Methods for Describing Sets of Data Chapter 2

Transcription

1 Methods for Describig Sets of Data Chapter. a. To fid the frequecy for each class, cout the umber of times each letter occurs. The frequecies for the three classes are: Class Frequecy X 8 Y 9 Z 3 Total 0 b. The relative frequecy for each class is foud by dividig the frequecy by the total sample size. The relative frequecy for the class X is 8/0.40. The relative frequecy for the class Y is 9/0.45. The relative frequecy for the class Z is 3/0.15. Class Frequecy Relative Frequecy X 8.40 Y 9.45 Z 3.15 Total c. The frequecy bar chart is: d. The pie chart for the frequecy distributio is: Methods for Describig Sets of Data 5

2 .4 a. The variable summarized i the table is Reaso for requestig the istallatio of the passeger-side o-off switch. The values this variable could assume are: Ifat, Child, Medical, Ifat & Medical, Child & Medical, Ifat & Child, ad Ifat & Child & Medical. Sice the resposes ame somethig, the variable is qualitative. b. The relative frequecies are foud by dividig the umber of requests for each category by the total umber of requests. For the category Ifat, the relative frequecy is 1,85/30, The rest of the relative frequecies are foud i the table below: Reaso Number of Relative Requests frequecies Ifat 1,85 1,85/30, Child 17,148 17,148/30, Medical 8,377 8,377/30, Ifat & Medical 44 44/30, Child & Medical /30, Ifat & Child 1,878 1,878/30, Ifat & Child & Medical /30, TOTAL 30, c. Usig MINITAB, a pie chart of the data is: Pie Chart of Reaso Child (17148, 56.5%) Child&Medica ( 903, 3.0%) If &Chd&Med ( 135, 0.4%) Ifat ( 185, 6.1%) If at&child ( 1878, 6.%) Ifat&Medic ( 44, 0.1%) Medical ( 8377, 7.6%) d. There are 4 categories where Medical is metioed as a reaso: Medical, Ifat & Medical, Child & Medical, ad Ifat & Child & Medical. The sum of the frequecies for these 4 categories is 8, ,459. The proportio listig Medical as oe of the reasos is 9,459/30, Chapter

3 .6 a. To fid relative frequecies, we divide the frequecies of each category by the total umber of icidets. The relative frequecies of the umber of icidets for each of the cause categories are: Maagemet System Number of Icidets Relative Frequecies Cause Category Egieerig & Desig 7 7 / Procedures & Practices 4 4 / Maagemet & Oversight / Traiig & Commuicatio / TOTAL 83 1 b. The Pareto diagram is: 35 Maagemet Syste Cause Category 30 5 Percet Eg&Des Proc&Pract Mgmt&Over Category Tr&Comm c. The category with the highest relative frequecy of icidets is Egieerig ad Desig. The category with the lowest relative frequecy of icidets is Traiig ad Commuicatio..8 a. The data collectio method was a survey. b. Sice the data were umbers (percetage of US labor ad materials), the variable is quatitative. Oce the data were collected, they were grouped ito 4 categories. Methods for Describig Sets of Data 7

4 c. Usig MINITAB, a pie chart of the data is: Pie Chart of Made i USA 100% (64, 60.4%) <50% ( 4, 3.8%) 75-99% (0, 18.9%) 50-74% (18, 17.0%) About 60% of those surveyed believe that Made i USA meas 100% US labor ad materials..10 Usig MINITAB, a bar chart of the frequecy of occurrece of the idustry types is: Chart of INDUSTRY Cout Aerospace/Defese Bakig Capital Goods Chemicals Coglomerates Costructio Cosumer Durables Diversified Fiacials Drugs/Biotechology Food Markets Food/Drik/Tobacco Health Care Hotels/Restaurats/Leisure Household/Persoal Products Isurace Materials Media Oil & Gas Retailig Semicoductors Services/Supplies Software & Services Techology Equipmet Telecommuicatios Trasportatio Utilities INDUSTRY 8 Chapter

5 .1 Usig MINITAB, the side-by-side bar charts are: Chart of 1999, 006 vs Use Yes No Do't kow Relative Frequecy Yes No Do't kow Uathorized Use of COmputer Systems The relative frequecy of uauthorized use of computer systems has decreased from 1999 to a. Usig MINITAB, the side-by-side graphs are: Chart of Exposure, Opportuity, Cotet, Faculty vs Stars Exposure Opportuity Frequecy 16 Cotet Faculty Stars From these graphs, oe ca see that very few of the top 30 MBA programs got 5-stars i ay criteria. I additio, about the same umber of programs got 4 stars i each of the 4 criteria. The biggest differece i ratigs amog the 4 criteria was i the umber of programs receivig 3-stars. More programs received 3-stars i Course Cotet tha i ay of the other criteria. Cosequetly, fewer programs received -stars i Course Cotet tha i ay of the other criteria. b. Sice this chart lists the rakigs of oly the top 30 MBA programs i the world, it is reasoable that oe of these best programs would be rated as 1-star o ay criteria. Methods for Describig Sets of Data 9

6 a. The origial data set has observatios. b. For the bottom row of the stem-ad-leaf display: The stem is 0. The leaves are 0, 1,. The umbers i the origial data set are 0, 1, ad. c. The dot plot correspodig to all the data poits is:.0. a. The measuremet class that cotais the highest proportio of respodets is oe. Sixty-oe percet of the respodets said that their compaies did ot outsource ay computer security fuctios. b. From the graph, 6% of the respodets idicated that they outsourced betwee 0% ad 40% of their computer security fuctios. c. The proportio of the 609 respodets who outsourced at least 40% of computer security fuctios is d. The umber of the 609 respodets who outsourced less tha 0% of computer security fuctios is ( )*609.88(609) Chapter

