solutions to problems (4-7)

Instructors Solution Manual

4.1Enumeration of the six parts: D1, D2, D3, A4, A5, A6D = Defective partA = Acceptable part

Sample Space:

D1 D2, D2 D3, D3 A5D1 D3, D2 A4, D3 A6D1 A4, D2 A5, A4 A5D1 A5, D2 A6, A4 A6D1 A6, D3 A4, A5 A6There are 15 members of the sample space

The probability of selecting exactly one defect out of two is:9/15 = .60

4.2X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9}, and Z = {1, 2, 3, 4, 7,} a)X Z = {1, 2, 3, 4, 5, 7, 8, 9} b)X Y = {7, 9} c)X Z = {1, 3, 7} d)X Y Z = {1, 2, 3, 4, 5, 7, 8, 9} e)X Y Z = {7} f)(X Y) Z = {1, 2, 3, 4, 5, 7, 8, 9} {1, 2, 3, 4, 7} = {1, 2, 3, 4, 7} g)(Y Z) (X Y) = {2, 4, 7} {7, 9} = {2, 4, 7, 9} h) X or Y = X Y = {1, 2, 3, 4, 5, 7, 8, 9}i)Y and Z = Y Z = {2, 4, 7}

4.3If A = {2, 6, 12, 24} and the population is the positive even numbers through 30, A = {4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30}

4.4 6(4)(3)(3) = 216

4.5Enumeration of the six parts: D1, D2, A1, A2, A3, A4D = Defective partA = Acceptable part

Sample Space:

D1 D2 A1, D1 D2 A2, D1 D2 A3,D1 D2 A4, D1 A1 A2, D1 A1 A3,D1 A1 A4, D1 A2 A3, D1 A2 A4,D1 A3 A4, D2 A1 A2, D2 A1 A3,D2 A1 A4, D2 A2 A3, D2 A2 A4,D2 A3 A4, A1 A2 A3, A1 A2 A4,A1 A3 A4, A2 A3 A4Combinations are used to counting the sample space because sampling is done without replacement.

6C3 = = 20

Probability that one of three is defective is: 12/20 = 3/5 .60 There are 20 members of the sample space and 12 of them have exactly 1 defective part.

4.6 107 = 10,000,000 different numbers

4.720C6 = = 38,760It is assumed here that 6 different (without replacement) employees are to be selected.

4.8 P(A) = .10, P(B) = .12, P(C) = .21, P(A C) = .05 P(B C) = .03 a) P(A C) = P(A) + P(C) - P(A C) = .10 + .21 - .05 = .26 b) P(B C) = P(B) + P(C) - P(B C) = .12 + .21 - .03 = .30 c) If A, B mutually exclusive, P(A B) = P(A) + P(B) = .10 + .12 = .22

4.9 D E F

A 5 81225

B10 6 420

C 8 2 515

23162160

a) P(A D) = P(A) + P(D) - P(A D) = 25/60 + 23/60 - 5/60 = 43/60 = .7167b) P(E B) = P(E) + P(B) - P(E B) = 16/60 + 20/60 - 6/60 = 30/60 = .5000c) P(D E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500d) P(C F) = P(C) + P(F) - P(C F) = 15/60 + 21/60 - 5/60 = 31/60 = .5167

4.10

E F

A.10.03.13

B.04.12.16

C.27.06.33

D.31.07.38

.72.281.00

a) P(A F) = P(A) + P(F) - P(A F) = .13 + .28 - .03 = .38 b) P(E B) = P(E) + P(B) - P(E B) = .72 + .16 - .04 = .84 c) P(B C) = P(B) + P(C) =.16 + .33 = .49 d) P(E F) = P(E) + P(F) = .72 + .28 = 1.000 4.11A = event of having flown in an airplane at least onceT = event of having ridden in a train at least once P(A) = .47 P(T) = .28 P (ridden either a train or an airplane) =P(A T) = P(A) + P(T) - P(A T) = .47 + .28 - P(A T) Cannot solve this problem without knowing the probability of the intersection. We need to know the probability of the intersection of A and T, the proportion who have ridden both or determine if these two events are mutually exclusive.

