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b 1
Chapterb 7b Homework
Questionb 1:b Whatb isb theb mean,b standardb deviation,b andb sampleb size?
Mean=b 4b hours
Standardb deviation=b 1.2b hoursb
Sampleb sizeb =b 16b employeesb
Questionb 2:b Completeb theb distribution.
a) Xb ~b Nb (4,b 1.2)
b) b Ub (1.2,b 16)b
Questionb 3:b Findb theb probabilityb thatb oneb reviewb willb takeb Yoonieb fromb 3.5b tob
4.25b hours.b
a)b
b)b b P(3.5<x<4.25)
=b z1b =b (x-b u)/b standardb deviation
=b (3.5b b 4)/b 1.2
=b (4.25b b 4)/b 1.2
=b -b 0.4167;b 0.2084
P(b -0.4167b <zb <b 0.2084)
P(3.5<x<4.25)b =b 0.2441
Questionb 17:b Findb theb probabilityb thatb theb sumb ofb theb 100b valuesb isb lessb thanb
3,900.
Meanb =b 39.01;b standardb deviation=b 0.5
Meanb sumsb =b (n)(x)b =b 39.01*b 100=b 3901
Pb xb <b 3900)b =b 0.4207
Questionb 19:b Findb theb sumb ofb ab z-b scoreb ofb -2.5
Zb =b (xb b mean)/b sd
-2.5b =b (xb b 39.01)/b 0.5
b 2
Xb =b (-1.25)b b 39.01
Xb =b 37.76*b 100
Xb =b 3776
Questionb 25:b Whatb isb Pb (Σxb >b 290)?
Meanb =b 12;b sdb =b 1b nb =b 25
=b 12(25)b =b 300
=b 1(25)b =b 25
Therefore,b Pb (Σxb >b 290)b =b 0.9772
Questionb 29:b Anb unknownb distributionb hasb ab meanb ofb 25b andb ab standardb deviationb
ofb six.b Letb Xb =b oneb objectb fromb thisb distribution.b Whatb isb theb sampleb sizeb ifb theb
standardb deviationb ofb ΣXb isb 42?
Meanb =b 25;b sdb =b 6
42=b (√n)(σ
x
)
√nb =b 42/b 6
N=b 7)**2
Nb =b 49
Questionb 34:b Whatb isb theb meanb ofb ΣX?
ΣXb =b 100*100
=b 10,000
Questionb 58:b Findb theb firstb quartileb forb theb sums
(6<b x<b 10);b Nb =b 50
a=b 6;b bb =b 10
meanb =b (16)/b 2b =b 8
sdb =b sqrtb ((10b b 6)^2)b /b 12
sdb =b 1.1547
sumb xb =b N(b 50*8,b (50**2)(1.1547)
N(400,b 8.13)
Zb =b 1.29
1.29b =b (q1b b 400)/b 8.13
b 3
10.48b =b q1b b 400
Q1b =b 410.49
Questionb 64:b Supposeb thatb ab categoryb ofb world-classb runnersb isb knownb tob runb ab
marathonb ofb 26b milesb inb anb averageb ofb 145b minutesb withb ab standardb deviationb ofb
14b minutes.b Considerb 49b ofb theb races.b Letb meb ofb averageb ofb theb racesb 49b races
Thisb isb theb averageb numberb ofb minutesb takenb inb theb race.b
80
th
b percentile;b 0.8
Zb scoreb =b 0.78814
Meanb =b 49b marathons
Sdb =b 14
0.78814b =b (xb b 49)/b 98
77.224b =b xb b 49b
Xb =b 126.2b b
Questionb 65:b Theb lengthb ofb songsb inb collectors’b iTunesb albumb collectionb isb
uniformlyb distributedb fromb twob tob 3.5b minutes.b Supposeb web randomlyb pickb fiveb
albumsb fromb theb collection.b Thereb areb ab totalb ofb 43b songsb onb theb fiveb albums
a) Theb lengthb ofb ab song,b inb minutesb inb theb collection
b) Ub (2,b 3.5)
c) Theb averageb lengthb inb minutesb ofb theb songsb fromb ab sampleb ofb fiveb albumsb fromb
theb collection
d) Z=b (xb b mean)/b sd/b sqrtb (n)
-0.675=b (q1b b 2.75)
Q1b =b 2.706
e) IQRb =b Q3b b Q
=b 2.794b b 2.706
=b 0.088
Questionb 66:b Inb 1940b theb averageb sizeb ofb ab USb farmb wasb 174b acres.b Let’sb sayb thatb
theb standardb deviationb wasb 55b acres.b Supposeb web randomlyb surveyb 38b farmersb
fromb 1940
a) Theb sizeb ofb farmsb inb acres
b) Theb averageb lengthb ofb farmsb fromb theb sampleb ofb 38b farmersb selected
c) Ub (
d) Zb =b (136b b 174)/b 55
Zb =b -b 0.690
Q1b =b 0,2451
b 4
Questionb 67:b Determineb whichb ofb theb followingb isb trueb andb whichb isb false,b thenb
a) Trueb theb meanb ofb ab samplingb distributionb ofb theb meansb isb approximatelyb theb
meanb ofb theb data
b) True.b Theb largerb theb sampleb theb closerb theb samplingb distributionb ofb theb meansb
becomesb normalb accordingb tob centralb limitb theorem.b
c) Theb standardb deviationb ofb theb samplingb distributionb ofb theb meansb willb decreaseb
makingb itb approximatelyb theb sameb asb theb standardb deviationb ofb xb asb theb sampleb
sizeb increases.b
Questionb 52:b Findb theb 80
th
b percentileb forb theb totalb lengthb ofb timeb 64b batteriesb last
Questionb 73:b Supposeb thatb theb durationb ofb ab particularb typeb ofb criminalb trialb isb
knownb tob haveb ab meanb ofb 21b daysb andb ab standardb deviationb ofb sevenb days.b Web
randomlyb ampleb nineb trialsb
a) Theb totalb numberb ofb daysb forb nineb trials
b) Nb (189,b 21)
c) 0.0432
d) 162.09;b ninetyb percentb ofb theb totalb nineb trialsb ofb thisb typeb willb lastb 162b daysb orb
more.
Questionb 81:b Theb 90
th
b percentileb sampleb averageb waitb timeb (inb minutes)b forb ab
sampleb ofb 100b ridersb is:b
90
th
b percentileb =0.9
=b (75b -0)/b 2b =b 37.5
Standardb deviationb =b sqrtb ((75-b 0)b **2)/b 10b =b 2.2
invNormb (0.90,b 37,5,b 2.2)
=b 40.3
Questionb 83:b What’sb theb approximateb probabilityb thatb theb averageb priceb forb 16b gasb
stationsb isb overb $b 4.69?
Meanb =b 4.59;b 4.69
Sdb =b 0.01
Xb =b 16
Zb scores=b (16b -4.59)/b 0.10b =b 1.141
Zb scoreb =b (16b -4.69)/b 0.10b =b 1.113b
Probabilityb =b 0.87286b b 0.87076
Probabilityb =b 0.0021b whichb isb almostb zero
b 5
Questionb 33:b isb Pb (Σxb <b 1,186)?
Meanb sum=b 400b *3b =b 1200
Meanb sdb =b sqrtb 0.7b *b 400b =b 334.66
=b normalb cdfb (1186,b e99,b (0.7*b 334.66)b
=b 0.15868
Questionb 34:b Whatb isb theb meanb ofb x?
Meanb =b 100;b sdb =b 100;b nb =100
Xb =b 1
Meanb sumb ofb xb =b (100*100)
Meanb sumb ofb xb =b 10,000
b
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