2: Describing Data with Numerical Measures

Transcription

1 : Describig Data with Numerical Measures. a The dotplot show below plots the five measuremets alog the horizotal axis. Sice there are two s, the correspodig dots are placed oe above the other. The approximate ceter of the data appears to be aroud. Dotplot 0 media mode mea 4 5 b The mea is the sum of the measuremets divided by the umber of measuremets, or x i x 5 5 To calculate the media, the observatios are first raked from smallest to largest: 0,,,, 5. The sice 5, the positio of the media is 0.5( + ), ad the media is the rd raked measuremet, or m. The mode is the measuremet occurrig most frequetly, or mode. c The three measures i part b are located o the dotplot. Sice the media ad mode are to the left of the mea, we coclude that the measuremets are skewed to the right.. a The mea is x i + + L+ 5 x b To calculate the media, the observatios are first raked from smallest to largest:,,, 4, 4, 5, 5, 6 Sice 8 is eve, the positio of the media is 0.5( + ) 4.5, ad the media is the average of the 4 th ad 5 th measuremets, or m ( 4+ 4) 4. c Sice the mea ad the media are equal, we coclude that the measuremets are symmetric. The dotplot show below cofirms this coclusio. Dotplot

2 . a x 58 x i b The raked observatios are:,, 4, 5, 5, 6, 6, 8, 9, 0. Sice 0, the media is halfway betwee m the 5 th ad 6 th ordered observatios, or ( ) c There are two measuremets, 5 ad 6, which both occur twice. Sice this is the highest frequecy of occurrece for the data set, we say that the set is bimodal with modes at 5 ad 6..4 a x i 40 x 54 x 785 b x i c The average premium cost i several differet cities is ot as importat to the cosumer as the average cost for a variety of cosumers i his or her geographical area..5 a Although there may be a few households who ow more tha oe DVD player, the majority should ow either 0 or. The distributio should be slightly skewed to the right. b Sice most households will have oly oe DVD player, we guess that the mode is. c The mea is x i + 0+ L+ 7 x To calculate the media, the observatios are first raked from smallest to largest: There are six 0s, thirtee s, four s, ad two s. The sice 5, the positio of the media is 0.5( + ), which is the raked measuremet, or m. The mode is the measuremet occurrig most frequetly, or mode. d The relative frequecy histogram is show below, with the three measures superimposed. Notice that the mea falls slightly to the right of the media ad mode, idicatig that the measuremets are slightly skewed to the right. Relative frequecy 0/5 5/5 0 0 media mea mode VCRs.6 a The stem ad leaf plot below was geerated by Miitab. It is skewed to the right. Stem-ad-Leaf Display: Reveues Stem-ad-leaf of Reveues N 0 Leaf Uit (5)

3 b The mea is x i L , 00. x 47, To calculate the media, otice that the observatios are already raked from smallest to largest. The sice 0, the positio of the media is 0.5( + ) 5.5, the average of the 5 th ad 6 th raked measuremets or m ( ) 9, c Sice the mea is strogly affected by outliers, the media would be a better measure of ceter for this data set..7 It is obvious that ay oe family caot have.5 childre, sice the umber of childre per family is a quatitative discrete variable. The researcher is referrig to the average umber of childre per family calculated for all families i the Uited States durig the 90s. The average does ot ecessarily have to be iteger-valued..8 a Similar to previous exercises. The mea is x i L x b To calculate the media, rak the observatios from smallest to largest. The positio of the media is 0.5( + ) 7.5, ad the media is the average of the 7 ad 8 th raked measuremet or m ( ) c Sice the mea is slightly larger tha the media, the distributio is slightly skewed to the right..9 The distributio of sports salaries will be skewed to the right, because of the very high salaries of some sports figures. Hece, the media salary would be a better measure of ceter tha the mea..0 a Similar to previous exercises. x 50 x i 5 0 b The raked observatios are show below: The positio of the media is 0.5( + ) 5.5 ad the media is the average of the 5 th ad 6 th observatio or c Sice there are o uusually large or small observatios to affect the value of the mea, we would probably report the mea or average time o task.. a Similar to previous exercises. x 4 x i The raked observatios are: The media is the average of the 8 th ad 9 th observatios or m ( + )/ ad the mode is the most frequetly occurrig observatio mode. 0

4 b Sice the mea is larger tha the media, the data are skewed to the right. c The dotplot is show below. The distributio is skewed to the right. 0 4 Starbucks a x 80 x i 8 0 b The raked data are: 00, 00, 050, 800, 750, 700, 670, 650,600, 500 ad the media is the average of the 5 th ad 6 th observatios or m 75 c Average cost would ot be as importat as may other variables, such as picture quality, soud quality, size, lowest cost for the best quality, ad may other cosideratios.. a x x i.4 5 b Create a table of differeces, ( x x ) ad their squares, ( x x ) The c x i xi i x ( x x ) Total 0.0 ( ) x x L i i. i (.4) + + (5.4).0 s The sample stadard deviatio is the positive square root of the variace or s s.8.67 d Calculate + + L The x i ( x ) ( ) i s.8 ad 4 4 The results of parts a ad b are idetical. s s.8.67.