4.12 P(L) = .75 P(M) = .78 P(M L) = .61a) P(M L) = P(M) + P(L) - P(M L) = .78 + .75 - .61 = .92 b) P(M L) but not both = P(M L) - P(M L) = .92 - .61 = .31 c) P(NM NL) = 1 - P(M L) = 1 - .92 = .08 Note: the neither/nor event is solved for here by taking the complement of the union.

4.13 Let C = have cable TV Let T = have 2 or more TV sets P(C) = .67, P(T) = .74, P(C T) = .55

a) P(C T) = P(C) + P(T) - P(C T) = .67 + .74 - .55 = .86b) P(C T but not both) = P(C T) - P(C T) = .86 - .55 = .31c) P(NC NT) = 1 - P(C T) = 1 - .86 = .14d) The special law of addition does not apply because P(C T) is not .0000. Possession of cable TV and 2 or more TV sets are not mutually exclusive.

4.14Let T = review transcript F = consider faculty referencesP(T) = .54 P(F) = .44 P(T F) = .35 a) P(F T) = P(F) + P(T) - P(F T) = .44 + .54 - .35 = .63 b) P(F T) - P(F T) = .63 - .35 = .28 c) 1 - P(F T) = 1 - .63 =.37 d) Faculty References

Y N

Transcript Y.35.19 .54

N.09.37 .46

.44.561.00

4.15 C D E F

A 51116 840

B 2 3 5 717

714211557

a) P(A E) = 16/57 = .2807 b) P(D B) = 3/57 = .0526 c) P(D E) = .0000 d) P(A B) = .0000

4.16 D E F

A.12.13.08.33

B.18.09.04.31

C.06.24.06.36

.36.46.181.00

a) P(E B) = .09 b) P(C F) = .06 c) P(E D) = .00

4.17Let D = Defective part

a) (without replacement)

P(D1 D2) = P(D1) P(D2 D1) = = .0122

b) (with replacement)

P(D1 D2) = P(D1) P(D2) = = .0144

4.18Let U = Urban I = care for Ill relatives P(U) = .78 P(I) = .15 P(IU) = .11 a) P(U I) = P(U) P(IU) P(U I) = (.78)(.11) = .0858

b) P(U NI) = P(U) P(NIU) but P(IU) = .11 So, P(NIU) = 1 - .11 = .89 and P(U NI) = P(U) P(NIU) = (.78)(.89) = .6942

c) U

Yes No

I Yes .15

No .85

.78 .22

The answer to a) is found in the YES-YES cell. To compute this cell, take 11% or .11 of the total (.78) people in urban areas. (.11)(.78) = .0858 which belongs in the YES-YES" cell. The answer to b) is found in the Yes for U and no for I cell. It can be determined by taking the marginal, .78, less the answer for a), .0858. d. P(NU I) is found in the no for U column and the yes for I row (1st row and 2nd column). Take the marginal, .15, minus the yes-yes cell, .0858, to get .0642.

4.19 Let S = stockholder Let C = college

P(S) = .43 P(C) = .37 P(CS) = .75

a) P(NS) = 1 - .43 = .57 b) P(S C) = P(S) P(CS) = (.43)(.75) = .3225 c) P(S C) = P(S) + P(C) - P(S C) = .43 + .37 - .3225 = .4775 d) P(NS NC) = 1 - P(S C) = 1 - .4775 = .5225 e) P(NS NC) = P(NS) + P(NC) - P(NS NC) = .57 + .63 - .5225 = .6775 f) P(C NS) = P(C) - P(C S) = .37 - .3225 = .0475

The matrix: C

Yes No

S Yes.3225.1075 .43

No.0475.5225 .57

.37 .631.00

4.20 Let F = fax machine Let P = personal computer

Given: P(F) = .10 P(P) = .52 P(PF) = .91a) P(F P) = P(F) P(P F) = (.10)(.91) = .091b) P(F P) = P(F) + P(P) - P(F P) = .10 + .52 - .091 = .529c) P(F NP) = P(F) P(NP F)

Since P(P F) = .91, P(NP F)= 1 - P(P F) = 1 - .91 = .09 P(F NP) = (.10)(.09) = .009d) P(NF NP) = 1 - P(F P) = 1 - .529 = .471e) P(NF P) = P(P) - P(F P) = .52 - .091 = .429

The matrix:

P NP

F.091.009.10

NF.429.471.90

.520.4801.00

4.21Let S = safety Let A = age

P(S) = .30P(A) = .39P(AS) = .87a) P(S NA) = P(S) P(NAS)

but P(NAS) = 1 - P(AS) = 1 - .87 = .13 P(S NA) = (.30)(.13) = .039b) P(NS NA) = 1 - P(S A) = 1 - [P(S) + P(A) - P(S A)] but P(S A) = P(S) P(AS) = (.30)(.87) = .261 P(NS NA) = 1 - (.30 + .39 - .261) = .571c) P(NS A) = P(NS) - P(NS NA)

but P(NS) = 1 - P(S) = 1 - .30 = .70 P(NS A) = .70 - 571 = .129

The matrix: A

Yes No

S Yes.261.039 .30

No.129.571 .70

.39 .611.00

4.22Let C = ceiling fans Let O = outdoor grillP(C) = .60 P(O) = .29P(C O) = .13a) P(C O)= P(C) + P(O) - P(C O) = .60 + .29 - .13 = .76b) P(NC NO) = 1 - P(C O)= 1 - .76 = .24c) P(NC O) = P(O) - P(C O) = .29 - .13 = .16d) P(C NO) = P(C) - P(C O) = .60 - .13 = .47

The matrix: O

Yes No

C Yes .13 .47 .60

No .16 .24 .40

.29 .711.00

4.23 E F G

A 15 12 835

B 11 17 1947

C 21 32 2780

D 18 13 1243

65 74 66205

a) P(GA) = 8/35 = .2286 b) P(BF) = 17/74 = .2297 c) P(CE) = 21/65 = .3231

d) P(EG) = .0000

4.24

C D

A .36 .44 .80

B .11 .09 .20

.47 .531.00

a) P(CA) = .36/.80 = .4500 b) P(BD) = .09/.53 = .1698 c) P(AB) = .0000

4.25 Calculator

Yes No

Computer Yes 46 3 49

No 11 15 26

57 18 75

Select a category from each variable and test P(V1V2) = P(V1).

For example,P(Yes ComputerYes Calculator) = P(Yes Computer)?

?

.8070 .6533 Since this is one example that the conditional does not equal the marginal in is matrix, the variable, computer, is not independent of the variable, calculator.

4.26 Let C = construction Let S = South Atlantic

83,384 total failures10,867 failures in construction 8,010 failures in South Atlantic 1,258 failures in construction and South Atlantic

a) P(S) = 8,010/83,384 = .09606b) P(C S) = P(C) + P(S) - P(C S) =

10,867/83,384 + 8,010/83,384 - 1,258/83,384 = 17,619/83,384 = .2113

c) P(CS) = = .15705

d) P(SC) = = .11576

e) P(NSNC) =

but NC = 83,384 - 10,867 = 72,517

and P(NC) = 72,517/83,384 = .869675

Therefore, P(NSNC) = (1 - .2113)/(.869675) = .9069

f) P(NSC) =

but P(C) = 10,867/83,384 = .1303 P(C S) = 1,258/83,384 = .0151

Therefore, P(NS C) = (.1303 - .0151)/.1303 = .8842

4.27Let E = EconomyLet Q = QualifiedP(E) = .46P(Q) = .37P(E Q) = .15a) P(EQ) = P(E Q)/P(Q) = .15/.37 = .4054b) P(QE) = P(E Q)/P(E) = .15/.46 = .3261c) P(QNE) = P(Q NE)/P(NE) but P(Q NE) = P(Q) - P(Q E) = .37 - .15 = .22

P(NE) = 1 - P(E) = 1 - .46 = .54

P(QNE) = .22/.54 = .4074d) P(NE NQ) = 1 - P(E Q) = 1 - [P(E) + P(Q) - P(E Q)]

= 1 - [.46 + .37 - .15] = 1 - (.68) = .32

The matrix: Q

Yes No

E Yes .15 .31 .46

No .22 .32 .54

.37 .631.00

4.28Let EM = email while on phoneLet TD = to-do lists during meetings

P(EM) = .54P(TDEM) = .20a) P(EM TD) = P(EM) P(TDEM) = (.54)(.20) = .1080b) P(not TDEM) = 1 - P(TDEM) = 1 - .20 = .80c) P(not TD EM) = P(EM) - P(EM TD) = .54 - .1080 = .4320 Could have been worked as: P(not TD EM) = P(EM) P(not TDEM) = (.54)(.80) = .4320