7 . a. Usig MINITAB, the stem-ad-leaf display of the data is: Stem-ad-Leaf Display: SCORE Stem-ad-leaf of SCORE N 169 Leaf Uit (100) b. From the stem-ad-leaf display, we see that there are oly 4 observatios with saitatio scores less tha the acceptable score of 86. The proportio of ships that have a accepted saitatio stadard would be (169 4) / c. The saitatio score of 84 is i bold i the stem-ad-leaf display i part a..4 a. Usig MINITAB, the frequecy histogram is: 30 Frequecy Legth Methods for Describig Sets of Data 11

8 b. Usig MINITAB, the frequecy histogram is: Frequecy Weight c. Usig MINITAB, the frequecy histogram is: Frequecy DDT Usig MINITAB, the two dot plots are: Dotplot for Arrive-Depart Yes. Most of the umbers of items arrivig at the work ceter per hour are i the 135 to 165 area. Most of the umbers of items departig the work ceter per hour are i the 110 to 140 area. Because the umber of items arrivig is larger tha the umber of items departig, there will probably be some sort of bottleeck. 1 Chapter

9 .8 a. Usig MINITAB, the three frequecy histograms are as follows (the same startig poit ad class iterval were used for each): Histogram of C1 N 5 Teth Performace Midpoit Cout * ***** ********** ****** ** * Histogram of C N 5 Thirtieth Performace Midpoit Cout * ********* ************ ** * Histogram of C3 N 5 Fiftieth Performace Midpoit Cout *** *************** **** ** * b. The histogram for the teth performace shows a much greater spread of the observatios tha the other two histograms. The thirtieth performace histogram shows a shift to the left implyig shorter completio times tha for the teth performace. I additio, the fiftieth performace histogram shows a additioal shift to the left compared to that for the thirtieth performace. However, the last shift is ot as great as the first shift. This agrees with statemets made i the problem. Methods for Describig Sets of Data 13

10 .30 a. A stem-ad-leaf display is as follows, where the stems are the uits place ad the leaves are the decimal places: Stem Leaves b. A little more tha half (6/49.53) of all compaies spet less tha moths i bakruptcy. Oly two of the 49 compaies spet more tha 6 moths i bakruptcy. It appears the, i geeral, the legth of time i bakruptcy for firms usig "prepacks" is less tha that of firms ot usig "prepacks." c. A dot diagram will be used to compare the time i bakruptcy for the three types of "prepack" firms: d. The circled times i part a correspod to compaies that were reorgaized through a leverage buyout. There does ot appear to be ay patter to these poits. They appear to be scattered about evely throughout the distributio of all times..3 Usig MINITAB, the stem-ad-leaf display for the data is: Stem-ad-leaf of Time N 5 Leaf Uit (7) The umbers i bold represet delivery times associated with customers who subsequetly did ot place additioal orders with the firm. Sice there were oly customers with delivery times of 68 days or loger that placed additioal orders, I would say the maximum tolerable delivery time is about 65 to 67 days. Everyoe with delivery times less tha 67 days placed additioal orders. 14 Chapter

11 .34 a. x b. x c. ( x 5) (3 5) + (8 5) + (4 5) + (5 5) + (3 5) + (4 5) + (6 5) 0 d. ( x ) (3 ) + (8 ) + (4 ) + (5 ) + (3 ) + (4 ) + (6 ) 71 e. ( x) ( ) a. x ( ) + ( 1) b. x ( ) + ( 1) c..38 a. b. c. d. ( ) 50 5 x x x 85 x x x x The media is the middle umber oce the data have bee arraged i order. If is eve, there is ot a sigle middle umber. Thus, to compute the media, we take the average of the middle two umbers. If is odd, there is a sigle middle umber. The media is this middle umber. A data set with five measuremets arraged i order is 1, 3, 5, 6, 8. The media is the middle umber, which is 5. A data set with six measuremets arraged i order is 1, 3, 5, 5, 6, 8. The media is the average of the middle two umbers which is Methods for Describig Sets of Data 15

12 x a. x Media (mea of 3rd ad 4th umbers, after orderig) Mode 3 x b. x Media 3 (7th umber, after orderig) Mode 3 x c. x Media 49 (mea of 5th ad 6th umbers, after orderig) Mode a. The sample mea is: xi i x The sample media is foud by fidig the average of the 13 th ad 14 th observatios oce the data are arraged i order. The 13 th ad 14 th observatios are 100 ad 105. The average of these two umbers (media) is: media 10.5 The mode is the observatio appearig the most. For this data set, the mode is 70, which appears 3 times. Sice the mea is larger tha the media, the data are skewed to the right. b. The sample mea is: xi i x The sample media is foud by fidig the average of the 13 th ad 14 th observatios oce the data are arraged i order. The 13 th ad 14 th observatios are 5 ad 5. The average of these two umbers (media) is: media 5 16 Chapter

13 The mode is the observatio appearig the most. For this data set, the mode is 6, which appears 6 times. Sice the mea ad media are about the same, the data are somewhat symmetric..46 a. The sample mea is: xi i x The sample average surface roughess of the 0 observatios is b. The media is foud as the average of the 10 th ad 11 th observatios, oce the data have bee ordered. The ordered data are: The 10 th ad 11 th observatios are.03 ad.05. The media is: The middle surface roughess measuremet is.04. Half of the sample measuremets were less tha.04 ad half were greater tha.04. c. The data are somewhat skewed to the left. Thus, the media might be a better measure of cetral tedecy tha the mea. The few small values i the data ted to make the mea smaller tha the media..48 a. Usig MINITAB, the stem-ad-leaf display is: Stem-ad-leaf of PAF N17 Leaf Uit () b. The media is the middle umber oce the data are arraged i order. The data arraged i order are: 0, 0, 0, 0, 0, 9, 1, 15, 4, 5, 31, 33, 40, 6, 70, 75, 77. The middle umber or the media is 4. c. The mea of the data is x x Methods for Describig Sets of Data 17