5 .4 The results will vary from studet to studet, depedig o their particular type of calculator. The results should agree with Exercise...5 a The rage is R 4. b c Calculate L+ 45. The x i ( x ) ( 7) x 7 x i.5 8 i s.679 ad s a The rage is R 6 5. b c Calculate + + L The s x i ( x ) ( ) i ad s s d The rage, R 5, is 5.55.stadard deviatios..7 a The rage is R b Calculate L The s x i ( x ) ( 8.56) x x i i ad s s c The rage, R., is stadard deviatios. s x a The rage is R b x i 7.7 c Calculate L The ad s s x i ( x ) ( 76.60) i 647, s a The rage of the data is R 6 5ad the rage approximatio with 0 is R s.67 b The stadard deviatio of the sample is s ( x ) ( ) i 0 0 s which is very close to the estimate for part a. c-e From the dotplot o the ext page, you ca see that the data set is ot moud-shaped. Hece you ca use Tchebysheff s Theorem, but ot the Empirical Rule to describe the data.

6 Dotplot a First calculate the itervals: x ± s 6 ± or to 9 x ± s 6 ± 6 or 0 to 4 x ± s 6 ± 9 or 7 to 45 Accordig to the Empirical Rule, approximately 68% of the measuremets will fall i the iterval to 9; approximately 95% of the measuremets will fall betwee 0 ad 4; approximately 99.7% of the measuremets will fall betwee 7 ad 45. b If o prior iformatio as to the shape of the distributio is available, we use Tchebysheff s Theorem. 0of the measuremets to fall i the iterval to 9; at least We would expect at least ( ) ( ) 4of the measuremets to fall i the iterval 0 to 4; at least ( ) measuremets to fall i the iterval 7 to of the. a The iterval from 40 to 60 represets µ ± σ 50 ± 0. Sice the distributio is relatively moudshaped, the proportio of measuremets betwee 40 ad 60 is 68% accordig to the Empirical Rule ad is show below. b Agai, usig the Empirical Rule, the iterval µ ± σ 50± (0) or betwee 0 ad 70 cotais approximately 95% of the measuremets.

7 c Refer to the figure below. Sice approximately 68% of the measuremets are betwee 40 ad 60, the symmetry of the distributio implies that 4% of the measuremets are betwee 50 ad 60. Similarly, sice 95% of the measuremets are betwee 0 ad 70, approximately 47.5% are betwee 0 ad 50. Thus, the proportio of measuremets betwee 0 ad 60 is d From the figure i part a, the proportio of the measuremets betwee 50 ad 60 is 0.4 ad the proportio of the measuremets which are greater tha 50 is Therefore, the proportio that are greater tha 60 must be Sice othig is kow about the shape of the data distributio, you must use Tchebysheff s Theorem to describe the data. a The iterval from 60 to 90 represets µ ± σ which will cotai at least 8/9 of the measuremets. b The iterval from 65 to 85 represets µ ± σ which will cotai at least /4 of the measuremets. c The value x 6.5 lies two stadard deviatios below the mea. Sice at least /4 of the measuremets are withi two stadard deviatio rage, at most /4 ca lie outside this rage, which meas that at most /4 ca be less tha 65.. a The rage of the data is R ad the approximate value of s is R s 0. b Calculate 7.6 ad 6.0, the sample mea is x i 7.6 x.76 0 ad the stadard deviatio of the sample is x i x i ( x ) ( 7.6) i s s 9 9 which is very close to the estimate from part a

8 .4 a The stem ad leaf plot geerated by Miitab shows that the data is roughly moud-shaped. Note however the gap i the ceter of the distributio ad the two measuremets i the upper tail. Stem-ad-Leaf Display: Weight Stem-ad-leaf of Weight N 7 Leaf Uit () b Calculate 8.4ad 0.607, the sample mea is x i 8.4 x.05 7 ad the stadard deviatio of the sample is x i x i ( x ) ( 8.4) i s s c The followig table gives the actual percetage of measuremets fallig i the itervals x ± ks for k,,. k x ± ks Iterval Number i Iterval Percetage.05 ± to.8 78%.05 ± to %.05 ± to % d The percetages i part c do ot agree too closely with those give by the Empirical Rule, especially i the oe stadard deviatio rage. This is caused by the lack of moudig (idicated by the gap) i the ceter of the distributio. e The lack of ay oe-poud packages is probably a marketig techique itetioally used by the supermarket. People who buy slightly less tha oe-poud would be draw by the slightly lower price, while those who eed exactly oe-poud of meat for their recipe might ted to opt for the larger package, icreasig the store s profit..5 Accordig to the Empirical Rule, if a distributio of measuremets is approximately moud-shaped, a approximately 68% or 0.68 of the measuremets fall i the iterval µ ± σ ±. or 9.7 to 4. b approximately 95% or 0.95 of the measuremets fall i the iterval µ ± σ ± 4.6 or 7.4 to 6.6 c approximately 99.7% or of the measuremets fall i the iterval µ ± σ ± 6.9 or 5. to 8.9 Therefore, approximately 0.% or 0.00 will fall outside this iterval. 5

