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# Quantitative Management BUSA 3060

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This 18 page Class Notes was uploaded by Miss Moriah Welch on Monday October 12, 2015. The Class Notes belongs to BUSA 3060 at Georgia Southwestern State University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/222076/busa-3060-georgia-southwestern-state-university in Business Administration at Georgia Southwestern State University.

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Date Created: 10/12/15

Chapter 6 Continuous Probability Distributions Learning Objectives 1 Understand the difference between how probabilities are computed for discrete and continuous random variables 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 Be able to compute probabilities using a normal probability distribution Understand the role of the standard normal distribution in this process Be able to compute probabilities using an exponential probability distribution Understand the relationship between the Poisson and exponential probability distributions Chapter 6 Solutions 1 f x 3 2 1 x 50 10 15 20 b PX 125 0 The probability of any single point is zero since the area under the curve above 7 P F any single point is zero P10 x 125 225 50 P120 lt X lt 15 230 60 0 10 20 30 40 PXlt 15 105 50 P12 g XS 18 106 60 E x 102015 2 833 Varx w ST 0 P S7 Continuous Probability Distributions 110 120 130 140 Mnutes PX 130 120 130 120 050 PX gt135 120 140 135 025 Ex w 130 minutes fx 15 10 5 x 0 1 2 3 P25 lt X lt 75 150 50 PX 30 130 30 PXgt60 140 40 Length of Interval 2612 2389 223 1 7 for 2389 S x S 2612 fx 223 0 elsewhere Note 1 223 0045 PXlt 250 0045250 2389 04995 Chapter 6 9 F ST 0 0 57 P Almost half drive the ball less than 250 yards PX 2 255 00452612 255 0279 P245 g X g 26000452602450675 PX 2 250 1 PXlt 250 1 04995 05005 The probability of anyone driving it 250 yards or more is 05005 With 60 players the expected number driving it 250 yards or more is 6005005 3003 Rounding I would expect 30 of these women to drive the ball 250 yards or more P12 X g 1205 058 40 PX 2 1202 088 64 Px lt1198 Px gt 1202 W v 0058 04 64 088 Therefore the probability is 04 64 68 P10000 g X lt 12000 2000 1 5000 40 The probability your competitor will bid lower than you and you get the bid is 40 P10000 g X lt 14000 4000 1 5000 80 A bid of 15000 gives a probability of 1 of getting the property Yes the bid that maximizes expected profit is 13000 The probability of getting the property with a bid of 13000 is P10000 g X lt 13000 3000 1 5000 60 The probability of not getting the property with a bid of 13000 is 40 The profit you will make if you get the property with a bid of 13000 is 3000 16000 13000 So your expected profit with a bid of 13000 is EP13000 6300040 1800 If you bid 15000 the probability of getting the bid is 1 but the profit if you do get the bid is only 1000 16000 15000 So your expected profit with a bid of 15000 is EP15000 1100000 1000 Continuous Probability Distributions 8 a i 10 I 70 80 90 100 110 120 130 9 a a 5 I 35 40 45 50 55 60 65 b 6826 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 I 3 2 l 0 1 2 3 a 3413 b 4332 c 4772 d 4938 Chapter 6 11 97 ST 0 97 ST 0 P S7 0 P F ST 3413 These probability values are read directly from the table of areas