14 d. The umber occurrig most frequetly is 0. The mode is 0. e. The mode correspods to the smallest umber. It does ot seem to locate the ceter of the distributio. Both the mea ad the media are i the middle of the stem-ad-leaf display. Thus, it appears that both of them locate the ceter of the data..50 a. The sample mea legth is: xi i x The average legth of the 144 fish is 4.81 cm. The media is the average of the middle two observatios oce they have bee ordered. The 7 d ad 73 rd observatios are 45 ad 45. The average of these two observatios is 45. Half of the fish legths are less tha 45 cm ad half are loger. The mode is 46 cm. This observatio occurred 1 times. b. The sample mea weight is: xi i x The average weight of the 144 fish is grams. The media is the average of the middle two observatios oce they have bee ordered. The 7 d ad 73 rd observatios are 989 ad The average of these two observatios is ,011 media 1000 Half of the fish weights are less tha 1000 grams ad half are heavier. There are modes, 886 ad Each of these observatios occurred 3 times. c. The sample mea DDT level is: xi i x The average DDT level of the 144 fish is 4.35 parts per millio. 18 Chapter

15 The media is the average of the middle two observatios oce they have bee ordered. The 7 d ad 73 rd observatios are 7.1 ad 7.. The average of these two observatios is media 7.15 Half of the fish DDT levels are less tha 7.15 parts per millio ad half are greater. The mode is 1. This observatio occurred 8 times. d. From the graph i Exercise.4a, the data are skewed to the left. This correspods to the relatioship betwee the mea ad the media. For data skewed to the left, the mea is less tha the media. For the fish legths, the mea is 4.81 ad the media is 45. e. From the graph i Exercise.4b, the data are slightly skewed to the right. This correspods to the relatioship betwee the mea ad the media. For data skewed to the right, the mea is more tha the media. For the fish weights, the mea is ad the media is f. From the graph i Exercise.4c, the data are skewed to the right. This correspods to the relatioship betwee the mea ad the media. For data skewed to the right, the mea is more tha the media. For the fish DDT levels, the mea is 4.35 ad the media is a. Due to the "elite" superstars, the salary distributio is skewed to the right. Sice this implies that the media is less tha the mea, the players' associatio would wat to use the media. b. The owers, by the logic of part a, would wat to use the mea..54 a. The sample mea is: x i 1 x i The sample media is foud by fidig the average of the 10 th ad 11 th observatios oce the data are arraged i order. The data arraged i order are: The 10 th ad 11 th observatios are 3 ad 4. The average of these two umbers (media) is: media 3.5 The mode is the observatio appearig the most. For this data set, the mode is 1, which appears 5 times. Methods for Describig Sets of Data 19

16 b. Elimiatig the largest umber which is 13 results i the followig: The sample mea is: x i 1 x i The sample media is foud by fidig the middle observatio oce the data are arraged i order. The data arraged i order are: The 10 th observatio is 3. The media is 3 The mode is the observatios appearig the most. For this data set, the mode is 1, which appears 5 times. By droppig the largest umber, the mea is reduced from 4 to The media is reduced from 3.5 to 3. There is o effect o the mode. c. The data arraged i order are: If we drop the lowest ad largest observatios we are left with: The sample 10% trimmed mea is: xi i x The advatage of the trimmed mea over the regular mea is that very large ad very small umbers that could greatly affect the mea have bee elimiated. 0 Chapter

17 .56 a. s b. s c. s ( x) x 1 ( x) x 1 ( x) x s ` s s a. Rage s ( x) x s b. Rage s ( x) x 1 c. Rage s ( x) x , , ,949.5 s 1, , s 1, This is oe possibility for the two data sets. x 1 x Data Set 1: 1, 1,,, 3, 3, 4, 4, 5, 5 Data Set : 1, 1, 1, 1, 1, 5, 5, 5, 5, 5 x x Therefore, the two data sets have the same mea. The variaces for the two data sets are: s 1 s ( ) x 30 x ( ) x 30 x Methods for Describig Sets of Data 1

18 The dot diagrams for the two data sets are show below..6 a. Rage s ( ) x 7 x s b. After addig 3 to each of the data poits, s Rage ( ) x x s c. After subtractig 4 from each of the data poits, Rage 1 ( 4) 3 s ( ) x ( 13) x s d. The rage, variace, ad stadard deviatio remai the same whe ay umber is added to or subtracted from each measuremet i the data set..64 a. The maximum age is 64. The miimum age is 39. The rage is b. The variace is: s i 1 xi i 1 x , c. The stadard deviatio is: s s d. Sice the stadard deviatio of the ages of the 50 most powerful wome i Europe is 10 years ad is greater tha that i the U.S. (5.75 years), the age data for Europe is more variable. Chapter

19 .66 a. The maximum weight is 1.1 carats. The miimum weight is.18 carats. The rage is carats. b. The variace is: s xi i x i i square carats c. The stadard deviatio is: s s carats d. The stadard deviatio. This gives us a idea about how spread out the data are i the same uits as the origial data..68 a. A worker's overall time to complete the operatio uder study is determied by addig the subtask-time averages. Worker A x 11 The average for subtask 1 is: x x 1 The average for subtask is: x 3 7 Worker A's overall time is Worker B x 13 The average for subtask 1 is: x x 9 The average for subtask is: x Worker B's overall time is b. Worker A s ( x) 7 11 x Worker B s ( x) 13 x c. The stadard deviatios represet the amout of variability i the time it takes the worker to complete subtask 1. Methods for Describig Sets of Data 3