9 .6 a The stem ad leaf plots are show below. The secod set has a slightly higher locatio ad spread. Stem-ad-Leaf Display: Method, Method Stem-ad-leaf of Method N 0 Stem-ad-leaf of Method N 0 Leaf Uit Leaf Uit (4) x b Method : Calculate x i 0.5 ad x i The x i 0.05 ad ( x ) ( 0.5) i s 0 s x Method : Calculate x i 0.8 ad x i The x i 0.08 ad ( x ) ( 0.8) i s 0 s 9 The results cofirm the coclusios of part a a The ceter of the distributio should be approximately halfway betwee 0 ad 9 or ( 0+ 9) 4.5. b The rage of the data is R Usig the rage approximatio, s R c Usig the data etry method the studets should fid x ad s.89, which are fairly close to our approximatios..8 a Similar to previous exercises. The itervals, couts ad percetages are show i the table. k x ± ks Iterval Number i Iterval Percetage ± to % ± to % ± to % b The percetages i part a do ot agree with those give by the Empirical Rule. This is because the shape of the distributio is ot moud-shaped, but flat..9 a Although most of the aimals will die at aroud days, there may be a few aimals that survive a very log time, eve with the ifectio. The distributio will probably be skewed right. b Usig Tchebysheff s Theorem, at least /4 of the measuremets should be i the iterval µ ± σ ± 7 or 0 to 04 days..0 a The value of x is µ σ 6 4. b The iterval µ σ is 6 ± ± should cotai approximately ( ) % of the survival times, of which 7% will be loger tha 68 days a d7% less tha 4 days. c The latter is clearly impossible. Therefore, the approximate values give by the Empirical Rule are ot accurate, idicatig that the distributio caot be moud-shaped. 6

10 . a We choose to use classes of legth.0. The tally ad the relative frequecy histogram follow. Class i Class Boudaries Tally f i Relative frequecy, f i / to < /70 to < 4 /70 4 to < 5 / to < 6 5 5/ to < 7 5 5/ to < 8 / to < 9 8 8/ to < 0 5 5/ to < 6 6/70 0 to < /70 to < 0 0 to < 4 /70 0/70 Relative frequecy 0/ TREES 0 5 x 54 b Calculate 70, x i 54ad x i 445. The x i 7.79 is a estimate of µ. 70 c The sample stadard deviatio is ( x ) ( ) i s The three itervals, x ± ks for k,, are calculated below. The table shows the actual percetage of measuremets fallig i a particular iterval as well as the percetage predicted by Tchebysheff s Theorem ad the Empirical Rule. Note that the Empirical Rule should be fairly accurate, as idicated by the moudshape of the histogram i part a. k x ± ks Iterval Fractio i Iterval Tchebysheff Empirical Rule 7.79 ± to / at least ± to / at least ± to /70.00 at least a Calculate R so that s R b Calculate 4, x i.55 ad x i.5. The ( x ) (.55) i.5 s which is fairly close to the approximate value of s from part a. 7 ad s

11 . a-b Calculate R so that s R c Calculate 0, x i 45 ad x i 58, 45. The ( x ) ( 45) i 58,45 s ad s which is fairly close to the approximate value of s from part b. d The two itervals are calculated below. The proportios agree with Tchebysheff s Theorem, but are ot to close to the percetages give by the Empirical Rule. (This is because the distributio is ot quite moud-shaped.) k x ± ks Iterval Fractio i Iterval Tchebysheff Empirical Rule 7.5 ± to /0.00 at least ± 9.0. to /0.00 at least a Aswers will vary. A typical histogram is show below. The distributio is skewed to the right..5.0 Relative frequecy Kids b Calculate 4, x i 50 ad x i 868. The xi 50 x.57, 4 s ( x ) ( 50) i ad s c The three itervals, x ± ks for k,, are calculated below. The table shows the actual percetage of measuremets fallig i a particular iterval as well as the percetage predicted by Tchebysheff s Theorem ad the Empirical Rule. Note that the Empirical Rule is ot very accurate for the first iterval, sice the histogram i part a is skewed. k x ± ks Iterval Fractio i Iterval Tchebysheff Empirical Rule.57 ±.85.7 to 6.4 / at least ± to /4.95 at least ± to. 4/4.976 at least a Calculate R.9.8. so that s R b I Exercise.7, we calculated x i 8.56 ad x i L The s ( x ) ( 8.56) i

12 ad s s , which is very close to our estimate i part a..6 a Aswers will vary. A typical stem ad leaf plot is geerated by Miitab. Stem-ad-Leaf Display: Passes Stem-ad-leaf of Passes N 8 Leaf Uit x 49 b Calculate 8, x i 49 ad x i 795. The x i 9.9, 8 s ( x ) ( 49) i ad s s c Calculate x± s 9.9 ± 0.04 or 9.5 to 9.4. From the origial data set, all 8 of the measuremets, or 00% fall i this iterval. x.7 a Calculate 5, x i ad x i 49. The x i.4 ad 5 b ( x ) ( ) i 49 s Usig the frequecy table ad the grouped formulas, calculate xf 0(4) + (5) + () + (4) The, as i part a, s i i x f 0 (4) + (5) + () + (4) 49 i i xf i i x.4 5 ( xf) ( ) i i fi Use the formulas for grouped data give i Exercise.7. Calculate 7, xi fi 79, ad xi fi 9. The, xf i i 79 x s ( xf) ( 79) i i fi ad s a The data i this exercise have bee arraged i a frequecy table. x i f i Usig the frequecy table ad the grouped formulas, calculate 9