for the standard 4332 normal probability distribution See Table 1 in Appendix B 4772 4938 4986 2967 4418 5000 1700 3300 09105000 5910 38495000 8849 5000 2612 2388 6879 0239 6640 8888 6985 1903 9599 8508 1091 Using the table of areas for the standard normal probability distribution the area of 4750 corresponds to Z 196 Using the table the area of 2291 corresponds to Z 61 Look in the table for an area of 5000 1314 3686 This provides Z 112 Look in the table for an area of 6700 5000 1700 This provides Z 44 Look in the table for an area of 5000 2119 2881 Since the value we are seeking is below the mean the zvalue must be negative Thus for an area of 2881 Z 80 Look in the table for an area of 9030 2 451539 Z 166 Look in the table for an area of 2052 2 102639 Z 26 Look in the table for an area of 494839 Z 256 Look in the table for an area of 1915 Since the value we are seeking is below the mean the zvalue must be negative Thus Z 50 Look in the table for an area of 5000 0100 4900 The area value in the table closest to 4900 provides the value Z 233 Look in the table for an area of 5000 0250 4750 This corresponds to Z 196 0 P 97 ST 0 P 37 ST 0 Continuous Probability Distributions Look in the table for an area of 5000 0500 4500 Since 4500 is exactly halfway between 4495 Z 164 and 4505 Z 165 we seleth 1645 However Z 164 or Z 165 are also acceptable answers Look in the table for an area of 5000 1000 4000 The area value in the table closest to 4000 provides the value Z 128 Convert mean to inches u 69 AtX 72 Z 7269 1 3 PXS 72 0500003413 08413 PXgt 72 108413 01587 AtX 60 Z 60 69 7 3 PX2 60 0500004986 09986 PXlt 60 109986 00014 AtX 70 Z M 033 3 PXS 70 0500001293 06293 AtX 66 7 6669 7 Z 7 71 3 PX 66 0500003413 01587 P66 g X g 70 PX g 70PX g 6606293 01587 04706 PX g 72 1 PXgt 72 1 01587 08413 Find PX 2 60 AtX 60 6049 069 Z 16 PX lt 60 0500002549 07549 PX 2 60 1 PX lt 6002451 FindPX 30 AtX 30 2 M 7119 16 PX 30 05000 03830 01170 Find zscore so that PZ 2 zscore 010 Z score 128 cuts off 10 in upper tail Chapter 6 Now solve for corresponding value of X x 7 49 16 128 X 49 16128 6948 So 10 of subscribers spend 6948 minutes or more reading T116 WallStr ct Journal We haveu 35 and a 8 z 1gg 8 PXgt 50 PZgt 188 1 PZlt 188 1 9699 0301 The rainfall exceeds 5 inches in 301 of the Aprils PXlt 30 PZlt 63 PZgt 63 1 PZlt 63 1 7357 2643 The rainfall is less than 3 inches in 2643 of the Aprils Z 128 cuts off approximately 10 in the upper tail of a normal distribution X 35 1288 4524 If it rains 4524 inches or more April will be classified as extremely wet Weusey27anda 8 2117272 8 PX 11 P2 2 5000 4772 0228 The probability a randomly selected subscriber spends less than 11 hours on the computer is 025 z 163 PXgt 40 PZgt 163 1 Hz 163 1 9484 0516 516 of subscribers spend over 40 hours per week using the computer A zvalue of 84 cuts off an area of 20 in the upper tail X 27 848 3372 A subscriber who uses the computer 3372 hours or more would be classified as a heavy user 97 ST 0 ST Continuous Probability Distributions From the normal probability tables a Z value of 205 cuts off an area of approximately 02 in the upper tail of the distribution Xu 2039 100 2051513075 A score of 131 or better should qualify a person for membership in Mensa Usey44184 and a 90 At400 zw46 90 At500 Z 500744184 mi65 90 P0 g zlt 