ap statistics quiz 6 model solutions · 2019. 12. 16. · ap statistics quiz 6 model solutions 1) b...

AP Statistics Quiz 6 Model Solutions 1) B Whether adding or subtracting random variables, variances add but standard deviations do not. 2) C ! = 1.75 and ! = 1.08 … 1.75 ± 2 1.08 = −0.41, 3.911 … P(x<4) = 0.09 + 0.36 + 0.35 + 0.13 = 0.93 3) D (D) is the essential difference between the binomial setting and the geometric setting. None of the other statements are true. ____________________________________________________________________________________________________________________ For number (4), let X = the number of adult tickets sold and let Y = the number of children’s tickets sold. 4a) ! = 10 ∙ ! = 10 28.3 = $283 ! = 10 ∙ ! = 10 5.3 = $53 4b) ! = 10 ∙ ! + 6 ∙ ! = 10 28.3 + 6(42.5) = $538 ! = (10 ∙ ! ) ! + (6 ∙ ! ) ! = ! = (10 ∙ 5.3) ! + (6 ∙ 8.1) ! = $71.91 4c) ! = ! − 300 = $538 − 300 = $238 ! = ! = $71.91 (not changed by subtracting a constant) ____________________________________________________________________________________________________________________ 5a) Let F = the number of people in a sample of 15 who subscribe to the “5-second rule”. F follows a binomial distribution with n = 15, p = 0.20, and k = 3. = = ! ∙ (1 − ) !!! = 3 = 3 (0.20) ! ∙ 0.80 !" = 0.2501 There is a 0.2501 probability that exactly 3 out of 15 people subscribe to the 5-second rule. 5b) < 4 = ≤ 3 = : 15, : 0.20, : 3 = 0.64816 There is a 0.2501 probability that less than 4 out of 15 people subscribe to the 5-second rule. 5c) ! = = 15 0.20 = 3 ! = (1 − ) = (15)(0.2)(0.8) = 1.549 If many, many random samples of 15 people are selected, the expected (or average) number of people who subscribe to the 5-second rule is about 3 people. On average, the number of people who subscribe to the 5-second rule in a random sample of 15 typically varies from the mean of 3 people by about 1.549 people. ____________________________________________________________________________________________________________________ 6) Let Y= the number of twelve-year olds to sample to find the first one who can pick out Colorado on a map. Y follows a Geometric distribution with p = 0.24 and k = 5. = = (1 − ) !!! ∙ = 5 = (0.76) ! ∙ 0.24 ≈ 0.0801 There is a 0.0801 probability that you must sample exactly 5 twelve-year olds until the first one who can pick out Colorado. 7a) The condition of independence has been violated because the students are sampled without replacement. This is a problem because the probability of selecting a student who takes AP Statistics changes if one or more previously sampled students take AP Statistics. 7b) The binomial distribution does a good job of estimating this probability anyways because the 10% Condition is met: sample of 50 students is less than or equal to 10% of the population of 1600 students. ≤ ! !" 50 ≤ ! !" (1600) 50 ≤ 160

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Page 1: AP Statistics Quiz 6 Model Solutions · 2019. 12. 16. · AP Statistics Quiz 6 Model Solutions 1) B Whether adding or subtracting random variables, variances add but standard deviations

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

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