13 The xf 0(0) + (5) + L+ 0() 5 i i x f 0 (0) + (5) + L+ 0 () 9 i i xf i i 5 x.04 5 ( xf) ( 5) i i fi 9 s ad s b-c The three itervals x ± ks for k,, are calculated i the table alog with the actual proportio of measuremets fallig i the itervals. Tchebysheff s Theorem is satisfied ad the approximatio give by the Empirical Rule are fairly close for k ad k. k x ± ks Iterval Fractio i Iterval Tchebysheff Empirical Rule.04 ± to / at least ± to 7.65 /5 0.9 at least ± to /5.00 at least The sorted data set, alog with the positios of the quartiles ad the quartiles themselves are show i the table. Sorted Data Set Positio of Q Positio of Q Lower quartile, Q Upper quartile, Q., 6,.,.8,.4,.76, (8).75(8) ,.7,.0,.,.,.8,.9, 4.4, 5., 6.5, 8.8.5().75() The data have already bee sorted. Fid the positios of the quartiles, ad the measuremets that are just above ad below those positios. The fid the quartiles by iterpolatio. Sorted Data Set Positio of Above Q Positio of Q Above Q ad below ad below,.5,,,..5(6).5 ad (6) 4.5 ad.. 0,.7,.8,.,.5() Noe.8.75() 9 Noe 8.9., 7, 8, 8.8, 8.9, 9, 0.,.0,.5,.4,.56,.58,.76,.80.5(9).5.0 ad (.05).5.75(9) ad (.8).750 Q.4 The ordered data are: 0,,, 4, 4, 5, 6, 6, 7, 7, 8 a With, the media is i positio 0.5( + ) 6.5, or halfway betwee the 6 th ad 7 th observatios. The lower quartile is i positio 0.5( + ).5 (oe-fourth of the way betwee the rd ad 4 th observatios) ad the upper quartile is i positio 0.75( + ) 9.75 (three-fourths of the way betwee the 9 th ad 0 th observatios). Hece, m ( ) Q Q (7 6) The the five-umber summary is Mi Q Media Q Max , + 0.5(4 ).5 ad

14 ad IQR Q Q x 57 b Calculate, x i 57 ad x i 7. The x i 4.75 ad the sample stadard deviatio is ( x ) ( ) i 57 7 s c For the smaller observatio, x 0, x x z-score.94 s.454 ad for the largest observatio, x 8, x x z-score. s.454 Sice either z-score exceeds i absolute value, oe of the observatios are uusually small or large..4 The ordered data are: 0,, 5, 6, 7, 8, 9, 0,,,, 4, 6, 9, 9 With 5, the media is i positio 0.5( + ) 8, so that m 0. The lower quartile is i positio 0.5( + ) 4 so that Q 6 ad the upper quartile is i positio 0.75( + ) so that Q 4. The the five-umber summary is Mi Q Media Q Max ad IQR Q Q The ordered data are:, 8,,, 4, 5, 5, 6, 6, 7, 8 For, the positio of the media is 0.5( + ) 0.5( + ) 6 ad m 5. The positios of the quartiles are 0.5( + ) ad 0.75( + ) 9, so that Q, Q 6, ad IQR 6 4. The lower ad upper feces are: Q.5IQR 6 6 Q +.5IQR The oly observatio fallig outside the feces is x which is idetified as a outlier. The box plot is show below. The lower whisker coects the box to the smallest value that is ot a outlier, x 8. The upper whisker coects the box to the largest value that is ot a outlier or x x 5 0 4

15 .45 The ordered data are:,, 4, 5, 6, 6, 6, 7, 8, 9, 9, 0, For, the positio of the media is 0.5( + ) 0.5( + ) 7 ad m 6. The positios of the quartiles are 0.5( + ).5 ad 0.75( + ) 0.5, so that Q 4.5, Q 9, ad IQR The lower ad upper feces are: Q.5IQR Q +.5IQR The value x lies outside the upper fece ad is a outlier. The box plot is show below. The lower whisker coects the box to the smallest value that is ot a outlier, which happes to be the miimum value, x. The upper whisker coects the box to the largest value that is ot a outlier or x x From Sectio.6, the 69 th percetile implies that 69% of all studets scored below your score, ad oly % scored higher..47 a The ordered data are show below: For 8, the positio of the media is 0.5( + ) 4.5 ad the positios of the quartiles are 0.5( + ) 7.5 ad 0.75 ( + ).75. The lower quartile is ¼ the way betwee the 7 th ad 8 th measuremets or Q (68 8) 0.5 ad the upper quartile is ¾ the way betwee the ad d measuremets or Q (8 6) 7.5. The the five-umber summary is Mi Q Media Q Max b Calculate IQR Q Q The the lower ad upper feces are: Q.5IQR Q +.5IQR st 4

16 The box plot is show below. Sice there are o outliers, the whiskers coect the box to the miimum ad maximum values i the ordered set Mercury c-d The boxplot does ot idetify ay of the measuremets as outliers, maily because the large variatio i the measuremets cause the IQR to be large. However, the studet should otice the extreme differece i the magitude of the first four observatios take o youg dolphis. These aimals have ot bee alive log eough to accumulate a large amout of mercury i their bodies..48 a See Exercise.4b. b For x.8, while for x. 4, x x.8.05 z-score.94 s 0.7 x x.4.05 z-score. s 0.7 The value x.4 would be cosidered somewhat uusual, sice its z-score exceeds i absolute value. c For 7, the positio of the media is 0.5( + ) 0.5(7 + ) 4 ad m.06. The positios of the quartiles are 0.5( + ) 7 ad 0.75( + ), so that Q 0.9, Q.7, ad IQR The lower ad upper feces are: Q.5IQR Q +.5IQR The box plot is show below. Sice there are o outliers, the whiskers coect the box to the miimum ad maximum values i the ordered set Weight...4 4

