chapter 3: data description section 3: measures of positions worksheet...

Chapter3:DataDescriptionSection3:MeasuresofPositionsWorksheet

1. Iftheaveragenumberofvacationdaysforaselectionofvariouscountrieshasameanof29.4daysandastandarddeviationof8.6,findthezscoresfortheaveragenumberofdaysineachofthesecountries.

a. Canada 26days𝑧 = 26−29.48.6 = −0.40*Roundoffto2decimalplacevalues.

b. Italy 42days

𝑧 = 42−29.48.6 = 1.47*Roundoffto2decimalplacevalues.

c. UnitedStates13days𝑧 = 13−29.48.6 = −1.91*Roundoffto2decimalplacevalues.

2. TheaverageonastateCDLlicenseexamis80withastandarddeviationof10.Findthe

correspondingzscoreforeachrawscore.a.75 b.90 c.80 d.60 e.95

a)𝑧 = !"!!"!"

= −0.50

b)𝑧 = !"!!"!"

= 1.00

c)𝑧 = !"!!"!"

= 0

d)𝑧 = !"!!"!"

= −2.00

e)𝑧 = !"!!"!"

= 1.50

Negativezscoremeansthedatavalueisbelowthemean.Wecanseethatthedatavalue75islessthanthemeanof80.The0.5tellsusthat75halfastandarddeviationawayfromthemean.Halfastandarddeviationis5units(halfofthestandarddeviation10)

Positivezscoremeansthedatavalueisabovethemean.Wecanseethatthedatavalue90ismorethanthemeanof80.The1.00tellsusthat90isonewholestandarddeviationawayfromthemean.Onewholestandarddeviationis10units.

zscoreof0meansthatthedatavalueisequaltothemean.

Negativezscore.Belowthemean.2.00meansthedatavalueis2standarddeviationsawayfromthemean.Onestandarddeviationis10,sotwostandarddeviationsis20.Weseethat60is20units–2standarddeviations–belowthemean80.

Positivezscore.Abovethemean.1.50meansthedatavalueisoneandahalfstandarddeviationsawayfromthemean.Onestandarddeviationis10,halfis5,so1.50standarddeviationis15.Weseethat95is15units–1.5standarddeviations–abovethemean80.

3. Theaverageteacherssalaryinaparticularstateis$54,166.Ifthestandarddeviationis$10,500,findthesalariescorrespondingtothefollowingzscores.a.2 b.-1 c.0 d.2.5 e.-1.6

4. Whichofthescoreindicatesthehighestrelativeposition?a. Ascoreof3.2onatestwith𝑋 = 4.6ands=1.5

b. Ascoreof630onatestwith𝑋 = 800ands=200

c. Ascoreof43onatestwith𝑋 = 50ands=5

Weusetheformula:𝑧 = !"#$%! !"#$

!"#$%#&% !"#$%&$'(

PluginthevaluesandsolveforX.Wecanalsouse:𝑋 = 𝑧𝑠𝑐𝑜𝑟𝑒 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛) +𝑚𝑒𝑎𝑛

a)2 = !!!"#$$!"!""

2(10500)=X–5416621000=X–54166X=$75166

b)−1 = !!!"#$$!"!""

-1(10500)=X–54166-10500=X–54166X=$43666

c)0 = !!!"#$$!"!""

0(10500)=X–541660=X–54166X=$54166

Fora)wecanseethatazscoreof2meansthevaluewillbe2standarddeviationsabove(sincezscoreis+)themean.1standarddeviationis$10500,so2standarddeviationsis$21000.Sothevalueis$21000abovethemean,hence$75166.b)-1zscoremeans1standarddeviationbelowthemean.Sothevalueis$10500belowthemean,hence$43666.c)0zscoremeansthedatavalueisequaltothemean.

𝑧 = !.!!!.!!.!

= −0.93

d)2.5 = !!!"#$$!"!""

2.5(10500)=X–5416626250=X–54166X=$80416

e)1.6 = !!!"#$$!"!""

-1.6(10500)=X–54166-16800=X–54166X=$37366

𝑧 = !"# !!""!""

= −0.85

𝑧 = !" !!"!

= −1.40

Althoughallthescoresarebelowthemean(neg.zscore),thehigherrelativepositionwouldbethescoreof630withameanof800sinceitisclosertothemean(only0.85standarddeviationsbelowthemean)comparedtotheother2scoresthatare0.93and1.40standarddeviationsbelowthemean.Iftherewasapositivezscoreamongthisset,thescorewiththepositivezscorewillhaveahigherrelativeposition.

