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STAT200(Shook)–Exam3–PracticeExamSolutions
1. C–Aconfidenceintervalusesthesampledatatoestimateapopulation
parameter.
2. A–Tofindtheprobabilitythataz-scoreisgreaterthan2,youwouldneedto
findtheareatotherightof2underthestandardnormalcurve(thezdistribution).
3. B–32/52
Probabilityofblackcard=26/52
Probabilityofredfacecard=6/52
*Theproblemasksfortheprobabilityofablackcard,afacecard(J,Q,orK),
orboth.Theprobabilityofablackfacecardisincludedintheprobabilityof
gettingablackcard.However,youalsoneedtoaccountforthefactthatthere
are6redfacecards(J,Q,K(Hearts)andJ,Q,K(Diamonds).
Totalprobabilityofeitherorboth=(26/52)+(6/52)=32/52
4. B–2.5
E(X)=np=(5)(0.50)=2.5
NotethattheexpectedvaluedoesnothavetobeapossiblevalueofX.
Therefore,youdonotneedtoround2.5to3.
5. D–0.9826,yousimplyneedtolookupthez-scoreof2.11inthestandard
normaltablebecauseyouarelookingfortheareatotheleftofthez-score.
6. D–80%
P(AorC)=P(A)+P(C)–P(AandC)
P(AandC)=0becauseAandCareindependenteventssoitisnotpossibleto
picktwotypesofshoes
P(AorC)=50%+30%–0%=80%
7. B–21/52
Spades=13cards
3to5non-spade=9cards
Totalcards=21
Probability=21/52
8. B–0.60
P(X³2)=0.45+0.10+0.05=0.60
9. A–Acensussurveyseveryoneinthepopulationsotheresultingvaluefrom
thesurveyisapopulationparameter.
10. B–Discreterandomvariablebecausethenumberoffootballgamesattended
hasacountablelistofpossibleoutcomes.
11. D–Thenumberofquestionsyouwouldgetcorrectonamultiplechoicetest
ifyourandomlyguessedonallquestions.Eachquestioniseithercorrector
incorrect,andyouhavethesameprobabilityofgettingeachquestioncorrect
orincorrectbecauseyouarerandomlyguessingoneachquestion.
12. A–0.0174
P(Z≥2.11)=1–0.9826=0.0174
13. B–0.75
P(A|B)=P(AandB)/P(B)
However,wedon’tknowP(AandB)sowewillneedtosolveforitfirst.
P(AorB)=P(A)+P(B)–P(AandB)
0.95=0.8+0.6–P(AandB)
P(AandB)=0.45
P(A|B)=P(AandB)/P(B)
P(A|B)=0.45/0.6=0.75
14. A–True
P(B|C)=P(BandC)/P(C)
Sincetheeventsaremutuallyexclusive,P(BandC)=0,whichmeans
P(B|C)=0.
15. B–Astatisticcanhaveonlyasinglevalueoncethepopulationisdefinedis
falsebecausedifferentsamplesfromthesamepopulationcanhavedifferent
samplestatistics.
16. A–Independentsamplesbecausetwocompletelyseparategroupsarebeing
used.
17. D–0.85
0.15+0.25+0.45=0.85
18. B–Falsebecauseeventsbeingindependentdoesnotmakethemmutually
exclusive.Forexample,SammakingthebaseballteamandSallymakingthe
basketballteamareindependenteventsbecauseonehasnoeffectonthe
other.Eventhoughtheseeventsareindependent,theyaren’tmutually
exclusivebecauseSamcouldmakeitonthebaseballteamandSallycould
alsomakeitonthebasketballteamatthesametime.
19. B–Thenumberoftimesa‘2’isrolledwhenthedieisrolledfivetimesis
binomialbecauseithasasetnumberoftrials(n=5),asuccess(rollinga‘2’)
andfailure(rollinganythingelse),andthetrialsareindependentwitha
constantprobabilityofsuccess(p=1/6).
20. C–Tofindtheareabetween-2and2,youwouldneedtosubtracttheareato
theleftof-2fromtheareatotheleftof2underthestandardnormalcurve
(z-distribution).