17 Sice the media lie is almost i the ceter of the box, the whiskers are early the same legths, the data set is relatively symmetric..49 a For 8, the positio of the media is 0.5( + ) 9.5 ad the positios of the quartiles are 0.5( + ) 4.75 ad 0.75 ( + ) 4.5. The lower quartile is ¼ the way betwee the 4 th ad 5 th measuremets ad the upper quartile is ¾ the way betwee the 4 th ad 5 th measuremets. The sorted measuremets are show below. Favre: 0,,, 4, 5, 5, 8, 9,,,,,,,, 5, 5, 6 McNabb: 9, 0,, 5, 5, 6, 6, 7, 8, 8, 8, 8, 9,,,, 4, 7 For Brett Favre, m ( + ).5, Q (5 4) 4.75 ad Q + 0.5( ). For Doova McNabb, m ( 8 + 8) 8, (5 5) 5 The the five-umber summaries are Mi Q Media Q Max Favre McNabb Q + ad Q + 0.5( ). b For Brett Favre, calculate IQR Q Q The the lower ad upper feces are: Q.5IQR Q +.5IQR For Doova McNabb, calculate IQR Q Q 5 6. The the lower ad upper feces are: Q.5IQR Q +.5IQR There are o outliers, ad the box plots are show below. Favre McNabb Completed passes c Aswers will vary. The Favre distributio is skewed left, while the Doova distributio is roughly symmetric, probably moud-shaped. The McNabb distributio is slightly more variable; Favre has a higher media umber of completed passes..50 Aswers will vary from studet to studet. The distributio is skewed to the right with three outliers (Truma, Clevelad ad F. Roosevelt)..5 a Just by scaig through the 5 measuremets, it seems that there are a few uusually large measuremets, which would idicate a distributio that is skewed to the right. b The positio of the media is 0.5( + ) 0.5(5 + ) ad m 4.4. The mea is x 960 x i

18 which is larger tha the media, idicate a distributio skewed to the right. c The positios of the quartiles are 0.5( + ) 6.5 ad 0.75( + ) 9.5, so that Q 8.7, Q 48.9, ad IQR The lower ad upper feces are: Q.5IQR Q +.5IQR The box plot is show below. There are three outliers i the upper tail of the distributio, so the upper whisker is coected to the poit x 69.. The log right whisker ad the media lie located to the left of the ceter of the box idicates that the distributio that is skewed to the right Visitors a The sorted data is: 65., 76.4, 78., 80.00, 04.94,. 5.47, 6.7, 8.66, 76.70, 09.70,.40 The positios of the media ad the quartiles are 0.5( + ) 6.5, 0.5( + ).5 ad 0.75( + ) 9.75, so that Q m ( ) / Q ad IQR The lower ad upper feces are: Q.5IQR Q +.5IQR There are o outliers, ad the box plot is show below Utility Bill

19 b Because of the log right whisker, the distributio is slightly skewed to the right..5 Aswers will vary. The studet should otice the outliers i the female group, that the media female temperature is higher tha the media male temperature. x a Calculate 4, x i 67 ad x i 964. The x i 6.4 ad 4 ( x ) ( ) i s 4.5 x 66 b Calculate 4, x i 66 ad x i The x i 6.4 ad 4 c ( x ) ( ) i s 4.4 The ceters are roughly the same; the Sumaid raisis appear slightly more variable..55 a The ordered sets are show below: Geeric Sumaid For 4, the positio of the media is 0.5( + ) 0.5(4 + ) 7.5 ad the positios of the quartiles are 0.5( + ).75 ad 0.75 ( + ).5, so that Geeric: m 6, Q 5, Q 7.5, ad IQR Sumaid: m 6, Q 4, Q 8, ad IQR b Geeric: Lower ad upper feces are: Q.5IQR Q +.5IQR Sumaid: Lower ad upper feces are: Q.5IQR Q +.5IQR The box plots are show below. There are o outliers. Sumaid Geeric Raisis

20 d If the boxes are ot beig uderfilled, the average size of the raisis is roughly the same for the two brads. However, sice the umber of raisis is more variable for the Sumaid brad, it would appear that some of the Sumaid raisis are large while others are small. The idividual sizes of the geeric raisis are ot as variable..56 a Calculate the rage as R 5 4. Usig the rage approximatio, s R b Calculate 5, x i 55.5 ad x i The x 55.5 x i 6. ad 5 ( x ) ( ) i s which is very close to the approximatio foud i part a. c Calculate x ± s 6.± 6.994or to.4. From the origial data, 4 measuremets or % of the measuremets fall i this iterval. This is close to the percetage give by the ( ) Empirical Rule..57 a The largest observatio foud i the data from Exercise.6 is., while the smallest is 0.. Therefore the rage is R b Usig the rage, the approximate value for s is: s R c Calculate 50, x i 48.4 ad x i The ( x ) ( ) i s a Refer to Exercise.57. Sice x i 48.4, the sample mea is x 48.4 x i The three itervals of iterest is show i the followig table, alog with the umber of observatios which fall i each iterval. k x ± ks Iterval Number i Iterval Percetage 8.68 ± to % 8.68 ± to % 8.68 ± to % b The percetages fallig i the itervals do agree with Tchebysheff s Theorem. At least 0 fall i the first iterval, at least fall i the secod iterval, ad at least fall i the third. The percetages are ot too close to the percetages described by the Empirical Rule (68%, 95%, ad 99.7%). c The Empirical Rule may be usuitable for describig these data. The data distributio does ot have a strog moud-shape (see the relative frequecy histogram i the solutio to Exercise.6), but is skewed to the right..59 The ordered data are show below