The matrix: TD

Yes No

EM Yes .1080 .4320 .54

No .46

1.00

4.29 Let H = hardware Let S = softwareP(H) = .37P(S) = .54P(SH) = .97

a) P(NSH) = 1 - P(SH) = 1 - .97 = .03b) P(SNH) = P(S NH)/P(NH) but P(H S) = P(H) P(SH) = (.37)(.97) = .3589 so P(NH S) = P(S) - P(H S) = .54 - .3589 = .1811

P(NH) = 1 - P(H) = 1 - .37 = .63

P(SNH) = (.1811)/(.63) = .2875c) P(NHS) = P(NH S)/P(S) = .1811//54 = .3354d) P(NHNS) = P(NH NS)/P(NS) but P(NH NS) = P(NH) - P(NH S) = .63 - .1811 = .4489

and P(NS) = 1 - P(S) = 1 - .54 = .46

P(NHNS) = .4489/.46 = .9759

The matrix: S

Yes No

H Yes .3589 .0111 .37

No .1811 .4489 .63

.54 .461.00

4.30Let R = agreed or strongly agreed that lack of role models was a barrier Let S = agreed or strongly agreed that gender-based stereotypes was a barrier

P(R) = .43P(S) = .46P(RS) = .77 P(not S) = .54

a.) P(not RS) = 1 - P(RS) = 1 - .77 = .23b.) P(not SR) = P(not S R)/P(R) but P(S R) = P(S) P(RS) = (.46)(.77) = .3542 so P(not S R) = P(R) - P(S R) = .43 - .3542 = .0758

Therefore, P(not SR) = (.0758)/(.43) = .1763c.) P(not Rnot S) = P(not R not S)/P(not S) but P(not R not S) = P(not S) - P(not S R) = .54 - .0758 = .4642

P(not Rnot S) = .4642/.54 = .8596

The matrix: S

Yes No

R Yes .3542 .0758 .43

No .1058 .4642 .57

.46 .541.00

4.31 LetA = product produced on Machine AB = product produces on Machine BC = product produced on Machine CD = defective product

P(A) = .10 P(B) = .40 P(C) = .50 P(DA) = .05P(DB) = .12P(DC) = .08

Event PriorConditional Joint Revised

P(Ei) P(DEi) P(D Ei)

A .10 .05 .005.005/.093=.0538

B .40 .12 .048.048/.093=.5161

C .50 .08 .040.040/.093=.4301

P(D)=.093

Revise:P(AD) = .005/.093 = .0538 P(BD) = .048/.093 = .5161 P(CD) = .040/.093 = .4301

4.32Let A = Alex fills the orderB = Alicia fills the orderC = Juan fills the orderI = order filled incorrectlyK = order filled correctly

P(A) = .30 P(B) = .45 P(C) = .25P(IA) = .20P(IB) = .12P(IC) = .05P(KA) = .80P(KB) = .88P(KC) = .95a)P(B) = .45

b)P(KC) = 1 - P(IC) = 1 - .05 = .95

c)


P(Ei) P(IEi) P(I Ei) P(EiI)

A .30 .20 .0600.0600/.1265=.4743

B .45 .12 .0540.0540/.1265=.4269

C .25 .05 .0125.0125/.1265=.0988

P(I)=.1265

Revised: P(AI) = .0600/.1265 = .4743 P(BI) = .0540/.1265 = .4269 P(CI) = .0125/.1265 = .0988

d)Event PriorConditional Joint Revised

P(Ei) P(KEi) P(K Ei) P(EiK)

A .30 .80 .2400.2400/.8735=.2748

B .45 .88 .3960.3960/.8735=.4533

C .25 .95 .2375.2375/.8735=.2719

P(K)=.8735

4.33 LetT = lawn treated by Tri-stateG = lawn treated by Green ChemV = very healthy lawnN = not very healthy lawn

P(T) = .72 P(G) = .28 P(VT) = .30 P(VG) = .20Event PriorConditional Joint Revised

P(Ei) P(VEi) P(V Ei) P(EiV)

A .72 .30 .216.216/.272=.7941

B .28 .20 .056.056/.272=.2059

P(V)=.272

Revised:P(TV) = .216/.272 = .7941 P(GV) = .056/.272 = .2059

4.34Let S = smallLet L = large

The prior probabilities are: P(S) = .70 P(L) = .30P(TS) = .18P(TL) = .82


P(Ei) P(TEi) P(T Ei) P(EiT)

S .70 .18 .1260.1260/.3720 = .3387

L .30 .82 .2460.2460/.3720 = .6613

P(T)=.3720

Revised:P(ST) = .1260/.3720 = .3387 P(LT) = .2460/.3720 = .6613

37.2% offer training since P(T) = .3720,.

