Chapter3:DataDescriptionSection2:MeasuresofVariationWorksheet.

1. Thenumberofsharkattacksanddeathsoverarecent5-yearperiodisshown.Findtherange,varianceandstandarddeviationforthedata.Thisisasample.Attacks 71 64 61 65 57 Deaths 1 4 4 7 4

Attacks:Usingtheformulaforthevariance:

𝑠! = !!! !

!!!

Firstfindthemean:𝑋 = 71+64+61+65+57

5 = 63.6

Variance:𝑠! = 71−63.6 2+ 64−63.6 2+ 61−63.6 2+ 65−63.6 2+ 57−63.6 2

5−1 = 107.24 = 26.8

*Roundingruleforvarianceandstandarddeviationwillbethesameasthemean:oneextradecimalplacevaluethatwhatoccursinthedataset.Standarddeviation:𝑠 = 𝑠! = 26.8 = 5.177 → 5.2Attacks:ShortcutMethodusing:

𝑠! = ! !! ! ! !

!(!!!)

𝑋! = 71! + 64! + 61! + 65! + 57! = 20332𝑋 = 71+ 64+ 61+ 65+ 57 = 318

Variance:𝑠! = 5 20332 − 318 2

5 5−1 = 101660−1011245 4 = 536

20 = 26.8 Standarddeviation:𝑠 = 𝑠! = 26.8 = 5.177 → 5.2

Deaths:Usingtheformulaforthevariance:

𝑠! = !!! !

!!!

Firstfindthemean:𝑋 = 1+4+4+7+4

5 = 4

Variance:𝑠! = 1−4 2+ 4−4 2+ 4−4 2+ 7−4 2+ 4−4 2

5−1 = 184 = 4.5

Standarddeviation:𝑠 = 𝑠! = 4 = 2.12 → 2.1Deaths:UsingShortcutMethod:

𝑠! = ! !! ! ! !

!(!!!)

𝑋! = 1! + 4! + 4! + 7! + 4! = 98𝑋 = 1+ 4+ 4+ 7+ 4 = 20

Variance:𝑠! = 5 98 − 20 2

5 5−1 = 490−4005 4 = 90

20 = 4.5Standarddeviation:𝑠 = 𝑠! = 4 = 2.12 → 2.1

2. Thethreedatasetshavethesamemeanandrangebutisthevariationthesame?Proveyouranswerbycomputingthestandarddeviation.Assumethedatawereobtainedfromsamples.

a. 5,7,9,11,13,15,17

b. 5,6,7,11,15,16,17

c. 5,5,5,11,17,17,17

3. Thenumberofincidentsinwhichpolicewereneededforasampleof10schoolsinAlleghenyCountyis7,37,3,8,48,11,6,0,10,3.Findtherange,varianceandstandarddeviation.

Range:48–0=48UsingShortcutMethod:

𝑠! = ! !! ! ! !

!(!!!)

𝑋! = 7! + 37! + 3! + 8! + 48! + 11! + 6! + 0! + 10! + 3! = 4061𝑋 = 7+ 37+ 3+ 8+ 48+ 11+ 6+ 0+ 10+ 3 = 133

Variance:𝑠! = 10 4061 − 133 2

10 10−1 = 40610−1768910 9 = 22921

90 = 254.677. . .→ 254.7Standarddeviation:𝑠 = 𝑠! = 254.6777778 = 15.9586… → 16.0

4. Thenormaldailyhightemperatures(indegreesFahrenheit)inJanuaryfor10selectedcitiesareasfollows.50,37,29,54,30,61,47,38,34,61Thenormalmonthlyprecipitation(ininches)forthesesame10citiesislistedhere.4.8,2.6,1.5,1.8,1.8,3.3,5.1,1.1,1.8,2.5Findtherange,varianceandstandarddeviationforbothsets.DailyTemperature:Usingtheformulaforthevariance:

𝑠! = !!! !

!!!

Firstfindthemean:𝑋 = 50+37+29+54+30+61+47+38+34+61

10 = 44.1Variance:

𝑠! =50−44.1 2+ 37−44.1 2+ 29−44.1 2+ 54−44.1 2+ 30−44.1 2+ 61−44.1 2+ 47−44.1 2

+ 38−44.1 2+ 34−44.1 2+ 61−44.1 2

10−9 = !"#$.!

!= 147.65 → 147.7

Standarddeviation:𝑠 = 𝑠! = 147.6555556 = 12.15136… → 12.2MonthlyPrecipitation:UsingShortcutMethod:

𝑠! = ! !! ! ! !

!(!!!)

𝑋! = 4.8! + 2.6! + 1.5! + 1.8! + 1.8! + 3.3! + 5.1! + 1.1! + 1.8! + 2.5! = 86.13𝑋 = 4.8+ 2.6+ 1.5+ 1.8+ 1.8+ 3.3+ 5.1+ 1.1+ 1.8+ 2.5 = 26.3

Variance:𝑠! = 10 86.13 − 26.3 2

10 10−1 = 861.3−691.6910 9 = 169.61

90 = 1.8845556. . .→ 1.88Standarddeviation:𝑠 = 𝑠! = 1.8845556 = 1.372791… → 1.37

5. Twelvebatteriesweretestedtoseehowmanyhourstheywouldlast.Thefrequencydistributionisshownhere.Findthevarianceandstandarddeviation.

Hours Frequency 𝑋! 𝑓 ∙ 𝑋! 𝑓 ∙ 𝑋!!1–3 1 !!!

!= 2 1 ∙ 2 =2 1 ∙ 2! = 1 ∙ 2 ∙ 2 =4

4–6 4 5 20 1007–9 5 8 40 32010–12 1 11 11 12113–15 1 14 14 196

12ßn 87ß (𝑓 ∙ 𝑋!) 741ß (𝑓 ∙ 𝑋!!)

Variance:

𝑠! = !( !∙!!! )! !∙!! !

!(!!!)

𝑠! = 12(741)− 87 2

12(12−1) = 8892−756912(11) = 10.022727… → 10.0

Standarddeviation:𝑠 = 𝑠! = 10.022727 = 3.166… → 3.2

6. Thedatarepresentthemurderrateper100,000individualsinasampleofselectedcitiesintheUnitedStates.Findthevarianceandstandarddeviation.

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

7. Thisfrequencydistributionrepresentsthedataobtainedfromasampleofwordprocessorrepairers.Thevaluesarethedaysbetweenservicecallson80machines.Findthevarianceandthestandarddeviation.Classboundaries Frequency 25.5–28.5 5 28.5–31.5 9 31.5–34.5 32 34.5–37.5 20 37.5–40.5 12 40.5–43.5 2

8. Inastudyofreactiontimestoaspecificstimulus,apsychologistrecordedthesedata(inseconds).Findthevarianceandstandarddeviation.

Classlimits Frequency 𝑋! 𝑓 ∙ 𝑋! 𝑓 ∙ 𝑋!!2.1–2.7 12 2.4 28.8 69.122.8–3.4 13 3.1 40.3 124.933.5–4.1 7 3.8 26.6 101.084.2–4.8 5 4.5 22.5 101.254.9–5.5 2 5.2 10.4 54.085.6–6.2 1 5.9 5.9 34.81

40 134.5 485.27Variance:

𝑠! = !( !∙!!! )! !∙!! !

!(!!!)

𝑠! = 40(485.27)− 134.5 2

40(40−1) = 19410.8−18090.2540(39) = 1320.55

1560 = 0.846506… → 0.85Standarddeviation:𝑠 = 𝑠! = 0.846506 = 0.9200578… → 0.92