20 d. Worker A s ( x) 1 x Worker B s ( x) 9 x e. I would choose workers similar to worker B to perform subtask 1. Worker B has a slightly higher average time o subtask 1 (A: x 30.14, B: x 30.43). But, Worker B has a smaller variability i the time it takes to complete subtask 1 (part b). He or she is more cosistet i the time eeded to complete the task. I would choose workers similar to Worker A to perform subtask. Worker A has a smaller average time o subtask (A: x 3, B: x 4.14). Worker A also has a smaller variability i the time eeded to complete subtask (part d)..70 Sice o iformatio is give about the data set, we ca oly use Chebyshev's Rule. a. Nothig ca be said about the percetage of measuremets which will fall betwee x s ad x + s. b. At least 3/4 or 75% of the measuremets will fall betwee x s ad x + s. c. At least 8/9 or 89% of the measuremets will fall betwee x 3s ad x + 3s..7 a. x s x ( x) 06 x s s 1.83 b. Iterval Number of Measuremets i Iterval Percetage x ± s, or (6.41, 10.07) 18 18/5.7 or 7% x ± s, or (4.58, 11.90) 4 4/5.96 or 96% x ± 3s, or (.75, 13.73) 5 5/5 1 or 100% c. The percetages i part b are i agreemet with Chebyshev's Rule ad agree fairly well with the percetages give by the Empirical Rule. 4 Chapter

21 d. Rage s rage/4 7/ The rage approximatio provides a satisfactory estimate of s 1.83 from part a..74 From Chebyshev s Theorem, we kow that at least ¾ or 75% of all observatios will fall withi stadard deviatios of the mea. From Exercise.47, x.631. From Exercise.66, s.77. This iterval is: x ± s.631± (.77).631 ±.5544 (.0766, ).76 a. From the iformatio give, we have x 375 ad s 5. From Chebyshev's Rule, we kow that at least three-fourths of the measuremets are withi the iterval: x ± s, or (35, 45) Thus, at most oe-fourth of the measuremets exceed 45. I other words, more tha 45 vehicles used the itersectio o at most 5% of the days. b. Accordig to the Empirical Rule, approximately 95% of the measuremets are withi the iterval: x ± s, or (35, 45) This leaves approximately 5% of the measuremets to lie outside the iterval. Because of the symmetry of a moud-shaped distributio, approximately.5% of these will lie below 35, ad the remaiig.5% will lie above 45. Thus, o approximately.5% of the days, more tha 45 vehicles used the itersectio..78 a. Sice the sample mea (18.) is larger tha the sample media (15), it idicates that the distributio of years is skewed to the right. I additio, the maximum umber of years is 50 ad the miimum is. If the distributio were symmetric, the mea ad media should be about halfway betwee these two umbers. Halfway betwee the maximum ad miimum values is 6, which is much larger tha either the mea or the media. b. The stadard deviatio ca be estimated by the rage divided by either 4 or 6. For this distributio, the rage is: Rage Largest smallest Dividig the rage by 4, we get a estimate of the stadard deviatio to be 48/4 1. Dividig the rage by 6, we get a estimate of the stadard deviatio to be 48/6 8. Thus, the stadard deviatio should be somewhere betwee 8 ad 1. For this problem, the stadard deviatio is s This value falls i the estimated rage of 8 to 1. Methods for Describig Sets of Data 5

22 c. First, we calculate the umber of stadard deviatios from the mea the value of 40 years is. To do this, we first subtract the mea ad the divide by the value of the stadard deviatio. 40 x Number of stadard deviatios is.05 s Usig Chebyshev's Rule, we kow that at most 1/k or 1/ 1/4 of the data will be more tha stadard deviatios from the mea. Thus, this would idicate that at most 5% of the Geeratio Xers respoded with 40 years or more. Next, we calculate the umber of stadard deviatios from the mea the value of 8 years is. Number of stadard deviatios is 8 x s Usig Chebyshev's Rule, we get o iformatio about the data withi 1 stadard deviatio of the mea. However, we kow the media (15) is more tha 8. By defiitio, 50% of the data are larger tha the media. Thus, at least 50% of the Geeratio Xers respoded with 8 years or more. No additioal iformatio ca be obtaied with the iformatio give..80 a. Usig MINITAB, the frequecy histogram for the time i bakruptcy is: 0 Frequecy Time i Bakrupt The Empirical Rule is ot applicable because the data are ot moud shaped. 6 Chapter

23 b. Usig MINITAB, the descriptive measures are: Descriptive Statistics: Time i Bakrupt Variable N Mea Media TrMea StDev SE Mea Time i Variable Miimum Maximum Q1 Q3 Time i From Chebyshev s Theorem, we kow that at least 75% of the observatios will fall withi stadard deviatios of the mea. This iterval is: x ± s.549 ± (1.88).549 ± ( 1.107, 6.05) c. There are 47 of the 49 observatios withi this iterval. The percetage would be (47/49)*100% 95.9%. This agrees with Chebyshev s Theorem (at least 75%0. It also agrees with the Empirical Rule (approximately 95%). d. From the above iterval we kow that about 95% of all firms filig for prepackaged bakruptcy will be i bakruptcy betwee 0 ad 6. moths. Thus, we would estimate that a firm cosiderig filig for bakruptcy will be i bakruptcy up to 6. moths..8 a. Sice it is give that the distributio is moud-shaped, we ca use the Empirical Rule. We kow that 1.84% is stadard deviatios below the mea. The Empirical Rule states that approximately 95% of the observatios will fall withi stadard deviatios of the mea ad, cosequetly, approximately 5% will lie outside that iterval. Sice a moud-shaped distributio is symmetric, the approximately.5% of the day's productio of batches will fall below 1.84%. b. If the data are actually moud-shaped, it would be extremely uusual (less tha.5%) to observe a batch with 1.80% zic phosphide if the true mea is.0%. Thus, if we did observe 1.8%, we would coclude that the mea percet of zic phosphide i today's productio is probably less tha.0%..84 a. Sice we do ot have ay idea of the shape of the distributio of SAT-Math score chages, we must use Chebyshev s Theorem. We kow that at least 8/9 of the observatios will fall withi 3 stadard deviatios of the mea. This iterval would be: x ± 3s 19 ± 3(65) 19 ± 195 ( 176, 14) Thus, for a radomly selected studet, we could be pretty sure that this studet s score would be ay where from 176 poits below his/her previous SAT-Math score to 14 poits above his/her previous SAT-Math score. b. Sice we do ot have ay idea of the shape of the distributio of SAT-Verbal score chages, we must use Chebyshev s Theorem. We kow that at least 8/9 of the observatios will fall withi 3 stadard deviatios of the mea. This iterval would be: x ± 3s 7 ± 3(49) 7 ± 147 ( 140, 154) Methods for Describig Sets of Data 7