4.35 Variable 1

D E

A 10 20 30

Variable 2 B 15 5 20

C 30 15 45

55 40 95

a) P(E) = 40/95 = .42105 b) P(B D) = P(B) + P(D) - P(B D) = 20/95 + 55/95 - 15/95 = 60/95 = .63158 c) P(A E) = 20/95 = .21053 d) P(BE) = 5/40 = .1250 e) P(A B) = P(A) + P(B) = 30/95 + 20/95 =

50/95 = .52632 f) P(B C) = .0000 (mutually exclusive) g) P(DC) = 30/45 = .66667

h) P(AB) = = .0000 (A and B are mutually exclusive)

i) P(A) = P(AD)??

Does 30/95 = 10/95 ??

Since, .31579 .18182, Variables 1 and 2 are not independent.

D E F G

A 3 9 7 1231

B 8 4 6 422

C 10 5 3 725

21 18 16 2378

4.36

a) P(F A) = 7/78 = .08974

b) P(AB) = = .0000 (A and B are mutually exclusive)

c) P(B) = 22/78 = .28205d) P(E F) = .0000 Mutually Exclusive e) P(DB) = 8/22 = .36364 f) P(BD) = 8/21 = .38095 g) P(D C) = 21/78 + 25/78 10/78 = 36/78 = .4615 h) P(F) = 16/78 = .20513

4.37

Age(years)

65

Gender Male .11 .20 .19 .12 .16 .78

Female .07 .08 .04 .02 .01 .22

.18 .28 .23 .14 .171.00

a) P(35-44) = .28 b) P(Woman 45-54) = .04 c) P(Man 35-44) = P(Man) + P(35-44) - P(Man 35-44) = .78 + .28 - .20 = .86 d) P( 52):

z = = 1.6 from Table A.5, Prob. = .4452

P( > 52) = .5000 - .4452 = .0548

b) P(< 51):

z = = 0.80 from Table A.5 prob. = .2881

P(< 51) = .5000 + .2881 = .7881

c) P(< 47):

z = = -2.40 from Table A.5 prob. = .4918

P(< 47) = .5000 - .4918 = .0082

d) P(48.5 < < 52.4):

z = = -1.20 from Table A.5 prob. = .3849

z = = 1.92

from Table A.5 prob. = .4726

P(48.5 < < 52.4) = .3849 + .4726 = .8575

e) P(50.6 < < 51.3):

z = = 0.48

from Table A.5, prob. = .1844

z = from Table A.5, prob. = .3508

P(50.6 < < 51.3) = .3508 - .1844 = .1644

7.14 = 23.45 = 3.8

a) n = 10, P( > 22):

z = = -1.21


P( > 22) = .3869 + .5000 = .8869

b) n = 4, P( > 26):

z = = 1.34 from Table A.5, prob. = .4099

P( > 26) = .5000 - .4099 = .0901

7.15 n = 36 = 278 P(< 280) = .86

.3600 of the area lies between = 280 and = 278. This probability is associated with z = 1.08 from Table A.5. Solving for :

z =

1.08 =

1.08 = 2

= = 11.11

7.16 n = 81 = 12 P( > 300) = .18

.5000 - .1800 = .3200 and from Table A.5, z.3200 = 0.92

Solving for :

z =

0.92 =

0.92 = 300 -

1.2267 = 300 - = 300 - 1.2267 = 298.77 7.17 a) N = 1,000 n = 60 = 75 = 6

P( < 76.5):


P(< 76.5) = .4772 + .5000 = .9772 b) N = 90n = 36 = 108 = 3.46

P(107 < < 107.7):

z = = -2.23 from Table A.5, prob. = .4871

z = = -0.67


P(107 < < 107.7) = .4871 - .2486 =.2385

c) N = 250 n = 100 = 35.6 = 4.89

P( > 36):