65 2422 P46 g zlt 0 1772 P400 S ZS 500 1772 2422 4194 The probability a worker earns between 400 and 500 is 4194 Must find the Z value that cuts off an area of 20 in the upper tail Using the normal tables we find Z 84 cuts off approximately 20 in the upper tail So Xu 2039 44184 8490 51744 Weekly earnings of 51744 or above will put a production worker in the top 20 2507 44184 N 90 At 250 z 7213 PXS 250 PZS 213 5000 4834 0166 The probability a randomly selected production worker earns less than 250 per week is 0166 z 72 Area to left is 5000 4772 0228 AtX 60 z w 2 Area to left is 0228 AtX 75 751 080 75 Areato left is 3085 P60 S X S 75 3085 0228 2857 Chapter 6 ST 0 P S7 90780 27 10 1 Area5000 34131587 Therefore 1587 of students Will not complete on time 60 1587 9522 We would expect 9522 students to be unable to complete the exam in time z 90275 n 7 2 s1im114185 n71 We Will use J as an estimate ofu and s as an estimate of fin parts b d below Rem ember the data are in thousands of shares At 800 Z 7 800790275 N 114185 790 PX 800 P2 901 PZ 90 1 8159 1841 The probability trading volume Will be less than 800 million shares is 1841 At1000 Z 71000790275 N 114185 85 PX21000 H22851 R28518023 1977 The probability trading volume Will exceed 1 billion shares is 1977 A Z value of 1645 cuts off an area of 05 in the upper tail Xu 2039 90275 16451141851090584 They should issue a press release any time share volume exceeds 1091 million Find PXgt 100 AtX 100 100110 Z 705 20 PXgt 100 P2 5 06915 Find PX 90 AtX 90 0 Z 90 110 20 FindP80 g X g 130 AtX 130 7 130 110 7 Z 20 AtX 80 807110 27 20 PXS 6 1 668 1 4724 5276 PXS 4 1 a 16065 3935 PX2 6 1PXS 6 15276 4724 Continuous Probability Distributions PX S 90 5000 3413 01587 PX g 130 08413 715 Area to left is 0668 P80 S X S 130 8413 0668 7745 P4 g XS 6 ng 6PXS 4 52763935 1341 PxSx01iequot 3 PXS 2 1 62 3 1 5134 4866 H212 3 1PXS 3 1 1 e PXS 5 1 65 3 11889 8111 P2 S XS 5 PXS 5PXS 2 PXlt 10 1 61020 3935 6391 3679 8111 4866 3245 PXgt 30 1PXS 3011emZ arm2 2231 P10 g x 30 ng 30PXS 10 1 31120 1 6r11120 6 0 6065 2231 3834 Chapter 6 7 P S7 0 ST 0 QBSESSSSSx 5 6 12 18 24 PX 12 1 6 13679 6321 PX 6 1 6612 16065 3935 PX2 30 lPXlt 30 1 1 63012 0821 50 hours PX 25 1 62550 16065 3935 PX 2 100 1 1 610050 1353 PXlt 2 1 62278 5130 PXgt 5 lPXS 5 1 1 c395 8e395 8 1655 PXgt 278 lPXS 278 1 1 6278 6391 3679 This may seem surprising since the mean is 278 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 ST 0 Continuous Probability Distributions If the average number of transactions per year follows the Poisson distribution the time between transactions follows the exponential distribution So iofa ear I 30 y and l 30 1 130 then fx 30 630 A month is 112 ofa year so P xgti liP xgi 1717e 30 2e 30 20821 12 12 The probability of no transaction during January is the same as the probability of no transaction during any month 0821 Since 12 month is 124 ofa year we compute P xgi 17673024 172865 7135 24 Let X sales price 1000s 1 fx 25 0 elsewhere for 200 g x g 225 PX 2 215 1 25 225 215 040 PXlt 210 1 25210 200 040 Ex 200 2252 212500 If she waits her expected sale price will be 2500 higher than if she sells it back to her company now However there is a 040 probability that she will get less It s a close call But the expected value approach to decision making would suggest she