21. C–EatingatChipotleandlivingoncampusaremutuallyexclusiveisfalse
becauseitispossibletobothliveoncampusandeatatChipotle.
P(AorB)=P(A)+P(B)–P(AandB)
P(AorB)=40%+25%–10%=55%
22. A–Continuousrandomvariablebecausetheoutcomecouldbeanywhere
withinarangeofvalues.Itisalwayspossibletotakeamoreprecise
measurementoftime.
23. C–0.7531
P(Z≤–1.1)=1–0.8643=0.1357
P(Z≤1.22)=0.8888
P(–1.1≤Z≤1.22)=0.8888–0.1357=0.7531
24. A–0.99
Specificity=100%-1%=99%=0.99
25. C–0.999
Sensitivity=100%-0.1%=99.9%=0.999
26. B–Binomialbecausethisvariablemeetseachofthebinomialvariable
conditionscoveredinchapter8.
27. B–Continuousrandomvariable
28. B–3/8
Forthisproblem,youneedtothinkaboutallpossibleoutcomesofflipping
thecointhreetimes.
H,H,H
H,H,T
H,T,H
T,H,H
H,T,T
T,H,T
T,T,H
T,T,T
Wecanseethatthereare8possibleoutcomesfromflippingacointhree
times.Threeofthoseoutcomesincludegettingheadsexactlyonetime.
3/8=0.375
29. A–P(X = 2) =
(∗*∗+∗,
((∗*)((∗*)
∗
( *
+
(
(1 − ),/* +
n=4
k=2
p=1/3
P(X = k) =
𝑛!
𝑝5 (1 − 𝑝)6/5 𝑘! (𝑛 − 𝑘)!
1∗2∗3∗4
1
P(X = 2) =
∗
(1 ∗ 2)(1 ∗ 2) 3
P(X = 2) =
*
1
(1 − ),/* 3
24
8
= 81 27
30. B–DandEbecausetheyaretheonlyrowswheretheprobabilitiesaddupto
100%,andtheydon’tincludeanynegativevalues.Probabilitiescanneverbe
negativevalues.Theymusthaveavaluebetween0and1,whichisthesame
asbetween0%and100%.
31. D–μbecausethisproblemdealswithasinglemean.
32. C–0.62
Totalparticipants=100
Totalmale=25+20+10=55
P(Male)=55/100=0.55
Totalgraduate=10+7=17
P(Grad)=17/100=0.17
Malegraduates=10
P(MaleandGrad)=10/100=0.10
P(AorB)=P(A)+P(B)–P(AandB)
P(MaleorGrad)=P(Male)+P(Grad)–P(MaleandGrad)
P(MaleorGrad)=0.55+0.17–0.10=0.62
33. A–Normal
34. B–False
P(AandB)=P(A)*P(B)
P(AandB)=0.4*0.2=0.08
35. A–Independentbecauseonegroup’sactionsdonotaffecttheothergroup’s
actions.
36. C–172.35to197.65
s.e.(𝑥)=s/ 𝑛
s.e.(𝑥)=40/ 10
s.e.(𝑥)=12.65
68%ofvaluesfallwithinonestandarderrorofthemean.
185±12.65=(172.35to197.65)
37. D–159.70to210.30
s.e.(𝑥)=s/ 𝑛
s.e.(𝑥)=40/ 10
s.e.(𝑥)=12.65
95%ofvaluesfallwithintwostandarddeviationofthemean.
185±(2)12.65=(159.70to210.30)
38. E–1becauseHomeronlypurchasedfourboxessothereisa100%hewill
find4orlessninjathrowingstars.
39. A–12%
WewanttoknowP(DisciplineandCollege)...sowecanuseourequationfor
P(AandB).
P(DisciplineandCollege)=P(Discipline)*P(College)=0.30*0.40=0.12
40. D–54%
We just found that P(Discipline and College) = 0.12. Next we need to
solve for P(No Discipline and College) using the same approach.
P(No discipline and College) = P(No Discipline) * P(College) = 0.70 * 0.60
= 0.42.