P( > 36) = .5000 - .3531 = .1469 d) N = 5000 n = 60 = 125 = 13.4

P( < 123):


P( < 123) = .5000 - .3770 = .1230

7.18 = 99.9 = 30 n = 38

a) P( < 90):

z = = -2. 03 from table A.5, area = .4788

P( < 90) = .5000 - .4788 = .0212

b) P(98 < < 105):

z = = 1.05

from table A.5, area = .3531

z = = -0.39


P(98 < < 105) = .3531 + .1517 = .5048

c) P( < 112):

z = = 2.49


P(< 112) = .5000 + .4936 = .9936

d) P(93 < < 96):

z = = -1.42


z = = -0.80


P(93 < < 96) = .4222 - .2881 = .1341

7.19 N = 1500 n = 100 = 177,000 = 8,500

P( > $185,000):

z = = 9.74


P( > $185,000) = .5000 - .5000 = .0000

7.20 = $65.12 = $21.45 n = 45 P( > ) = .2300

Prob. lies between and = .5000 - .2300 = .2700 from Table A.5, z.2700 = 0.74

Solving for :

z =

0.74 =

2.366 = - 65.12 and = 65.12 + 2.366 = 67.486

7.21 = 50.4 = 11.8 n = 42

a) P( > 52):

z = = 0.88

from Table A.5, the area for z = 0.88 is .3106

P(> 52) = .5000 - .3106 = .1894

b) P( < 47.5):

z = = -1.59

from Table A.5, the area for z = -1.59 is .4441

P( < 47.5) = .5000 - .4441 = .0559

c) P( < 40):

z = = -5.71


P( < 40) = .5000 - .5000 = .0000

d) 71% of the values are greater than 49. Therefore, 21% are between the sample mean of 49 and the population mean, = 50.4.

The z value associated with the 21% of the area is -0.55

z.21 = -0.55

z =

-0.55 =

= 16.4964

7.22 p = .25

a) n = 110 P(< .21):


P(< .21) = .5000 - .3340 = .1660

b) n = 33 P( > .24):

z = = -0.13


P( > .24) = .5000 + .0517 = .5517

c) n = 59 P(.24 < < .27):


z = = 0.35


P(.24 < < .27) = .0714 + .1368 = .2082

d) n = 80 P(> .30):

z = = 1.03


P( > .30) = .5000 - .3485 = .1515

e) n = 800 P( > .30):

z = = 3.27


P( > .30) = .5000 - .4995 = .0005

7.23 p = .58 n = 660

a) P(> .60):

z = = 1.04


P(> .60) = .5000 - .3508 = .1492

b) P(.55 < < .65):

z = = 3.64


z = = 1.56


P(.55 < < .65) = .4998 + .4406 = .9404

c) P( > .57):

z = = -0.52 from table A.5, area = .1985

P( > .57) = .1985 + .5000 = .6985

d) P(.53 < < .56):

z = = -1.04 z = = -2.60

from table A.5, area for z = -1.04 is .3508 from table A.5, area for z = -2.60 is .4953

P(.53 < < .56) = .4953 - .3508 = .1445

e) P( < .48):

z = = -5.21


P( < .48) = .5000 - .5000 = .0000

7.24p = .40 P( > .35) = .8000

P(.35 < < .40) = .8000 - .5000 = .3000

from Table A.5, z.3000 = -0.84

Solving for n:

z =

-0.84 = =

8.23 = n = 67.73 68

7.25p = .28n = 140P( < ) = .3000

P( < < .28) = .5000 - .3000 = .2000

from Table A.5, z.2000 = -0.52

Solving for : z =

-0.52 =

-.02 = - .28

= .28 - .02 = .26 7.26 P(x > 150): n = 600 p = .21 x = 150

= = .25

z = = 2.41


P(x > 150) = .5000 - .4920 = .0080

7.27p = .48 n = 200

a) P(x < 90):

= = .45

z = = -0.85


P(x < 90) = .5000 - .3023 = .1977

b) P(x > 100):

= = .50

z = = 0.57


P(x > 100) = .5000 - .2157 = .2843

c) P(x > 80):

