Chapter3:DataDescriptionSection1:MeasuresofCentralTendencyWorksheet

1. Thereportedhightemperatures(indegreesFahrenheit)forselectedworldcitiesonanOctoberdayareshownbelow.Findthemean,median,mode,andmidrange.Whichmeasureofcentraltendencydoyouthinkbestdescribesthesedata?62 72 66 70 83 61 62 85 72 6474 71 42 38 91 66 77 90 74 6364 68 42

a. Mean∑𝑥 = 62+ 74+ 64+ 72+ 71+ 68+ 66+ 42+ 42+ 70+ 38+ 83+ 91+ 61+ 66+ 62+77+ 85+ 90+ 72+ 74+ 64+ 63 = 1557(Addupallthedatavalues)𝑛 = 23ß#ofdatavalues.

𝑋 =𝑥𝑛 =

155723 = 67.695 → 67.7℉

(Thisisasamplesoweuse𝑋ratherthan𝜇).Datasetisinwhole#s,soweROUNDOFFtooneextradecimalplacevalue(tenths).

b. Median38,42,42,61,62,62,63,64,64,66,66,68,70,71,72,72,74,74,77,83,85,90,91

middlevalue𝑀𝐷 = 68℉

*Thereisanoddnumberofdatavalues,sothemedianwillbeanactualdatavalue.Datamustinorderfirst.Thenwefindthemiddlevalue.

c. Mode42,62,64,66,72,74(multimodal)

*Thesevalueshavethehighestfrequencyof2.Datasetismultimodalbecausemorethanonedatavaluehasthehighestfrequencyof2.d. Midrange

𝑀𝑅 =38+ 91

2 =1292 = 64.5℉

Addthelowestandhighestdatavalue.Thendivideby2.

2. Theamountofgarbageinmillionsoftonscollectedovera16-yearperiodisshown.Findthemean,median,mode,andmidrange.Whichmeasureofcentraltendencybestdescribesthedata?

29.7 48 58.4 37.9 47.3 57.2 55.8 43.5 32.9 53.746.1 50.1 36 52.8 46.4 52.7

a. Mean∑𝑥 = 29.7+ 46.1+ 48+ 50.1+ 58.4+ 36+ 37.9+ 52.8+ 47.3+ 46.4+ 57.2+ 52.7+55.8+ 43.5+ 32.9+ 53.7 = 748.5(Addupallthedatavalues)𝑛 = 16ß#ofdatavalues.

𝑋 =𝑥𝑛 =

748.516 = 46.781 → 46.78 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑡𝑜𝑛𝑠

(Thisisasamplesoweuse𝑋ratherthan𝜇).Datasethasatmostonedecimalplace(tenths),soweROUNDOFFtooneextradecimalplacevalue(hundredths).

b. Median

29.7,32.9,36,37.9,43.5,46.1,46.4,47.3,48,50.1,52.7,52.8,53.7,55.8,57.2,58.4Middlevalues.Medianwillbebetweenthesetwovalues.

𝑀𝐷 =47.3+ 48

2 = 47.65 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑡𝑜𝑛𝑠

c. ModeNoModeAllthedatavalueshaveafrequencyof1.Ifnovalueoccursmorethanonce,donotuse“0”forthemode,since“0”mightactuallybeadatavalue.

d. Midrange

𝑀𝑅 =29.7+ 58.4

2 =88.12 = 44.05 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑡𝑜𝑛𝑠

Addthelowestandhighestdatavalue.Thendivideby2.

3. Findthemean,median,mode,andmidrangeforthefollowingdata.

Stem Leaf 4 0023579 5 389 6 222458 7 0123 8 9

a. Mean

∑𝑥 = 40+ 40+ 42+ 43+ 45+ 47+ 49+ 53+ 58+ 59+ 62+ 62+ 62+ 64+ 65+68+ 70+ 71+ 72+ 73+ 89 = 1234(Addupallthedatavalues)𝑛 = 21ß#ofdatavalues.

𝑋 =𝑥𝑛 =

123421 = 58.76 → 58.8

(Thisisasamplesoweuse𝑋ratherthan𝜇).Datasetisinwhole#,soweROUNDOFFtooneextradecimalplacevalue(tenths).

b. Median

62

c. Mode62(unimodal)62hasthehighestfrequencyof3.Eventhough40hasafrequencyof2,62hasthehigherfrequencyof3.Datasetisunimodalbecausethereisonlyonevalueforthemode.

d. Midrange

𝑀𝑅 =40+ 89

2 =1292 = 64.5

4. Thenumberofhighwaymilespergallonofthe10worstvehiclesisshown.Findthemean,median,modeandmidrangeforthefollowingdata.12.53 11.75 15.00 14.02 16.27 17.94 16.08 17.00 13.90 18.00

a. Meanb. Medianc. Moded. Midrange

5. Thesedatarepresentthenetworth(inmillionsofdollars)of45nationalcorporations.Findthemeanandmodalclass.

Classlimits Frequency 𝑋! 𝑓 ∙ 𝑋! 10–20 2 !"!!"

!=15 2 ∙ 15 =30

21–31 8 !"!!"!

=26 8 ∙ 26 = 208

32–42 15 37 555 43–53 7 48 336 54–64 10 59 590 65–75 3 70 210

45ßn 1929ß (𝑓 ∙ 𝑋!)

𝑋 =(𝑓 ∙ 𝑋!)𝑛 =

192945 = 42.86 → $42.9 𝑚𝑖𝑙𝑙𝑖𝑜𝑛

*Classlimitsareinwhole#s,someanwillberoundedofftooneextradecimalplacevalue(tenths).ModalClass:32–42Modalclassistheclass(es)withthehighestfrequency.

6. Thirtyautomobilesweretestedforfuelefficiency(inmilespergallon).Thisfrequencydistributionwasobtained.Findthemeanandmodalclass.

Classboundaries Frequency 𝑋! 𝑓 ∙ 𝑋! 7.5–12.5 3 !.!!!".!

!=10 3 ∙ 10 =30

12.5–17.5 5 15 75 17.5–22.5 15 20 300 22.5–27.5 5 25 125 27.5–32.5 2 30 60

30ßn 590ß (𝑓 ∙ 𝑋!)

𝑋 =(𝑓 ∙ 𝑋!)𝑛 =

59030 = 19. 6 → 19.7 𝑚𝑖𝑙𝑒𝑠 𝑝𝑒𝑟 𝑔𝑎𝑙𝑙𝑜𝑛

*Classlimitsareinwhole#sinceboundarieshaveonedecimalplacevalue.Someanwillberoundedofftooneextradecimalplacevalue(tenths)thantheclasslimits.Modalclass:17.5–22.5or18-22

1. Thedatarepresentthemurderrateper100,000individualsinasampleofselectedcitiesintheUnitedStates.Findthemeanandmodalclass.

Class Frequency 5–11 8 12–18 5 19–25 7 26–32 1 33–39 1 40–46 3

7. Thefuelcapacityingallonsof50randomlyselectedcarsisshown.Findthemeanandmodalclass.

Class Frequency 𝑋! 𝑓 ∙ 𝑋! 10–12 6 !"!!"

!=11 6 ∙ 11 =66

13–15 4 14 56 16–18 14 17 238 19–21 15 20 300 22–24 8 23 184 25–27 2 26 52 28–30 1 29 29

50 925

𝑋 =(𝑓 ∙ 𝑋!)𝑛 =

92550 = 18.5 → 18.5 𝑔𝑎𝑙𝑙𝑜𝑛𝑠

ModalClass:19-21