Highervs.BetterRelativePosition:Higherrelativepositionhasahigherzscore.Betterrelativepositiondependsonthesituation.With#4,thebetterrelativepositionwouldbe(b)thescoreof630sincethezscoreishigherthantheothersandwearetalkingaboutscoresonanexam.Supposewetalkaboutdebt.Theaveragedebtis$13000.Ahigherzscorewouldnotbeagoodthing.Wewoulddefinitelywantalowerzscorepreferablybelowtheaverage.

5. Theaverageweeklyearningsindollarsforvariousindustriesarelistedbelow.Findthepercentilerankofeachvalue.804 736 959 489 777 623 597 524 228

6. Forthedatafrom#5,whatvaluecorrespondstothea. 40thpercentile?

b. 25thpercentile?

7. Findthepercentilerankofeachtestscoreinthedataset.12 28 28 35 42 47 47 47 50

8. Thefollowingdatasetrepresentsthevaluesinbillionsofdollarsofthedamageof10hurricanes.

1.1 1.7 1.9 2.1 2.2 2.5 3.3 6.2 6.8 20.3Findthevaluethatcorrespondstothe

a. 50thpercentile

b. 30thpercentile

228,489,524,597,623,736,777,804,959

228 → 𝑝 = !!!.!(!)!

∙ 100 = 5.5 → 6𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

489 → 𝑝 = !!!.!(!)!

∙ 100 = 16. 6! → 17𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

524 → 𝑝 = !!!.!(!)!

∙ 100 = 27. 7! → 28𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

597 → 𝑝 = !!!.!(!)!

∙ 100 = 38. 8! → 39𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

623 → 𝑝 = !!!.!(!)!

∙ 100 = 50 → 50𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

736 → 𝑝 = !!!.!(!)!

∙ 100 = 61. 1! → 61𝑠𝑡 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

777 → 𝑝 = !!!.!(!)!

∙ 100 = 72. 2! → 72𝑛𝑑 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

804 → 𝑝 = !!!.!(!)!

∙ 100 = 83. 3! → 83𝑟𝑑 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

959 → 𝑝 = !!!.!(!)!

∙ 100 = 94. 4! → 94𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

𝑐 = !(!")!""

= 3.6 → 4 → (4𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 )597

𝑐 = !(!")!""

= 2.25 → 3 → (3𝑟𝑑 𝑣𝑎𝑙𝑢𝑒 ) 524

Thepercentilesin#5andthedatathatcorrespondstothepercentilesin#6areestimates.You’llnoticethatfor#5,597isinthe39thpercentilewhereas#6,thedatathatcorrespondstothe40thpercentileis597.#6,noticethatcwhencalculateddidnotcomeouttobeawhole#.SoweroundUPtothenextwhole#.Thisisyournewc.Thenlookforthecthvalue.

12 → 𝑝 = !!!.!(!)!

∙ 100 = 5. 5! → 6𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

28 → 𝑝 = !!!.!(!)!

∙ 100 = 22. 2! → 22𝑛𝑑 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

35 → 𝑝 = !!!.!(!)!

∙ 100 = 38.8!!!!!! → 39𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

42 → 𝑝 = !!!.!(!)!

∙ 100 = 50 → 50𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

47 → 𝑝 = !!!.!(!)!

∙ 100 = 72. 2! → 72𝑛𝑑 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

50 → 𝑝 = !!!.!(!)!

∙ 100 = 94. 4! → 94𝑡ℎ 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒

𝑐 = !"(!")!""

= 5 → 𝐿𝑜𝑜𝑘 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 5𝑡ℎ 𝑎𝑛𝑑 6𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 → !.!!!.!!

= 2.35

𝑐 = !"(!")!""

= 3 → 𝐿𝑜𝑜𝑘 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 3𝑟𝑑 𝑎𝑛𝑑 4𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 → !.!!!.!!

= 2

Noticethatcwhencalculatedcameouttoawhole#.Becausecequaledawhole#,welookforthevaluebetween(midpoint)thecthvalueandthe(c+1)thvalue.Forexample:ifccameouttobeexactly6,thenwewouldfindthevaluebetween(midpoint)the6thandthe(6+1)thor7thvalue.

c. 80thpercentile

d. 20thpercentile

9. Forthefollowingdatasets,findthequartilesandcheckforoutliers.a. 16,18,22,19,3,21,17,20

b. 24,32,54,31,16,18,19,14,17,20

c. 321,343,350,327,200

𝑐 = !"(!")!""

= 8 → 𝐿𝑜𝑜𝑘 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 8𝑡ℎ 𝑎𝑛𝑑 9𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 → !.!!!.!!