Now we can use the equation P(A or B) = P(A) + P(B) - P(A and B)
You want to think of A as No discipline and college and B as Discipline
and College. If we define A and B this way they are mutually exclusive
events since it is not possible to have no discipline and to have discipline.
That means that the P(A and B) = 0 because the events are mutually
exclusive.
So that means P(A or B) = 0.42 + 0.12 – 0 = 0.54
41. D–AandBarenotmutuallyexclusive,andtheyarenotindependent.Thisis
becausebotheventshavethecommonvalueof4.
42. B–AmongallMacBookowners,theprobabilityofselectingajunior.
43. A–Amongalljuniors,theprobabilityofselectingaMacBookowner.
44. B–0.30/0.40
P(S|B)=P(SandB)/P(B)
P(S|B)=0.30/0.40
45. A–0.1
P(AandB)=P(A)+P(B)–P(AorB)
P(AandB)=0.4+0.3–0.6
P(AandB)=0.1
46. A–0.300
Total#ingroup=60+90+150+100+70+30=500
StudentwhouseTwitterandhave6ormoreclasses=150
Probability=150/500=0.30
47. C–0.200
TotalTwitterusers=60+90+150=300
Twitteruserswith4orlessclasses=60
Probability=60/300=0.20
48. C–pbecausetheproblemdealswithproportions.
49. D–70%ofmenwhodonothavetesticularcancerwilltestnegative
50. B–Itisreasonabletoconcludethereisadifferencebetweengenders
becausetheconfidenceintervalsdonotoverlap.
51. D–Oneminusspecificity.Specificityisfoundbytakingoneminusthe
probabilityofafalsepositive.Thustheprobabilityofafalsepositiveisone
minusspecificity.
52. D–Theprobabilityofanegativetestresultwhenyoudonothavethe
disease.
53. B–(4425+4428–4200)/4685.Wehavetosubtract4200oncebecausethe
valueisincludedinboththetotalthatactuallyhavethediseaseandthetotal
ofnegativeresults.Ifwedon’tsubtractit,wewillbecountingthevaluetwice.
54. D–28/260.Thenumberofpeoplewhohavethediseasebutgetanegative
resultdividedbythetotalnumberofpeoplewhoactuallyhavethedisease.
55. C–Neitherbecausethisisadiscrete,waitingtimevariable.
56. D–12/16.Since,theproblemisaskingforthecumulativedistribution
functionyouwanttofindP(X≤2)soyouaddtheprobabilitiesfrom0,1,and
2.
57. A–1/16.Youknowtheprobabilitiesofalloftheoutcomesmustaddupto
100%.Ifyouadduptheprobabilitiesfor0,1,2,and3,youwillget15/16.So
theunknownprobabilitymuchbe1/16.
58. B–1.2
E(X)=6(0.2)=1.2
59. C–3.1
E(X)=1(0.15)+2(0.10)+3(0.30)+4(0.40)+5(0.05)=3.1
60. C–0.25becauseyouaddtheprobabilitiesfor1and2becauseitwantsto
knowabouttheprobabilityfor2fightsorless.
61. B–Discrete
62. C–Findthecumulativeprobabilityof6successesforabinomialvariable
withn=10andp=¼.Youknowyouwantthecumulativeprobability
becausetheproblemasksabout6orfewer.Iftheproblemhadaskedforthe
probabilityofgettingexactly6correct,thenanswerEwouldhavebeen
correct.
63. D–No,thenumberoftrialsisnotfixed.
64. B–Negative1
x=25
μ=30
σ=5
z=(25–30)/5=–1
65. A–2
x=40
μ=30
σ=5
z=(40–30)/5=2
66. A–0.6915
x=75
μ=70
σ=10
z=(75–70)/10=0.5
Lookup0.50instandardnormaltable=0.6915
TheproblemaskedfortheproportionofstudentswhoscoredLOWERthan
CassidysowealooingforP(Z≤0.50),whichis0.6915.
67. B–(1–0.6915)
x=75
μ=70
σ=10
z=(75–70)/10=0.5
Lookup0.50instandardnormaltable=0.6915
TheproblemaskedfortheproportionofstudentswhoscoredHIGHERthan
CassidysowealooingforP(Z>0.50),whichis1–0.6915.
