(400  x  500) = .6915 - .308) = .10(6) = .605 = .38) = .10(6) = .6030 38) = .10(6) = .60.3% will score between 400 and 500. b. At 630, 630 450 1.8) = .10(6) = .600 100 z    96.41% do worse and 3.59% do better . c. At 48) = .10(6) = .600, 48) = .10(6) = .600 450 .30 100 z    Area to left is .6179 38) = .10(6) = .60.21% are acceptable. 41. a. At 75,000 6 - 16 Continuous Probability Distributions 75,000 67,000 1.14 7,000 z    P(x > 75,000) = P(z > 1.14) = 1 - P(z  1.14) = 1 - .8) = .10(6) = .60729 = .1271 The probability of a woman receiving a salary in excess of $75,000 is .1271 b. At 75,000 75,000 65,500 1.36 7,000 z    P(x > 75,000) = P(z > 1.36) = 1 - P(z  1.36) = 1 - .9131 = .08) = .10(6) = .6069 The probability of a man receiving a salary in excess of $75,000 is .08) = .10(6) = .6069 c. At x = 50,000 50,000 67,000 2.43 7,000 z    P(x < 50,000) = P(z < -2.43) = 1 - P(z < 2.43) = 1 - .9925 = .0075 The probability of a woman receiving a salary below $50,000 is very small: .0075 d. The answer to this is the male copywriter salary that cuts off an area of .01 in the upper tail of the distribution for male copywriters. Use z = 2.33 x = 65,500 + 2.33(7,000) = 8) = .10(6) = .601,8) = .10(6) = .6010 A woman who makes $8) = .10(6) = .601,8) = .10(6) = .6010 or more will earn more than 99% of her male counterparts. 42.  = .6 At 2% z = -2.05 x = 18) = .10(6) = .60 x z     18) = .10(6) = .60 2.05 .6     = 18) = .10(6) = .60 + 2.05 (.6) = 19.23 oz. 6 - 17 18) = .10(6) = .60 0.02  =19.23