Asignment Sum Solutionn q.A

8/3/2019 Asignment Sum Solutionn q.A

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3.3 MedianArrange terms in ascending order:073, 167, 199, 213, 243, 345, 444, 524, 609, 682There are 10 terms.Since there are an even number of terms, the median is the average of the

two middle terms:

Median = 243 + 345 = 588 = 294

Using the formula, the median is located at the n + 1 th term.2

n = 10 therefore 10 + 1 = 11 = 5.5th term.2 2

The median is located halfway between the 5th and 6th terms.

5

th

term = 243 6

th

term = 345Halfway between 243 and 345 is the median = 2942 2

3.6 Rearranging the data into ascending order:

11, 13, 16, 17, 18, 19, 20, 25, 27, 28, 29, 30, 32, 33, 34

25.5)15(100

35==i

P35 is located at the 5 + 1 = 6th term

P35 = 19

25.8)15(100

55==i


P55 = 27

Q1 = P25

75.3)15(100

25

==i

Q1 = P25 is located at the 3 + 1 = 4th term

Q1 = 17


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Q2 = Median

The median is located at the termthth

82

115=

+

Q2 = 25

Q3 = P75

25.11)15(100

75==i

Q3 = P75 is located at the 11 + 1 = 12

th term

Q3 = 30

3.11 x x - (x-)26 6-4.2857 = 1.7143 2.93882 2.2857 5.22444 0.2857 .08169 4.7143 22.22461 3.2857 10.79583 1.2857 1.65305 0.7143 .5102

x = 30 x-= 14.2857 (x -)2 = 43.4284

2857.4730 ===

Nx

a.) Range = 9 - 1 = 8

b.) M.A.D. = ==

7

2857.14

N

x 2.041

c.) 2 =7

4284.43)( 2 =N

x = 6.204

d.) = 204.6)(2

=

N

x = 2.491

e.) 1, 2, 3, 4, 5, 6, 9


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Q1 = P25

i = )7(100

25= 1.75

Q1 is located at the 1 + 1 = 2th

term, Q1 = 2

Q3 = P75:

i = )7(100

75= 5.25

Q3 is located at the 5 + 1 = 6th term, Q3 = 6

IQR = Q3 - Q1 = 6 - 2 = 4

f.) z =491.22857.46

= 0.69

z =491.2

2857.42= -0.92

z =491.2

2857.44= -0.11

z =491.2

2857.49= 1.89

z =491.2

2857.41= -1.32

z =491.2

2857.43= -0.52

z =491.2

2857.45= 0.29

3.1614, 15, 18, 19, 23, 24, 25, 27, 35, 37, 38, 39, 39, 40, 44,46, 58, 59, 59, 70, 71, 73, 82, 84, 90

Q1 = P25

i = )25(100

25= 6.25



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Q1 = 25

Q3 = P75

i = )25(

100

75= 18.75

P75 is located at the 18 + 1 = 19

th term

Q3 = 59

IQR = Q3 - Q1 = 59 - 25 = 34

3.19

xxx

2)( xx

7 1.833 3.3615 3.833 14.694

10 1.167 1.36112 3.167 10.0289 0.167 0.0288 0.833 0.694

14 5.167 26.6943 5.833 34.028

11 2.167 4.69413 4.167 17.3618 0.833 0.694

6 2.833 8.028106 32.000 121.665

12

106=

=

n

xx = 8.833

a) MAD =12

32=

n

xx= 2.667

b) s2 =11

665.121

1

)(2

=

n

xx= 11.06

c) s = 06.112 =s = 3.326

d) Rearranging terms in order:

3 5 6 7 8 8 9 10 11 12 13 14

Q1 = P25: i = (.25)(12) = 3


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Q1 = the average of the 3rd and 4th terms:

Q1 = (6 + 7)/2 = 6.5

Q3 = P75: i = (.75)(12) = 9

Q3 = the average of the 9th and 10th terms:

Q3 = (11 + 12)/2 = 11.5

IQR = Q3 - Q1 = 11.5 - 9 = 2.5

e.) z =

326.3

833.86= - 0.85

f.) CV =833.8

)100)(326.3(= 37.65%

3.27 Mean

Class f M fM0 - 2 39 1 392 - 4 27 3 814 - 6 16 5 806 - 8 15 7 105

8 - 10 10 9 9010 - 12 8 11 8812 - 14 6 13 78

f=121 fM=561

= 121561

=

f

fM= 4.64

Mode: The modal class is 0 2.The midpoint of the modal class = the mode = 1

3.30 Class f M fM fM2

5 - 9 20 7 140 9809 - 13 18 11 198 2,178

13 - 17 8 15 120 1,80017 - 21 6 19 114 2,16621 - 25 2 23 46 1,058

f=54 fM= 618 fm2= 8,182


6/24

s2 =

53

7.70718182

53

54

)618(8182

1

)(22

2

=

=

n

n

fMfM

= 20.9

s = 9.202 =S = 4.57

3.31 Class f M fM fM218 - 24 17 21 357 7,49724 - 30 22 27 594 16,03830 - 36 26 33 858 28,31436 - 42 35 39 1,365 53,23542 - 48 33 45 1,485 66,82548 - 54 30 51 1,530 78,03054 - 60 32 57 1,824 103,96860 - 66 21 63 1,323 83,34966 - 72 15 69 1,035 71,415

f= 231 fM= 10,371 fM2= 508,671

a.) Mean:231

371,10=

=

=

f

fM

n

fMx = 44.9

b.) Mode. The Modal Class = 36-42. The mode is the class midpoint =39

c.) s2 =

230

5.053,43

230

231

)371,10(671,508

1

)(22

2

=

=

n

n

fMfM

=

187.2

d.) s = 2.187 = 13.7

3.32

a.) Mean

Class f M fM fM2

0 - 1 31 0.5 15.5 7.751 - 2 57 1.5 85.5 128.252 - 3 26 2.5 65.0 162.50


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3 - 4 14 3.5 49.0 171.504 - 5 6 4.5 27.0 121.505 - 6 3 5.5 16.5 90.75

f=137 fM=258.5 fM2=682.25

= 137 5.258=

ffM = 1.89

b.) Mode: Modal Class = 1-2. Mode = 1.5

c.) Variance:

2

=137

137

)5.258(25.682

)( 222

=

NN

fMfM

=

1.4197

d.) standard Deviation:

= 4197.12 = = 1.19155.10 n = 16 p = .40

P(x > 9): from Table A.2:

x Prob9 .084

10 .03911 .01412 .00413 .001

.142

P(3 < x < 6):

x Prob

3 .0474 .1015 .1626 .198

.508

n = 13 p = .88


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P(x = 10) = 13C10(.88)10(.12)3 = 286(.278500976)(.001728) = .1376

P(x = 13) = 13C13(.88)13(.12)0 = (1)(.1897906171)(1) = .1898

Expected Value = = n p = 13(.88) = 11.445.13n = 15 p = .20

a) P(x = 5) = 15C5(.20)5(.80)10 =

3003(.00032)(.1073742) = .1032

b) P(x > 9): Using Table A.2

P(x = 10) + P(x = 11) + . . . + P(x = 15) =

.000 + .000 + . . . + .000 = .000

c) P(x = 0) = 15C0(.20)0(.80)15 =

(1)(1)(.035184) = .0352

d) P(4 < x < 7): Using Table A.2

P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) =

.188 + .103 + .043 + .014 = .348

e)

5.20 = 5.6 days3 weeks


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a) Prob(x=0 = 5.6):

from Table A.3 = .0037

b) Prob(x=6 = 5.6):from Table A.3 = .1584

c) Prob(x > 15 = 5.6):

x Prob.15 .000516 .000217 .0001

x > 15 .0008

Because this probability is so low, if it actually occurred, theresearcher would

actually have to question the Lambda value as too low for this period.

5.24 n = 100,000 p = .00004

Prob(x > 7n = 100,000 p = .00004):

= = n p = 100,000(.00004) = 4.0

Since n > 20 and n p < 7, the Poisson approximation to this binomial

problem isclose enough.