= 6.5

𝑐 = !"(!")!""

= 2 → 𝐿𝑜𝑜𝑘 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 2𝑛𝑑 𝑎𝑛𝑑 3𝑟𝑑 𝑣𝑎𝑙𝑢𝑒 → !.!!!.!!

= 1.8

3,16,17,18,19,20,21,22Q2=18.5(medianforthewholedataset)Q1=16.5(medianamongdatavalueslessthanQ2,18.5)Q3=20.5(medianamongdatavaluesgreaterthanQ2,18.5)

IQR=Q3–Q1=20.5–16.5=4IQR(1.5)=4(1.5)=6Takethis6andsubtractfromQ1.àQ1–(IQR)(1.5)=16.5–4(1.5)=16.5–6=10.5Takethis6andaddtoQ3.àQ3–(IQR)(1.5)=20.5+4(1.5)=20.5+6=26.5Anydatavalueslessthan10.5oranydatavaluesgreaterthan26.5isanoutlier.3isanoutlier.

14,16,17,18,19,20,24,31,32,54Q2=19.5(medianforthewholedataset)Q1=17(medianamongdatavalueslessthanQ2,19.5)Q3=31(medianamongdatavaluesgreaterthanQ2,19.5)

IQR=Q3–Q1=31–17=14IQR(1.5)=14(1.5)=21Takethis21andsubtractfromQ1.àQ1–(IQR)(1.5)=17–14(1.5)=17-21=-4Takethis21andaddtoQ3.àQ3–(IQR)(1.5)=31+14(1.5)=31+21=52Anydatavalueslessthan-4oranydatavaluesgreaterthan52isanoutlier.54isanoutlier.

200,321,327,343,350Q2=327(medianforthewholedataset)Q1=260.5(medianamongdatavalueslessthanQ2,327)Q3=346.5(medianamongdatavaluesgreaterthanQ2,327)

IQR=Q3–Q1=346.5–260.5=86IQR(1.5)=86(1.5)=129Takethis129andsubtractfromQ1.àQ1–(IQR)(1.5)=260.5–86(1.5)=260.5–129=131.5Takethis129andaddtoQ3.àQ3–(IQR)(1.5)=346.5+86(1.5)=346.5+129=475.5Anydatavalueslessthan131.5oranydatavaluesgreaterthan475.5isanoutlier.Nooutlier.

d. 88,72,97,84,86,85,100

e. 145,119,122,118,125,116

f. 14,16,27,18,13,19,36,15,20

72,84,85,86,88,97,100Q2=86(medianforthewholedataset)Q1=84(medianamongdatavalueslessthanQ2,86)Q3=97(medianamongdatavaluesgreaterthanQ2,86)

IQR=Q3–Q1=97–84=13IQR(1.5)=13(1.5)=19.5Takethis19.5andsubtractfromQ1.àQ1–(IQR)(1.5)=84–13(1.5)=84–19.5=64.5Takethis19.5andaddtoQ3.àQ3–(IQR)(1.5)=97+13(1.5)=97+19.5=116.5Anydatavalueslessthan64.5oranydatavaluesgreaterthan116.5isanoutlier.Nooutlier.

116,118,119,122,125,145Q2=120.5(medianforthewholedataset)Q1=118(medianamongdatavalueslessthanQ2,120.5)Q3=125(medianamongdatavaluesgreaterthanQ2,120.5)

IQR=Q3–Q1=125–118=7IQR(1.5)=7(1.5)=10.5Takethis10.5andsubtractfromQ1.àQ1–(IQR)(1.5)=118–7(1.5)=118–10.5=107.5Takethis10.5andaddtoQ3.àQ3–(IQR)(1.5)=125+7(1.5)=125+10.5=135.5Anydatavalueslessthan107.5oranydatavaluesgreaterthan135.5isanoutlier.145isanoutlier.

13,14,15,16,18,19,20,27,36Q2=18(medianforthewholedataset)Q1=14.5(medianamongdatavalueslessthanQ2,18)Q3=23.5(medianamongdatavaluesgreaterthanQ2,18)

IQR=Q3–Q1=23.5–14.5=9IQR(1.5)=9(1.5)=13.5Takethis13.5andsubtractfromQ1.àQ1–(IQR)(1.5)=14.5–9(1.5)=14.5–13.5=1Takethis13.5andaddtoQ3.àQ3–(IQR)(1.5)=23.5+9(1.5)=23.5+13.5=37Anydatavalueslessthan1oranydatavaluesgreaterthan37isanoutlier.Nooutlier.