= = .40

z = = -2.26


P(x > 80) = .5000 + .4881 = .9881

7.28p = .19 n = 950

a) P( > .25):

z = = 4.71

from Table A.5, area = .5000

P( > .25) = .5000 - .5000 = .0000

b) P(.15 < < .20):

z = = -3.14

z = = 0.79

from Table A.5, area for z = -3.14 is .4992

from Table A.5, area for z = 0.79 is .2852

P(.15 < < .20) = .4992 + .2852 = .7844

c) P(133 < x < 171):

= = .14 = = .18

P(.14 < < .18):

z = = -3.93z = = -0.79

from Table A.5, the area for z = -3.93 is .49997 the area for z = -0.79 is .2852

P(133 < x < 171) = .49997 - .2852 = .21477

7.29 = 76, = 14

a) n = 35, P( > 79):

z = = 1.27


P( > 79) = .5000 - .3980 = .1020

b) n = 140, P(74 < < 77):

z = = -1.69z = = 0.85

from table A.5, area for z = -1.69 is .4545 from table A.5, area for 0.85 is .3023

P(74 < < 77) = .4545 + .3023 = .7568

c) n = 219, P( < 76.5):

z = = 0.53 from table A.5, area = .2019

P( < 76.5) = .5000 + .2019 = .7019

7.30 p = .46

a) n = 60

P(.41 < < .53):

z = = 1.09 from table A.5, area = .3621

z = = -0.78


P(.41 < < .53) = .3621 + .2823 = .6444

b) n = 458P(< .40):

z = = -2.58


P( < .40) = .5000 - .4951 = .0049

c) n = 1350 P(> .49):

z = = 2.21


P(> .49) = .5000 - .4864 = .0136

7.31Under 18 250(.22) = 55 18 25 250(.18) = 45 26 50 250(.36) = 90 51 65 250(.10) = 25 over 65 250(.14) = 35 n = 250

7.32p = .55 n = 600 x = 298

= = .497

P(< .497):

z = = -2.61

from Table A.5, Prob. = .4955

P(< .497) = .5000 - .4955 = .0045

No, the probability of obtaining these sample results by chance from a population that supports the candidate with 55% of the vote is extremely low (.0045). This is such an unlikely chance sample result that it would cause the researcher to probably reject her claim of 55% of the vote.

7.33 a) Roster of production employees secured from the human resources department of the company.

b) Alpha/Beta store records kept at the headquarters of their California division or merged files of store records from regional offices across the state.

c) Membership list of Maine lobster catchers association.

7.34 = $ 17,755 = $ 650 n = 30 N = 120

P( < 17,500):

z = = -2.47


P( < 17,500) = .5000 - .4932 = .0068

7.35Number the employees from 0001 to 1250. Randomly sample from the random number table until 60 different usable numbers are obtained. You cannot use numbers from 1251 to 9999.

7.36 = $125 n = 32 = $110 2 = $525

P( > $110):

z = = -3.70 from Table A.5, Prob.= .5000

P( > $110) = .5000 + .5000 = 1.0000

P( > $135):

z = = 2.47

from Table A.5, Prob.= .4932

P( > $135) = .5000 - .4932 = .0068

P($120 < < $130):

z = = -1.23

z = = 1.23


P($120 < < $130) = .3907 + .3907 = .7814

7.37 n = 1100

a) x > 810, p = .73

=

z = = 0.48


P(x > 810) = .5000 - .1844 = .3156

b) x < 1030, p = .96,

= = .9364

z = = -3.99


P(x < 1030) = .5000 - .49997 = .00003

c) p = .85

P(.82 < < .84):

z = = -2.79


z = = -0.93


P(.82 < < .84) = .4974 - .3238 = .1736

7.381)The managers from some of the companies you are interested in studying do not belong to the American Managers Association.2)The membership list of the American Managers Association is not up-to-date.3)You are not interested in studying managers from some of the companies belonging to the American Management Association.4)The wrong questions are asked.5)The manager incorrectly interprets a question.6)The assistant accidentally marks the wrong answer.7)The wrong statistical test is used to analyze the data.8)An error is made in statistical calculations.9)The statistical results are misinterpreted.

7.39Divide the factories into geographic regions and select a few factories to represent those regional areas of the country. Take a random sample of employees from each selected factory. Do the same for distribution centers and retail outlets. Divide the United States into regions of areas. Select a few areas. Take a random sample from each of the selected area distribution centers and retail outlets.