Prob(x > 7 = 4):

Using Table A.3 x Prob.7 .05958 .0298


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9 .013210 .005311 .001912 .000613 .0002

14 .0001x > 7 .1106

Prob(x>10 = 4):

Using Table A.3 x Prob.11 .001912 .000613 .0002

14 .0001x > 10 .0028

Since getting more than 10 is a rare occurrence, thisparticulargeographicregionappears to have a higher average rate than other regions. Aninvestigationofparticular characteristics of this region might be warranted.

5.29 N = 17 A = 8 n = 4

a) P(x = 0) =417

4908

C

CC =

2380

)126)(1(= .0529

b) P(x = 4) =417

0948

C

CC =

2380

)1)(70(= .0294

c) P(x = 2 non computer) =417

2829

C

CC =

2380

)28)(36(= .4235

5.30 N = 20 A = 16 white N - A = 4 red n = 5

a) Prob(x = 4 white) =520

14416

C

CC =

15504

)4)(1820(= .4696


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b) Prob(x = 4 red) =520

11644

C

CC =

15504

)16)(1(= .0010

c) Prob(x = 5 red) = 520

01654

C

CC

= .0000 because 4C5 is impossible

determine.The participant cannot draw 5 red beads if there are only 4 to draw from

5.32 N = 16 A = 4 defective n = 3

a) Prob(x = 0) =560

)220)(1(

316

31204=

C

CC= .3929

b) Prob(x = 3) =560

)1)(4(

316

01234=

C

CC= .0071

c) Prob(x > 2) = Prob(x=2) + Prob(x=3) =316

11224

C

CC

+ .0071 (from

part b.) =

560

)12)(6(

+ .0071 = .1286 + .0071 = .1357

d) Prob(x < 1) = Prob(x=1) + Prob(x=0) =


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316

21214

C

CC + .3929 (from part a.) =

560

)66)(4(+ .3929 = .4714 + .

3929 = .86436.3 a = 2.80 b = 3.14

=2

14.380.2

2=+ba = 2.97

=12

80.214.3

12

=

ab= 0.10

P(3.00 < x < 3.10) =80.214.3

00.310.3

= 0.2941

6.4 a = 11.97 b = 12.03

Height = 97.1103.12

11

=ab = 16.667

P(x > 12.01) =97.1103.12

01.1203.12

= .3333

P(11.98 < x < 12.01) =97.1103.12

98.1101.12

= .5000

6.5 = 2100 a = 400 b = 3800

=12

4003800

12

=

ab= 981.5

Height =12

40038001 =

ab= .000294

P(x > 3000) =3400

800

4003800

30003800=

= .2353

P(x > 4000) =.0000

P(700 < x < 1500) =3400

800

4003800

7001500=

= .2353

6.8 = 22 = 4

a) Prob(x > 17):


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z =4

2217 =

x= -1.25

area betweenx = 17 and = 22 from table A.5 is .3944

Prob(x > 17) = .3944 + .5000 = .8944

b) Prob(x < 13):

z =4

2213=

x= -2.25

from table A.5, area = .4878

Prob(x < 13) = .5000 - .4878 = .0122

c) P(25 $2000) = .5000 - .3212 = .1788

b) Prob(owes money) = Prob(x < 0):


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z =725

13320 =

x= -1.84

from Table A.5, the z = -1.84 yields: .4671

Prob(x < 0) = .5000 - .4671 = .0329

c) Prob($100 < x < $700):

z =725

1332100 =

x= -1.70


z =

725

1332700 =

x= -0.87


Prob($100 < x < $700) = .4554 - .3078 = .1476

6.11 = $30,000 = $9,000

a) Prob($15,000 < x < $45,000):

z =000,9

000,30000,45 =

x= 1.67

From Table A.5, z = 1.67 yields: .4525

z =000,9

000,30000,15 =

x= -1.67

From Table A.5, z = -1.67 yields: .4525

Prob($15,000 < x < $45,000) = .4525 + .4525 = .9050

b) Prob(x > $50,000):

z =000,9

000,30000,50 =

x= 2.22


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From Table A.5, z = 2.22 yields: 4868