21 Sice 50, the positio of the media is 0.5( + ) 5.5 ad the positios of the lower ad upper quartiles are 0.5( + ).75 ad 0.75( + ) 8.5. The m ( ) Q , (.4.).5 ad Q (.5.6).85. The IQR The lower ad upper feces are: Q.5IQR Q +.5IQR ad the box plot is show below. There is oe outlier, x.. The distributio is skewed to the right TIME a For 4, the positio of the media is 0.5( + ) 7.5 ad the positios of the quartiles are 0.5( + ).75 ad 0.75 ( + ).5. The lower quartile is ¾ the way betwee the rd ad 4 th measuremets or Q ( ) 0.65 ad the upper quartile is ¼ the way betwee the th ad th measuremets or Q (..).475 The the five-umber summary is Mi Q Media Q Max b Calculate IQR Q Q The the lower ad upper feces are: Q.5IQR Q +.5IQR

22 The box plot is show below. Sice there are o outliers, the whiskers coect the box to the miimum ad maximum values i the ordered set Prices c Calculate 4, x i.55, x i.5. The x.55 x i ad 4 ( x ) ( ) i.55.5 s The z-score for x.9 is x x z.56 s which is somewhat ulikely..6 First calculate the itervals: x ± s 0.7 ± 0.0 or 0.6 to 0.8 x ± s 0.7 ± 0.0 or 0.5 to 0.9 x ± s 0.7± 0.0 or 0.4 to 0.0 a If o prior iformatio as to the shape of the distributio is available, we use Tchebysheff s Theorem. 0of the measuremets to fall i the iterval 0.6 to 0.8; at least We would expect at least ( ) ( ) 4of the measuremets to fall i the iterval 0.5 to 0.9; at least ( ) 8 9of the measuremets to fall i the iterval 0.4 to 0.0. b Accordig to the Empirical Rule, approximately 68% of the measuremets will fall i the iterval 0.6 to 0.8; approximately 95% of the measuremets will fall betwee 0.5 to 0.9; approximately 99.7% of the measuremets will fall betwee 0.4 ad 0.0. Sice moud-shaped distributios are so frequet, if we do have a sample size of 0 or greater, we expect the sample distributio to be moud-shaped. Therefore, i this exercise, we would expect the Empirical Rule to be suitable for describig the set of data. c If the chemist had used a sample size of four for this experimet, the distributio would ot be moud-shaped. Ay possible histogram we could costruct would be o-moud-shaped. We ca use at most 4 classes, each with frequecy, ad we will ot obtai a histogram that is eve close to moudshaped. Therefore, the Empirical Rule would ot be suitable for describig 4 measuremets..6 Sice it is ot obvious that the distributio of amout of chloroform per liter of water i various water sources is moud-shaped, we caot make this assumptio. Tchebysheff s Theorem ca be used, however, ad the ecessary itervals ad fractios fallig i these itervals are give i the table. 49

23 k x ± ks Iterval Tchebysheff 4 ± 5 9 to 87 at least 0 4 ± 06 7 to 40 at least ± 59 5 to 9 at least The followig iformatio is available: 400, x 600, s 4900 The stadard deviatio of these scores is the 70, ad the results of Tchebysheff s Theorem follow: k x ± ks Iterval Tchebysheff 600 ± to 670 at least ± to 740 at least ± 0 90 to 80 at least 0.89 If the distributio of scores is moud-shaped, we use the Empirical Rule, ad coclude that approximately 68% of the scores would lie i the iterval 50 to 670 (which is x ± s ). Approximately 95% of the scores would lie i the iterval 460 to a Calculate 0, x i 68.5, x i The x 68.5 x i 6.85 ad 0 ( x ) ( ) i s b The z-score for x 8.5 is x x z.64 s.008 This is ot a uusually large measuremet. c The most frequetly recorded measuremet is the mode or x 7 hours of sleep. d For 0, the positio of the media is 0.5( + ) 5.5 ad the positios of the quartiles are 0.5( + ).75 ad 0. 75( + ) 8.5. The sorted data are: 5, 6, 6, 6.75, 7, 7, 7, 7.5, 8, 8.5 The m ( ) Q , (6 6) 6 ad Q (8 7.5) The IQR ad the lower ad upper feces are: Q.5IQR Q +.5IQR

24 There are o outliers (cofirmig the results of part b) ad the box plot is show below Hours of sleep a Max 7, Mi 0. ad the rage is R b Aswers will vary. A typical histogram is show below. The distributio is slightly skewed to the left..5.0 Percet mpg c Calculate 0, x i 479., x i 5.8. The x 479. x i.96 0 d ( x ) ( ) i s 0 9 The sorted data is show below: The z-scores for x 0. ad x 7 are x x ad x z z x.85 s.64 s.64 Sice either of the z-scores are greater tha i absolute value, the measuremets are ot judged to be outliers. e The positio of the media is 0.5( + ) 0.5 ad the media is m ( )/ 4. f The positios of the quartiles are 0.5( + ) 5.5 ad 0.75( + ) The Q (..9).95 ad Q ( )

25 .66 Refer to Exercise.65. Calculate IQR The lower ad upper feces are: Q.5IQR Q +.5IQR There are o outliers, which cofirms the coclusio i Exercise.65. The box plot is show below. 0 4 mpg a The rage is R 7 40 ad the rage approximatio is s R b Calculate 0, x i 59, x i 604. The x 59 x i ( x ) ( ) i s The sample stadard deviatio calculated above is of the same order as the approximated value foud i part a. c The ordered set is: 40, 49, 5, 54, 59, 6, 67, 69, 70, 7 Sice 0, the positios of m, Q, ad Q are 5.5,.75 ad 8.5 respectively, ad m ( ) 60, Q (5 49) 5.5, Q ad IQR The lower ad upper feces are: Q.5IQR Q +.5IQR ad the box plot is show below. There are o outliers ad the data set is slightly skewed left Bacteria