24 Thus, for a radomly selected studet, we could be pretty sure that this studet s score would be ay where from 140 poits below his/her previous SAT-Verbal score to 154 poits above his/her previous SAT-Verbal score. c. A chage of 140 poits o the SAT-Math would be a little less tha stadard deviatios from the mea. A chage of 140 poits o the SAT-Verbal would be a little less tha 3 stadard deviatios from the mea. Sice the 140 poit chage for the SAT-Math is ot as big a chage as the 140 poit o the SAT-Verbal, it would be most likely that the score was a SAT-Math score..86 a. z x x s (sample) stadard deviatios above the mea. 5 b. z c. z x μ (populatio).5 stadard deviatios above the mea. σ x μ (populatio) 0 stadard deviatios above the mea. σ 5 d. z x x s (sample).5 stadard deviatios below the mea The 50th percetile of a data set is the observatio that has half of the observatios less tha it. Aother ame for the 50th percetile is the media..90 Sice the elemet 40 has a z-score of ad 90 has a z-score of 3, 40 μ ad 3 90 μ σ σ σ 40 μ 3σ 90 μ μ σ 40 μ + 3σ 90 μ 40 + σ By substitutio, 40 + σ + 3σ 90 5σ 50 σ 10 By substitutio, μ 40 + (10) 60 Therefore, the populatio mea is 60 ad the stadard deviatio is The percetile rakig of the age of 5 years would be 100% 73.5% 6.5%. 8 Chapter

25 .94 a. From Exercise.77, x ad s The z-score for a observatio of 78 is: x x z 3.50 s 4.83 This z-score idicates that a observatio of 78 is 3.5 stadard deviatios below the mea. Very few observatios will be lower tha this oe. b. The z-score for a observatio of 98 is: x x z 0.63 s 4.83 This z-score idicates that a observatio of 98 is.63 stadard deviatios above the mea. This score is ot a uusual observatio i the data set..96 a. From the problem, μ.7 ad σ.5 z x - μ σ zσ x μ x μ + zσ For z.0, x.7 +.0(.5) 3.7 For z 1.0, x.7 1.0(.5). For z.5, x.7 +.5(.5).95 For z.5, x.7.5(.5) 1.45 b. For z 1.6, x.7 1.6(.5) 1.9 c. If we assume the distributio of GPAs is approximately moud-shaped, we ca use the Empirical Rule. From the Empirical Rule, we kow that.05 or.5% of the studets will have GPAs above 3.7 (with z ). Thus, the GPA correspodig to summa cum laude (top.5%) will be greater tha 3.7 (z > ). We kow that.16 or 16% of the studets will have GPAs above 3. (z 1). Thus, the limit o GPAs for cum laude (top 16%) will be greater tha 3. (z > 1). We must assume the distributio is moud-shaped. Methods for Describig Sets of Data 9

26 .98 a. Sice the data are approximately moud-shaped, we ca use the Empirical Rule. O the blue exam, the mea is 53% ad the stadard deviatio is 15%. We kow that approximately 68% of all studets will score withi 1 stadard deviatio of the mea. This iterval is: x ± s 53 ± (15) (38, 68) About 95% of all studets will score withi stadard deviatios of the mea. This iterval is: x ± s 53 ± (15) 53 ± 30 (3, 83) About 99.7% of all studets will score withi 3 stadard deviatios of the mea. This iterval is: x ± 3s 53 ± 3(15) 53 ± 45 (8, 98) b. Sice the data are approximately moud-shaped, we ca use the Empirical Rule. O the red exam, the mea is 39% ad the stadard deviatio is 1%. We kow that approximately 68% of all studets will score withi 1 stadard deviatio of the mea. This iterval is: x ± s 39 ± (1) (7, 51) About 95% of all studets will score withi stadard deviatios of the mea. This iterval is: x ± s 39 ± (1) 39 ± 4 (15, 63) About 99.7% of all studets will score withi 3 stadard deviatios of the mea. This iterval is: x ± 3s 39 ± 3(1) 39 ± 36 (3, 75) c. The studet would have bee more likely to have take the red exam. For the blue exam, we kow that approximately 95% of all scores will be from 3% to 83%. The observed 0% score does ot fall i this rage. For the blue exam, we kow that approximately 95% of all scores will be from 15% to 63%. The observed 0% score does fall i this rage. Thus, it is more likely that the studet would have take the red exam..100 The 5th percetile, or lower quartile, is the measuremet that has 5% of the measuremets below it ad 75% of the measuremets above it. The 50th percetile, or media, is the measuremet that has 50% of the measuremets below it ad 50% of the measuremets above it. The 75th percetile, or upper quartile, is the measuremet that has 75% of the measuremets below it ad 5% of the measuremets above it. 30 Chapter