Prob(x > $50,000) = .5000 - .4868 = .0132

c) Prob($5,000 < x < $20,000):

z =000,9

000,30000,5 =

x= -2.78

From Table A.5, z = -2.78 yields: .4973

z =000,9

000,30000,20 =

x= -1.11

From Table A.5, z = -1.11 yields .3665

Prob($5,000 < x < $20,000) = .4973 - .3665 = .1308

d) 90.82% of the values are greater than x = $7,000.

Then x = $7,000 is in the lower half of the distribution and .9082 - .5000 =

.4082 lie between x and .

From Table A.5, z = -1.33 is associated with an area of .4082.

Solving for :

z =

x

-1.33 =

000,30000,7

= 17,293.23

e) = $9,000. If 79.95% of the costs are less than $33,000, x =$33,000 is inthe upper half of the distribution and .7995 - .5000 = .2995 of the

values liebetween $33,000 and the mean.

From Table A.5, an area of .2995 is associated with z = 0.84


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Solving for :

z =

x

0.84 = 000,9000,33

= $25,440

6.12 = 200, = 47 Determine x

a) 60% of the values are greater than x:

Since 50% of the values are greater than the mean, = 200, 10% or .1000 lie

between x and the mean. From Table A.5, the z value associated withan area

of .1000 is z = -0.25. The z value is negative since x is below the mean.Substituting z = -0.25, = 200, and = 47 into the formula and

solving for x:

z =

x

-0.25 =47

200x

x = 188.25

b) x is less than 17% of the values.

Since x is only less than 17% of the values, 33% (.5000- .1700) or .3300lie


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between x and the mean. Table A.5 yields a z value of 0.95 for an areaof

.3300. Using this z = 0.95, = 200, and = 47, x can be solved for:

z =

x

0.95 =47

200X

x = 244.65

c) 22% of the values are less than x.

Since 22% of the values lie below x, 28% lie between x and the mean

(.5000 - .2200). Table A.5 yields a z of -0.77 for an area of .2800.Using the z

value of -0.77, = 200, and = 47, x can be solved for:

z =

x

-0.77 =47

200x

x = 163.81

d) x is greater than 55% of the values.

Since x is greater than 55% of the values, 5% (.0500) lie between x andthe

mean. From Table A.5, a z value of 0.13 is associated with an area of .05.

Using z = 0.13, = 200, and = 47, x can be solved for:

z =

x

0.13 =47

200x

x = 206.11

6.22 n = 300 p = .53

= 300(.53) = 159


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= )47)(.53(.300= qpn = 8.645

Test: + 3 = 159 + 3(8.645) = 133.065 to 184.935

which lies between 0 and 300. It is okay to use the normal distribution asan approimation on parts a) and b).

a) Prob(x > 175 transmission)

correcting for continuity: x = 175.5

z =645.8

1595.175 = 1.91

from A.5, the area for z = 1.91 is .4719

Prob(x > 175) = .5000 - .4719 = .0281

b) Prob(165 < x < 170)

correcting for continuity: x = 164.5; x = 170.5

z =645.8

1595.170 = 1.33

z =645.8

1595.164 = 0.64

from A.5, the area for z = 1.33 is .4082the area for z = 0.64 is .2389

Prob(165 < x < 170) = .4082 - .2389 = .1693

For parts c and d: n = 300 p = .60

= 300(.60) = 180

= )40)(.60(.300= qpn = 8.485

Test: + 3 = 180 + 3(8.485) = 180 + 25.455

154.545 to 205.455 lies between 0 and 300

It is okay to use the normal distribution toapproimate c) and d)