26 .68 The results of the Empirical Rule follow: k x ± ks Iterval Empirical Rule 40 ± 5 45 to 45 approximately ± 0 40 to 40 approximately ± to 45 approximately Notice that we are assumig that attedace follows a moud-shaped distributio ad hece that the Empirical Rule is appropriate..69 If the distributio is moud-shaped, the almost all of the measuremets will fall i the iterval µ ± σ, which is a iterval 6σ i legth. That is, the rage of the measuremets should be approximately 6σ. I this case, the rage is , so that σ They are probably referrig to the average umber of times that me ad wome go campig per year..7 The stem legths are approximately ormal with mea 5 ad stadard deviatio.5. a I order to determie the percetage of roses with legth less tha.5, we must determie the proportio of the curve which lies withi the shaded area i the figure below. Usig the Empirical Rule, the proportio of the area betwee.5 ad 5 is half of 0.68 or 0.4. Hece, the fractio below.5 would be or 6% % 4% 47.5% x 0 b Refer to the figure show above. Agai we use the Empirical Rule. The proportio of the area betwee.5 ad 5 is half of 0.68 or 0.4, while the proportio of the area betwee 5 ad 0 is half of 0.95 or The total area betwee.5 ad 0 is the or 8.5%..7 a The rage is R ad the rage approximatio is s R b Calculate 5, x i 04, x i 8,807. The x 04 x i ( x ) ( ) i 04 8,807 s c Accordig to Tchebysheff s Theorem, with k, at least /4 or 75% of the measuremets will lie withi k stadard deviatios of the mea. For this data, the two values, a ad b, are calculated as x± s 6.07 ± (7.0) 7.07 ± 4.0 or a 0.87 ad b

27 .7 The diameters of the trees are approximately moud-shaped with mea 4 ad stadard deviatio.8. a The value x 8.4 lies two stadard deviatios below the mea, while the value x.4 is three stadard deviatios above the mea. Use the Empirical Rule. The fractio of trees with diameters betwee 8.4 ad 4 is half of 0.95 or 0.475, while the fractio of trees with diameters betwee 4 ad.4 is half of or The total fractio of trees with diameters betwee 8.4 ad.4 is b The value x 6.8 lies oe stadard deviatio above the mea. Usig the Empirical Rule, the fractio of trees with diameters betwee 4 ad 6.8 is half of 0.68 or 0.4, ad the fractio of trees with diameters greater tha 6.8 is a The rage is R ad the rage approximatio is s R b Calculate 5, x i 75, x i 7. The x 75 x i.67 5 ( x ) ( ) i 75 7 s c Calculate the iterval x± s.67 ± (.75).67 ± 7.47 or 4.0 to 9.4. Referrig to the origial data set, the fractio of measuremets i this iterval is 4/ a It is kow that duratio times are approximately ormal, with mea 75 ad stadard deviatio 0. I order to determie the probability that a commercial lasts less tha 5 secods, we must determie the fractio of the curve which lies withi the shaded area i the figure below. Usig the Empirical Rule, the fractio of the area betwee 5 ad 75 is half of 0.95 or Hece, the fractio below 5 would be b The fractio of the curve area that lies above the 55 secod mark may agai be determied by usig the Empirical Rule. Refer to the figure i part a. The fractio betwee 55 ad 75 is 0.4 ad the fractio above 75 is 0.5. Hece, the probability that a commercial lasts loger tha 55 secods is

28 .76 a The relative frequecy histogram for these data is show below..7.6 Relative frequecy x 6 8 b Refer to the formulas give i Exercise.7. Usig the frequecy table ad the grouped formulas, calculate 00, x f 66, x f 4. The i i i i xf i i 66 x ( xf) ( 66) i i fi 4 s ad s c The three itervals, x ± ks for k, are calculated i the table alog with the actual proportio of measuremets fallig i the itervals. Tchebysheff s Theorem is satisfied ad the approximatio give by the Empirical Rule are fairly close for k ad k. k x ± ks Iterval Fractio i Iterval Tchebysheff Empirical Rule 0.66 ±.78. to.44 95/ at least ± to / at least a The percetage of colleges that have betwee 45 ad 05 teachers correspods to the fractio of measuremets expected to lie withi two stadard deviatios of the mea. Tchebysheff s Theorem states that this fractio will be at least ¾ or 75%. b If the populatio is ormally distributed, the Empirical Rule is appropriate ad the desired fractio is calculated. Referrig to the ormal distributio show below, the fractio of area lyig betwee 75 ad 90 is 0.4, so that the fractio of colleges havig more tha 90 teachers is We must estimate s ad compare with the studet s value of 0.6. I this case, 0 ad the rage is R The estimated value for s is the 55