27 .10 a. Media is approximately 4. b. Q L is approximately 3 (Lower Quartile) Q U is approximately 6 (Upper Quartile) c. IQR Q U Q L d. The data set is skewed to the right sice the right whisker is loger tha the left, there is oe outlier, ad there are two potetial outliers. e. 50% of the measuremets are to the right of the media ad 75% are to the left of the upper quartile. f. There are two potetial outliers, 1 ad 13. There is oe outlier, a. From the problem, x 5.33 ad s 9.. The highest salary is 75 (thousad). The z-score is z x x s Therefore, the highest salary is.46 stadard deviatios above the mea. The lowest salary is 35.0 (thousad). The z-score is z x x s Therefore, the lowest salary is 1.88 stadard deviatios below the mea. The mea salary offer is 5.33 (thousad). The z-score is z x x s The z-score for the mea salary offer is 0 stadard deviatios from the mea. No, the highest salary offer is ot uusually high. For ay distributio, at least 8/9 of the salaries should have z-scores betwee 3 ad 3. A z-score of.46 would ot be that uusual. Methods for Describig Sets of Data 31

28 b. Usig MINITAB, the box plot is: Sice o salaries are outside the ier feces, oe of them are potetially faulty observatios..106 Usig MINITAB, the side-by-side box plots are: AGE GROUP 3 From the boxplots, there appears to be oe outlier i the third group..108 a. First, we will compute the mea ad stadard deviatio. The sample mea is: xi i x The sample variace is: s xi i x 393 i 5943 i Chapter

29 The stadard deviatio is: s s Sice this data set is highly skewed, we will use stadard deviatios from the mea as the cutoff for outliers. Z-scores with values greater tha i absolute value are cosidered outliers. A observatio with a z-score of would have the value: x x x 5.4 z (7.44) x x 5.4 x s 7.44 A observatio with a z-score of - would have the value: x x x 5.4 z (7.44) x 5.4 s x 5.4 x 9.48 Thus ay observatio that is greater tha to or less tha would be cosidered a outlier. I this data set there would be 4 outliers: 1, 1, 5, 48. b. Deletig these 4 outliers, we will recalculate the mea, media, variace, ad stadard deviatio. The media for the origial data set is the middle umber oce they have bee arraged i order ad is the 38 th observatio which is 3. The ew mea is: xi i 1 78 x The ew sample variace is: s xi i x 78 i 13 i The ew stadard deviatio is: s s The ew media is the 36 th observatio oce the data have bee arraged i order ad is 3. I the origial data set, the mea is 5.4, the stadard deviatio is 7.44, ad the media is 3. I the revised data set, the mea is 3.9, the stadard deviatio is 3.861, ad the media is 3. The mea has bee decreased, the stadard deviatio has bee almost halved, but the media stays the same. Methods for Describig Sets of Data 33

30 .110 For Perturbed Itrisics, but o Perturbed Projectios: xi i x s i 1 xi i 1 x 8.1 i s s The z-score correspodig to a value of 4.5 is x x z 3.63 s.79 Sice this z-score is greater tha 3, we would cosider this a outlier for perturbed itrisics, but o perturbed projectios. For Perturbed Projectios, but o Perturbed Itrisics: xi i x s i 1 xi i 1 x 15.8 i s s The z-score correspodig to a value of 4.5 is x x z s Sice this z-score is less tha -3, we would cosider this a outlier for perturbed projectios, but o perturbed itrisics. Sice the z-score correspodig to 4.5 for the perturbed projectios, but o perturbed itrisics is smaller tha that for perturbed itrisics, but o perturbed projectios, it is more likely that the that the type of camera perturbatio is perturbed projectios, but o perturbed itrisics. 34 Chapter

31 .11 Usig MINITAB, a scatterplot of the data is: Var Var1.114 Usig MINITAB, the scatterplot of the data is: Lawyers Offices 10 As the umber of offices icreases, the umber of lawyers also teds to icrease..116 a. Usig MINITAB, the scatterplot is: th th 30 It appears that as the completio time for the 10 th trial icreases, the completio time for the 30 th trial decreases. 40 Methods for Describig Sets of Data 35

32 b. Usig MINITAB, the scatterplot is: th th 30 It appears that as the completio time for the 10 th trial icreases, the completio time for the 50 th trial icreases. 40 c. Usig MINITAB, the scatterplot is: th th 15 0 It appears that as the completio time for the 30 th trial icreases, the completio time for the 50 th trial icreases. 36 Chapter

33 .118 Usig MINITAB, the scatterplot of the data is: 7 Scatterplot of Mass vs Time Mass Time There is evidece to idicate that the mass of the spill teds to dimiish as time icreases. As time is gettig larger, the mass is decreasig..10 The mea is sesitive to extreme values i a data set. Therefore, the media is preferred to the mea whe a data set is skewed i oe directio or the other..1 a. If we assume that the data are about moud-shaped, the ay observatio with a z-score greater tha 3 i absolute value would be cosidered a outlier. From Exercise 1.11, the z-score correspodig to 50 is 1, the z-score correspodig to 70 is 1, ad the z-score correspodig to 80 is. Sice oe of these z-scores is greater tha 3 i absolute value, oe would be cosidered outliers. b. From Exercise 1.11, the z-score correspodig to 50 is, the z-score correspodig to 70 is, ad the z-score correspodig to 80 is 4. Sice the z-score correspodig to 80 is greater tha 3, 80 would be cosidered a outlier. c. From Exercise 1.11, the z-score correspodig to 50 is 1, the z-score correspodig to 70 is 3, ad the z-score correspodig to 80 is 4. Sice the z-scores correspodig to 70 ad 80 are greater tha or equal to 3, 70 ad 80 would be cosidered outliers. d. From Exercise 1.11, the z-score correspodig to 50 is.1, the z-score correspodig to 70 is.3, ad the z-score correspodig to 80 is.4. Sice oe of these z-scores is greater tha 3 i absolute value, oe would be cosidered outliers. Methods for Describig Sets of Data 37

34 .14 a. x x x 34 x ( ) x 34 x s s b. x ( 3) ( 3) + ( 6) 9 x ( 1) ( 3) ( 3) + ( 6) 71 x 9 x -$1.5 6 ( ) x ( 9) x s 11.5 dollars squared s 11.5 $3.39 c x x x.065 x.415% 5 ( ) x.065 s x % squared s % d. (a) Rage (b) Rage $4 ($-6) $ (c) Rage % % % % %.7375% σ rage/4 0/ Chapter