68. B–(1–0.9332)
x=55
μ=70
σ=10
z=(55–70)/10=–1.5
Ourstandardnormaltableonlyincludespositivezvaluessowewillneedto
lookup1.50instandardnormaltable=0.9332
Thevaluewefoundinthestandardnormaltable,0.9332,istheP(Z<1.50).
However,wewanttofindP(Z<–1.50).Sincethenormaldistributionis
perfectlysymmetric,weknowtheareatotherightof1.50isthesamething
astheareatotheleftof–1.50.
P(Z<–1.50)=1–0.9332
69. A–0.9332
x=55
μ=70
σ=10
z=(55–70)/10=–1.5
Ourstandardnormaltableonlyincludespositivezvaluessowewillneedto
lookup1.50instandardnormaltable=0.9332
Thevaluewefoundinthestandardnormaltable,0.9332,istheP(Z<1.50).
However,wewanttofindP(Z>–1.50).Sincethenormaldistributionis
perfectlysymmetric,weknowtheareatotheleftof1.50isthesamethingas
theareatotherightof–1.50.
P(Z>–1.50)=0.9332
70. B–Astatisticcanhaveonlyasinglevalueoncethepopulationisdefinedis
falsebecauseeachsampletakenfromthesamepopulationcanhavea
differentstatistic.
71. C–Apopulationparameterhasacorrespondingsamplingdistributionis
falsebecausesamplestatisticshavesamplingdistributions,notpopulation
parameters.
72. B–Mutuallyexclusive(disjointed)eventsbecausethetwoeventscannot
happenatthesametimeandtheycovertheentiresamplespace.Theevents
arenotindependentbecauseifyouknowthetutorisfemale,itchangesthe
probabilitythatthetutorismaleto0%.
73. B–Mutuallyexclusive(disjointed)eventsbecausethetwoeventscannot
happenatthesametimeandtheycovertheentiresamplespace.Theevents
arenotindependentbecauseifyouknowthetutorisfemale,itchangesthe
probabilitythatthetutorisKennyto0%.
74. A–Theprobabilitythatthetestisnegativewhenthepersondoeshavethe
disease.Itisonlypossibletohaveafalsenegativeresultifyoudohavethe
disease.Ifyouhaveadisease,youwilleithergetacorrectresultthatyou
havethedisease,oryouwillgetafalsenegativesayingyoudonothavethe
disease.
75. D–μ=themeanamountspentonalcoholbyallfinancemajors
76. B–Pairedsamplesbecauseeachsetoftwinsisamatchedpair.
77. D–665
Thefirstthingtodoisdeterminethez-scorethatcorrespondstothe95th
percentile.Youdothisbylookingup0.9500inthemiddleofthez-table.You
willseethat0.9500isn’tlistedonthez-table;however,wecanseethatit
fallsbetween1.64and1.65sowearegoingtouse1.645asourz-score.You
wouldstillbeabletofindthecorrectanswerusing1.64or1.65.Thismeans
thevaluewearesolvingforwillbe1.645standarddeviationsabovethe
mean.
Standarddeviation=100
100x1.645=164.5
Thismeansthatthe95thpercentilewillbe164.5pointsabovethemeanof
500.
95thpercentile=500+164.5=664.5
78. B–0.18.Makesuretolookatthewordingofthisproblemcarefully.90isthe
numberofpeopleoutof500thatrecognizedtheDJ’sname;however,that
isn’twhattheproblemisaskingfor.Itwantsthemeanofthesampling
distributionforthesampleproportionofpeoplewhoknewtheDJ’sname,
whichis0.18.
79. B–Thenumberofstudentstallerthan6ftinarandomsampleof5students
becauseitistheonlyanswerchoicethatmeetsalloftherequirementsfora
binomialexperiment.
80. A–Independentsamplesbecausetherearetwocompletelyseparate
samples.
81. B–Pairedsamplebecausewearelookingatthedifferencebetweentheright
andleftindexfingersonthesameindividual.