29 s R which is less tha 0.6. It is importat to cosider the magitude of the differece betwee the rule of thumb ad the calculated value. For example, if we were workig with a stadard deviatio of 00, a differece of 0.4 would ot be great. However, the studet s calculatio is twice as large as the estimated value. Moreover, two stadard deviatios, or (0.6) 0.56, already exceeds the rage. Thus, the value s 0.6 is probably icorrect. The correct value of s is ( x ) i s Notice that two (Sosa ad McGuire) of the four players have relatively symmetric distributios. The whiskers are the same legth ad the media lie is close to the middle of the box. The variability of the distributios is similar for all four players, but Barry Bods has a distributio with a log right whisker, meaig that there may be a uusually large umber of homers durig oe of his seasos. The distributio for Babe Ruth is slightly differet from the others. The media lie to the right of middle idicates a distributio skewed to the left; that there were a few seasos i which his homeru total was uusually low. I fact, the media umber of homerus for the other three players are all about 4-5, while Babe Ruth s media umber of homerus is closer to a Use the iformatio i the exercise. For 00, IQR 6.5, ad the upper fece is Q +.5IQR For 00, IQR 0.5, ad the upper fece is Q +.5IQR b The upper fece is differet i 00, so that the record umber of homers, x 7 is o loger a outlier, although it is still the most homers ever hit i a sigle seaso!.8 a Calculate 50, x 48 x i 48, so that x i b The positio of the media is.5( + ) 5.5 ad m (4 + 4)/ 4. c Sice the mea is larger tha the media, the distributio is skewed to the right. d Sice 50, the positios of Q ad Q are.5(5).75 ad.75(5) 8.5, respectively The Q ( 0).75, Q 7 +.5(9 7) 7.5 ad IQR The lower ad upper feces are: Q.5 IQR Q +.5IQR ad the box plot is show below. There are o outliers ad the data is skewed to the right Age (Years)

30 .8 Each bulleted statemet produces a percetile. x ideal family size. The value x is the 5 d percetile. x umber of times per week you reheat leftovers. The value x is the 0 th percetile. x time util a prescriptio is filled. The value x 5 is the 50 th percetile..8 Aswers will vary. Studets should otice that the distributio of baselie measuremets is relatively moud-shaped. Therefore, the Empirical Rule will provide a very good descriptio of the data. A measuremet which is further tha two or three stadard deviatios from the mea would be cosidered uusual..84 a Calculate 5, x i 04.9, x i The x 04.9 x i b ( x ) ( ) i s 5 4 The ordered data set is show below: c The z-scores for x.5 ad x 5.7 are x x ad x z z x.9 s.78 s.78 Sice either of the z-scores are greater tha i absolute value, the measuremets are ot judged to be uusually large or small..85 a For 5, the positio of the media is 0.5( + ) ad the positios of the quartiles are 0.5( + ) 6.5 ad 0.75 ( + ) 9.5. The m 4., Q (.7 +.8) /.75 ad Q ( ) / 4.75.The the five-umber summary is b-c Calculate Mi Q Media Q Max IQR Q Q The the lower ad upper feces are: Q.5IQR Q +.5IQR There are o uusual measuremets, ad the box plot is show below Times

31 d Aswers will vary. A stem ad leaf plot, geerated by Miitab, is show below. The data is roughly moud-shaped. Stem-ad-Leaf Display: Times Stem-ad-leaf of Times N 5 Leaf Uit (7) a Whe the applet loads, the mea ad media are show i the upper left-had corer: x 6.6 ad m 6.0 b Whe the largest value is chaged to x, x 7.0 ad m 6.0. c Whe the largest value is chaged to x, x.0 ad m 6.0. The mea is larger whe there is oe uusually large measuremet. d Extremely large values cause the mea to icrease, but ot the media..87 a-b As the value of x gets smaller, so does the mea. c The media does ot chage util the gree dot is smaller tha x 0, at which poit the gree dot becomes the media. d The largest ad smallest possible values for the media are 5 m a Whe the applet loads, the mea ad media are show i the upper left-had corer: x.6 ad m.0 b Whe the smallest value is chaged to x 5, x. ad m.0. c Whe the smallest value is chaged to x 5, x 7. ad m.0. The mea is smaller whe there is oe uusually small measuremet. d x 9.0 e The largest ad smallest possible values for the media are m 4. f Extremely small values cause the mea to decrease, but ot the media..89 Aswers will vary from studet to studet. Studets should otice that, whe the estimators are compared i the log ru, the stadard deviatio whe dividig by is closer to σ 9.. Whe dividig by, the estimate is closer to a Aswers will vary from studet to studet. Studets should otice that, whe the estimators are compared i the log ru, the stadard deviatio whe dividig by is closer to σ 9.. Whe dividig by, the estimate is closer to 7.5. b Whe the sample size is larger, the estimate is ot as far from the true value σ 9.. The differece betwee the two estimators is less oticeable..9 The box plot shows a distributio that is skewed to the left, but with oe outlier to the right of the other observatios ( x 50 )..9 The box plot shows a distributio that is slightly skewed to the right, with o outliers. The studet should estimate values for m, Q, ad Q that are close to the true values: m, Q 8.75, ad Q

32 Case Study: The Boys of Summer The Miitab computer package was used to aalyze the data. I the pritout below, various descriptive statistics as well as histograms ad box plots are show. 5 Histogram of Natioal League Battig Champios 0 Frequecy AVERAGE Histogram of America League Battig Champios 6 4 Frequecy AVERAGE Descriptive Statistics: AVERAGE Variable LEAGUE N N* Mea SE Mea StDev Miimum Q Media AVERAGE Variable LEAGUE Q Maximum AVERAGE Notice that the mea percetage of hits is almost the same for the two leagues, but that the America League () is slightly more variable. Natioal LEAGUE America AVERAGE The box plot shows that there are two outliers i the Natioal League (0). 4 I summary, except for the two outliers, there is very little differece betwee the two leagues