35 .18 Usig MINITAB, a pie chart of the data is: Pie Chart of defect true 9.8% Category false true false 90.% A respose of true meas the software cotaied defective code. Thus, oly 9.8% of the modules cotaied defective software code..130 The z-score would be: x x z 1.06 s Sice this value is ot very big, this is ot a uusual value to observe..13 a. The variable of iterest is opiio of book reviews. The values could be would ot recommed, cautious or very little recommedatio, little or o preferece, favorable/recommeded, ad outstadig/sigificat cotributio. Sice these resposes are ot umerical, the variable is quatitative. b. Most of the books (63%) received a "favorable/recommeded" review. About the same percetage of books received the followig reviews: "cautious or very little recommedatio" (10%), "little or o preferece" (9%), ad "outstadig/sigificat cotributio" (1%). Oly 5% of the books received "would ot recommed" reviews. c. If the top two categories are added together, the percet recommeded is 75% (actually slightly higher tha 75%). This agrees with the study..134 a. To display the status, we use a pie chart. From the pie chart, we see that 58% of the Beaie babies are retired ad 4% are curret. Methods for Describig Sets of Data 39

36 b. Usig Miitab, a histogram of the values is: Most (40 of 50) Beaie babies have values less tha $100. Of the remaiig 10, 5 have values betwee $100 ad $300, 1 has a value betwee $300 ad $500, 1 has a value betwee $500 ad $700, have values betwee $700 ad $900, ad 1 has a value betwee $1900 ad $100. c. A plot of the value versus the age of the Beaie Baby is as follows: From the plot, it appears that as the age icreases, the value teds to icrease..136 a. Usig MINITAB, the stem-ad-leaf display is: Stem-ad-leaf of C1 Leaf Uit 0.10 N (5) Chapter

37 b. The leaves that represet those brads that carry the America Detal Associatio seal are circled above. c. It appears that the cost of the brads approved by the ADA ted to have the lower costs. Thirtee of the twety brads approved by the ADA, or (13/0) 100% 65% are less tha the media cost..138 a. Usig MINITAB, the summary statistics are: Descriptive Statistics: Marketig, Egieerig, Accoutig, Total Variable N Mea Media TrMea StDev SE Mea Marketi Egieer Accouti Total Variable Miimum Maximum Q1 Q3 Marketi Egieer Accouti Total b. The z-scores correspodig to the maximum time guidelies developed for each departmet ad the total are as follows: Marketig: z x x s Egieerig: z Accoutig: z x x s x x s Total: z x x s c. To fid the maximum processig time correspodig to a z-score of 3, we substitute i the values of z,, ad s ito the z formula ad solve for x. z x x x x zs x x + zs s Marketig: x (.58) Noe of the orders exceed this time. Egieerig: x (3.84) Noe of the orders exceed this time. These both agree with both the Empirical Rule ad Chebyshev's Rule. Methods for Describig Sets of Data 41

38 Accoutig: x (6.6) Oe of the orders exceeds this time or 1/50.0. Total: x (6.8) Oe of the orders exceeds this time or 1/50.0. These both agree with Chebyshev's Rule but ot the Empirical Rule. Both of these last two distributios are skewed to the right. d. Marketig: x (.58) Two of the orders exceed this time or / Egieerig: x (3.84) Two of the orders exceed this time or / Accoutig: x (6.6) Three of the orders exceed this time or 3/ Total: x (6.8) Two of the orders exceed this time or / All of these agree with Chebyshev's Rule but ot the Empirical Rule. e. No observatios exceed the guidelie of 3 stadard deviatios for both Marketig ad Egieerig. Oe observatio exceeds the guidelie of 3 stadard deviatios for both Accoutig (#3, time 30.0 days) ad Total (#3, time 36. days). Therefore, oly (1/10) 100% of the "lost" quotes have times exceedig at least oe of the 3 stadard deviatio guidelies. Two observatios exceed the guidelie of stadard deviatios for both Marketig (#31, time 11.0 days ad #48, time 10.0 days) ad Egieerig (#4, time 13.0 days ad #49, time 14.4 days). Three observatios exceed the guidelie of stadard deviatios for Accoutig (#0, time.0 days; #3, time 30.0 days; ad #36, time 18. days). Two observatios exceed the guidelie of stadard deviatios for Total (#0, time 30. days ad #3, time 36. days). Therefore, (7/10) 100% 70% of the "lost" quotes have times exceedig at least oe the stadard deviatio guidelies. We would recommed the stadard deviatio guidelie sice it covers 70% of the lost quotes, while havig very few other quotes exceed the guidelies..140 a. First, costruct a relative frequecy distributio for the departmets. Class Departmet Frequecy Relative Frequecy 1 Productio Maiteace Sales R & D Admiistratio TOTAL Chapter

39 The Pareto diagram is: From the diagram, it is evidet that the departmets with the worst safety record are Maiteace ad Productio. b. First, costruct a relative frequecy distributio for the type of ijury i the maiteace departmet. Class Ijury Frequecy Relative Frequecy 1 Bur Back strai Eye damage Cuts Broke arm Broke leg Cocussio Hearig loss.065 TOTAL The Pareto diagram is: From the Pareto diagram, it is evidet that cuts is the most prevalet type of ijury. Burs ad back strai are the ext most prevalet types of ijuries..14 a. Usig MINITAB, the descriptive statistics are: Descriptive Statistics: MPG Variable N Mea Media TrMea StDev SE Mea MPG Variable Miimum Maximum Q1 Q3 MPG Methods for Describig Sets of Data 43

