APStatistics Quiz6ModelSolutions1)BWhetheraddingorsubtractingrandomvariables,variancesaddbutstandarddeviationsdonot.2)C𝜇! = 1.75and𝜎! = 1.08…1.75± 2 1.08 = −0.41, 3.911 …P(x<4)=0.09+0.36+0.35+0.13=

0.933)D(D)istheessentialdifferencebetweenthebinomialsettingandthegeometricsetting.Noneofthe

otherstatementsaretrue.____________________________________________________________________________________________________________________Fornumber(4),letX=thenumberofadultticketssoldandletY=thenumberofchildren’sticketssold.4a)𝜇! = 10 ∙ 𝜇! = 10 28.3 = $283 𝜎! = 10 ∙ 𝜎! = 10 5.3 = $534b) 𝜇! = 10 ∙ 𝜇! + 6 ∙ 𝜇! = 10 28.3 + 6(42.5) = $538

𝜎! = (10 ∙ 𝜎!)! + (6 ∙ 𝜎!)! = 𝜎! = (10 ∙ 5.3)! + (6 ∙ 8.1)! = $71.914c)𝜇! = 𝜇! − 300 = $538− 300 = $238

𝜎! = 𝜎! = $71.91(notchangedbysubtractingaconstant)____________________________________________________________________________________________________________________5a)LetF=thenumberofpeopleinasampleof15whosubscribetothe“5-secondrule”.

Ffollowsabinomialdistributionwithn=15,p=0.20,andk=3.𝑃 𝐹 = 𝑘 = 𝑛

𝑘 𝑝! ∙ (1− 𝑝)!!! 𝑃 𝐹 = 3 = 3 (0.20)! ∙ 0.80 !" = 0.2501Thereisa0.2501probabilitythatexactly3outof15peoplesubscribetothe5-secondrule.

5b)𝑃 𝐹 < 4 = 𝑃 𝐹 ≤ 3 = 𝑏𝑖𝑛𝑜𝑚𝑐𝑑𝑓 𝑡𝑟𝑖𝑎𝑙𝑠: 15,𝑝: 0.20, 𝑥: 3 = 0.64816

Thereisa0.2501probabilitythatlessthan4outof15peoplesubscribetothe5-secondrule.5c)𝜇! = 𝑛𝑝 = 15 0.20 = 3

𝜎! = 𝑛𝑝(1− 𝑝) = (15)(0.2)(0.8) = 1.549Ifmany,manyrandomsamplesof15peopleareselected,theexpected(oraverage)numberofpeoplewhosubscribetothe5-secondruleisabout3people.Onaverage,thenumberofpeoplewhosubscribetothe5-secondruleinarandomsampleof15typicallyvariesfromthemeanof3peoplebyabout1.549people.

____________________________________________________________________________________________________________________6)LetY=thenumberoftwelve-yearoldstosampletofindthefirstonewhocanpickoutColorado

onamap.YfollowsaGeometricdistributionwithp=0.24andk=5.𝑃 𝑌 = 𝑘 = (1− 𝑝)!!! ∙ 𝑝𝑃 𝑌 = 5 = (0.76)! ∙ 0.24 ≈ 0.0801Thereisa0.0801probabilitythatyoumustsampleexactly5twelve-yearoldsuntilthefirstonewhocanpickoutColorado.

7a)Theconditionofindependencehasbeenviolatedbecausethestudentsaresampledwithout

replacement.ThisisaproblembecausetheprobabilityofselectingastudentwhotakesAPStatisticschangesifoneormorepreviouslysampledstudentstakeAPStatistics.

7b)Thebinomialdistributiondoesagoodjobofestimatingthisprobabilityanywaysbecausethe

10%Conditionismet:sampleof50studentsislessthanorequalto10%ofthepopulationof1600students. 𝑛 ≤ !

!"𝑁 50 ≤ !

!"(1600) 50 ≤ 160