7.40N = 12,080n = 300

k = N/n = 12,080/300 = 40.27 Select every 40th outlet to assure n > 300 outlets. Use a table of random numbers to select a value between 0 and 40 as a starting point.

7.41p = .54 n = 565

a) P(x > 339):

= = .60

z = = 2.86


P(x > 339) = .5000 - .4979 = .0021

b) P(x > 288):

= = .5097

z = = -1.45


P(x > 288) = .5000 + .4265 = .9265

c) P( < .50):

z = = -1.91


P( < .50) = .5000 - .4719 = .0281

7.42 = $550n = 50 = $100

P( < $530):

z = = -1.41

from Table A.5, Prob.=.4207 P(x < $530) = .5000 - .4207 = .0793

7.43 = 56.8n = 51 = 12.3

a) P( > 60):

z = = 1.86 from Table A.5, Prob. = .4686

P( > 60) = .5000 - .4686 = .0314

b) P( > 58):

z = = 0.70 from Table A.5, Prob.= .2580

P( > 58) = .5000 - .2580 = .2420

c) P(56 < < 57):

z = = -0.46z = = 0.12

from Table A.5, Prob. for z = -0.46 is .1772

from Table A.5, Prob. for z = 0.12 is .0478

P(56 < < 57) = .1772 + .0478 = .2250

d) P(< 55):

z = = -1.05


P( < 55) = .5000 - .3531 = .1469

e) P( < 50):

z = = -3.95 from Table A.5, Prob.= .5000

P( < 50) = .5000 - .5000 = .0000 7.45 p = .73 n = 300

a) P(210 < x < 234):

= = .70 = = .78

z = = -1.17

z = = 1.95 from Table A.5, the area for z = -1.17 is .3790 the area for z = 1.95 is .4744

P(210 < x < 234) = .3790 + .4744 = .8534

b) P(> .78):

z = = 1.95


P(> .78) = .5000 - .4744 = .0256

c) p = .73 n = 800P(> .78):

z = = 3.19


P(> .78) = .5000 - .4993 = .0007

7.46 n = 140 P(x > 35):

= = .25 p = .22

z = = 0.86


P(x > 35) = .5000 - .3051 = .1949

P(x < 21):

= = .15

z = = -2.00


P(x < 21) = .5000 - .4772 = .0228

n = 300 p = .20

P(.18 < < .25):

z = = -0.87


z = = 2.17from Table A.5, the area for z = 2.17 is .4850

P(.18 < < .25) = .3078 + .4850 = .7928

7.47By taking a sample, there is potential for obtaining more detailed information. More time can be spent with each employee. Probing questions can be asked. There is more time for trust to be built between employee and interviewer resulting in the potential for more honest, open answers.

With a census, data is usually more general and easier to analyze because it is in a more standard format. Decision-makers are sometimes more comfortable with a census because everyone is included and there is no sampling error. A census appears to be a better political device because the CEO can claim that everyone in the company has had input.

7.48 p = .75 n = 150 x = 120

P(> .80):

z = = 1.41


P( > .80) = .5000 - .4207 = .0793

7.49 a) Switzerland: n = 40 = $ 30.67 = $ 3

P(30 < < 31):

from Table A.5, the area for z = 0.70 is .2580 the area for z = -1.41 is .4207

P(30 < < 31) = .2580 + .4207 = .6787

b) Japan: n = 35 = $ 20.20 = $3

P( > 21):


P( > 21) = .5000 - .4429 = .0571

c) U.S.: n = 50 = $ 23.82 = $ 3

P( < 22.75):


P(< 22.75) = .5000 - .4941 = .0059

7.50a)Age, Ethnicity, Religion, Geographic Region, Occupation, Urban-Suburban-Rural, Party Affiliation, Genderb)Age, Ethnicity, Gender, Geographic Region, Economic Classc)Age, Ethnicity, Gender, Economic Class, Educationd)Age, Ethnicity, Gender, Economic Class, Geographic Location

7.51 = $281 n = 65 = $47

P( > $273):

z = = -1.37from Table A.5 the area for z = -1.37 is .4147

P( > $273) = .5000 + .4147 = .9147

solutions to problems (4-7)

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