40 The mea is ad the stadard deviatio is.177. Both of these measures are measured i the same uits as the origial data, which are miles per gallo. b. Sice the sample mea is a good estimate of the populatio mea, the maufacturer should be satisfied. The sample mea is which is greater tha 40. c. The rage of the data set is Usig Chebyshev's Rule, the rage should cover approximately 6 stadard deviatios. Thus, a good estimate of the stadard deviatio would be 10/ Usig the Empirical Rule, the rage should cover approximately 4 stadard deviatios. Thus, a good estimate of the stadard deviatio would be 10/4.5 The give stadard deviatio is.177 which is betwee these two estimates. Thus, it is a reasoable value. d. Usig MINITAB, the frequecy histogram is (the relative frequecy histogram would have the same shape): 9 Frequecy MPG Yes, the data appear to be moud-shaped. e. Because the data are moud-shaped, we ca use the Empirical Rule. We would expect approximately 68% of the data withi the iterval x ± s, approximately 95% of the data withi the iterval x ± s, ad approximately all of the data withi the iterval x ± 3s. f. The iterval x ± s is ±.177 or (37.879, 4.33). Twety-seve of the observatios fall i this iterval or 7/36.75 or 75%. This umber is a little larger tha 68%. The iterval x ± s is ± (.177) or (35.70, ). Thirty-four of the observatios fall i this iterval or 34/36.94 or 94%. This umber is very close to 95%. The iterval x ± 3s is ± 3(.177) or (33.55, ). Thirty-six of the observatios fall i this iterval or 36/ or 100%. This umber is the same as all of the observatios. 44 Chapter

41 .144 a. Both the height ad width of the bars (peauts) chage. Thus, some readers may ted to equate the area of the peauts with the frequecy for each year. b. The frequecy bar chart is: Methods for Describig Sets of Data 45

42 The Ketucky Milk Case (To accompay Chapters 1 ) There are may thigs that could be icluded i a report about the possibility of collusio. I have cocetrated o the icumbecy rates, bid levels ad dispersio, ad average wiig bids. With the data available, o compariso of market share ca be made sice there was so much missig data. Actually, with the data available, the exact aalysis caot be made, sice oly the wiig bid iformatio is provided. Thus, we have o idea what the losig bids were. I will preset what I thik is a reasoable solutio. This is by o meas the oly solutio to the case. May other presetatios could also be used. Icumbecy Rates The icumbecy rate is the percet of the school districts that are wo by the same vedor who wo the previous year. A table cotaiig the icumbecy rates is icluded as well as a plot. Notice i the plot that the icumbecy rates i the Tri-couty market is higher tha that i the Surroudig market. From 1985 through 1988, the icumbecy rate for the Tri-couty market was ever lower tha.93, while i the same period i the Surroudig market, the icumbecy rate was ever higher tha.730. This implies the possibility of collusio i the Tri-couty market. Surroudig Market Tri-couty Market Year Number of Districts Same Vedors Icumbecy Rate Number of Districts Same Vedors Icumbecy Rate The Ketucky Milk Case

43 The plot of the icumbecy rates is: Bid Levels ad Dispersio Sice we oly have access to the wiig bids i each of the school districts, we caot make a true aalysis of the bid levels ad dispersios. As a compromise, I have used the wiig bids of the two dairies i questio Trauth ad Meyer. I have looked at oly the wiig bids of these two dairies i both the Tri-couty market ad i the Surroudig market. If there was o collusio, the the wiig bids ad the dispersios of the wiig bids should be similar i the two markets for the two dairies. I looked at the box plots of the wiig bids of the two dairies i each market for each type of milk: whole white, lowfat white ad lowfat chocolate. I have icluded oly a few of the box plots as illustratios. Those icluded are for 1985 ad The Ketucky Milk Case 47

44 1985 Wiig Bids: WHOLE LOWFAT LOWFAT OBS MARKET WINNER WHITE WHITE CHOCOLATE 1 SUR MEYER SUR TRAUTH SUR TRAUTH SUR TRAUTH SUR MEYER SUR TRAUTH SUR MEYER TRI TRAUTH TRI TRAUTH TRI MEYER TRI TRAUTH TRI MEYER TRI MEYER TRI MEYER TRI TRAUTH TRI TRAUTH TRI MEYER TRI TRAUTH TRI MEYER TRI TRAUTH Box Plots for Whole White Milk Boxplots for Whole White Milk WWBID SURRO UND MARKET TRI-COUNTY 48 The Ketucky Milk Case

45 Box Plots for Lowfat White Milk Boxplots for Lowfat White Milk LFWBID SURRO UND MARKET TRI-COUNTY Box Plots for Lowfat Chocolate Milk 1985 Boxplots for Lowfat Chocolate Milk LFCBID SURRO UND MARKET TRI-COUNTY The Ketucky Milk Case 49

46 For each type of milk, the mea ad media wiig bids for the Tri-couty market were higher tha the correspodig wiig bids i the Surroudig market. Also, the dispersio, idicated by the width of the boxes ad the legth of the whiskers, for the Surroudig market is larger tha for the Tri-couty market i most cases. This is idicative of collusio i the Tri-couty market. This same patter also existed i Wiig Bids: WHOLE LOWFAT LOWFAT OBS MARKET WINNER WHITE WHITE CHOCOLATE 1 SUR TRAUTH SUR TRAUTH SUR TRAUTH SUR MEYER SUR TRAUTH SUR TRAUTH SUR TRAUTH SUR TRAUTH TRI TRAUTH TRI TRAUTH TRI MEYER TRI TRAUTH TRI MEYER TRI MEYER TRI MEYER TRI TRAUTH TRI TRAUTH TRI MEYER TRI TRAUTH TRI MEYER TRI TRAUTH Box Plots for Whole White Milk 1986 Boxplots for Whole White Milk WWBID SURRO UND MARKET TRI-COUNTY 50 The Ketucky Milk Case