should wait For a normal distribution the mean and the median are equal u 63000 Find the zscore that cuts off 10 in the lower tail Z score 128 Solving for g 7128 T X 7 63 000 000 Chapter 6 0 P X 63000 128 15000 43800 The lower 10 of mortgage debt is 43800 or less Find PXgt 80000 AtX 80000 Z 80000 763000 1 13 15000 39 PX gt 80000 10000 8708 01292 Find the zscore that cuts off 5 in the upper tail Z score 1645 Solve for X X7 63 000 1645 Th 000 X 63000 1645 15000 87675 The upper 5 ofmortgage debt is in excess of 87675 Hdcfcct 1P985 g x 1015 1P1 g z 1 l 6826 3174 Expected number ofdefects 10003174 3174 Hdcfcct 1P985 g x 1015 1P3 g 2 g 3 l 9972 0028 Expected number of defects 10000028 28 Reducing the process standard deviation causes a substantial reduction in the number of defects At11 Z 123 Continuous Probability Distributions X 7 7123 1800 2071 U U 22033 7 1800 72071 Therefore a 7 7123 b wym Areatoleftis500032553745 22033 w195 Area to left is9744 22033 P2000 g X g 2500 9744 3745 5999 c Z 188 X 2071 188 22033 165678 37 1 10000 a 1500 a AtX 12000 27 12000710000 1500 133 Area to left is 9082 PXgt 12000 100009082 0918 b At95 Z 1645 X 10000 150 Therefore X 10000 16451500 12468 1 95 005 K 10000 12468 12468 tubes should be produced 38 a AtX 200 z 2 Area4772 PXgt 200 54772 0228 b Expected Profit Expected Revenue Expected Cost Chapter 6 39 a 7 0 200 150 50 Find P80000 g X 150000 AtX 150000 7 150000 7126681 27 3OOOO 078 PX g 150000 07823 AtX 80000 Z 80000 7126681 7156 30000 PX g 80000 5000 4406 00594 P80000 g X 150000 07823 00594 07229 Find PXlt 50000 AtX 50000 50000 7126681 7256 30000 PX lt 50000 5000 4948 00052 Find the ZSCOI C cutting off 95 in the left tail Solve for X Z score 1645 X7 126 681 1645 30000 X 126681 1645 30000 176031 The probability is 095 that the number of lostjobs will not exceed 176031 At 400 z M 7500 100 Area to left is 3085 At 500 z M 500 100 Area to left is 6915 P400 X g 500 6915 3085 3830 383 will score between 400 and 500 At 630 Continuous Probability Distributions z M180 100 9641 do worse and 359 do better c At 480 z M 30 100 Area to left is 6179 3821 are acceptable 41 a At 75000 Z 7 75 000 7 67 000 7000 114 PXgt 75000 Pzgt 114 1 PZ 114 1 8729 1271 The probability of a woman receiving a salary in excess of 75000 is 1271 b At 75000 Z 71007060601500 m136 PXgt 75000 PZgt 136 1 PZ 136 1 9131 0869 The probability of a man receiving a salary in excess of 75000 is 0869 c AtX 50000 Z 50 007060607 000 m 7243 PXlt 50000 PZlt 243 1 PZlt 243 1 9925 0075 The probability of a woman receiving a salary below 50000 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 233 X 65500 2337000 81810 A woman who makes 81810 or more will earn more than 99 of her male counterparts 42 c7 6 At 2 Chapter 6 ST 0 ST 0 P 0 7 P 18205 6 1923 oz 18 1923 The mean filling weight must be 1923 oz PX 15 1 651536 16592 3408 PX 45 1 63945 12865 7135 Therefore P15 x 45 71353408 3727 PX 2 60 1 1 63960361889 lPXlt60 4hours X 14 6quot forX2 0 PX2 1 lPXlt 1 1 1 63914 7788 PXgt 8 lPXS 8 684 1353 fxequot12 forX 2 0 P5 g x 10 Rx 10Rx 5 1639112l 6512 5654 3408 2246 PXgt 1 1PX 1 15654 4346 l 05 therefore u 2 minutes mean time between telephone calls u Note 30 seconds 5 minutes PX 5 1 652 17788 2212 PX 1 1 612 16065 3935 PX2 5 lPXlt 5 1065 0821

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