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c) Prob(155 < x < 170 personnel):

correcting for continuity: x = 154.5; x = 170.5

z =485.8

1805.170

= -1.12

z =485.8

1805.154 = -3.01

from A.5, the area for z = -1.12 is .3686the area for z = -3.01 is .4987

Prob(155 < x < 170) = .4987 - .3686 = .1301

d) Prob(x < 200 personnel):

correcting for continuity: x = 199.5

z =485.8

1805.199 = 2.30

from A.5, the area for z = 2.30 is .4893

Prob(x < 200) = .5000 + .4893 = .98936.23 p = .16 n = 130

Conversion to normal dist.: = n(p) = 130(.16) = 20.8

= )84)(.16(.130= qpn = 4.18

a) Prob(x > 25):

Correct for continuity: x = 25.5

z =18.4

8.205.25 = 1.12


Prob(x > 20) = .5000 - .3686 = .1314

b) Prob(15


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z =18.4

8.205.14 = -1.51

z =

18.4

8.205.23 = 0.65

from table A.5, area for z = -1.51 is .4345area for z = 0.65 is .2422

Prob(15 < x < 23) = .4345 + .2422 = .6767

c) Prob(x < 12):

correct for continuity: x = 11.5

z =18.4

8.205.11 = -2.22

from table A.5, area for z = -2.22 is .4868

Prob(x < 12) = .5000 - .4868 = .0132

d) Prob(x = 22):

correct for continuity: 21.5 to 22.5

z =18.4

8.205.21 = 0.17

z =18.4

8.205.22 = 0.41

from table A.5, area for 0.17 = .0675area for 0.41 = .1591

Prob(x = 22) = .1591 - .0675 = .0916

7.18 = 99.9 = 30 n = 38

a) Prob( x < 90):

z =38

30

9.9990

1

=

N

nN

n

x

= -2.03


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Prob( x < 90) = .5000 - .4788 = .0212

b) Prob(98 < x < 105):

z =38

30

9.99105

1

=

N

nN

n

x

= 1.05


z =

38

30

9.9998

1

=

N

nN

n

x

= -0.39


Prob(98 < x < 105) = .3531 + .1517 = .5048

c) Prob( x < 112):

z =38

30

9.99112

1

=

N

nN

n

x

= 2.49


Prob( x < 112) = .5000 - .4936 = .0064

d) Prob(93 < x < 96):

z =3830

9.9993

1

=

NnN

n

x

= -1.42



22/24

z =38

30

9.9996

1

=

N

nN

n

x

= -0.80


Prob(93 < x < 96) = .4222 - .2881 = .1341

7.20 = $65.12 = $21.45 n = 45

Prob( x > 0x ) = .2300

Prob. x lies between 0x and = .5000 - .2300 = .2700

from Table A.5, z.2700 = 0.74

Solving for 0x :

z =

n

x

0

0.74 =

45

45.21

12.650 x

2.366 = 0x - 65.12

0x = 65.12 + 2.366 = 67.486

7.27 P = .48 n = 200

a) Prob(x < 90):

p =200

90= .45


23/24

z =

200

)52)(.48(.

48.45. =

n

QP

Pp

= -0.85

from Table A.5, the area for z = -0.85 is .3023

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

b) Prob(x > 100):

p =200

100= .50

z =

200

)52)(.48(.

48.50. =

n

QP

Pp

= 0.57

from Table A.5, the area for z = 0.57 is .2157

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

c) Prob(x > 80):

p =200

80= .40

z =

200

)52)(.48(.

48.40.

=

n

QP

Pp

= -2.26

from Table A.5, the area for z = -2.26 is .4881

Prob(x > 80) = .5000 + .4881 = .98817.28 P = .19 n = 950

a) Prob( p > .25):

z =

950

)89)(.19(.19.25. =

n

QPPp

= 4.71

from Table A.5, area = .5000

Prob( p > .25) = .5000 - .5000 = .0000


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b) Prob(.15 < p < .20):

z =

950

)81)(.19(.

19.15. =

n

QP

Pp

= -3.14

z =

950

)89)(.19(.

19.20. =

n

QP

Pp

= 0.79

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

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

Prob(.15 < p < .20) = .4992 + .2852 = .7844

c) Prob(133 < x < 171):

1p =

950

133= .14 2p =

950

171= .18

Prob(.14 < p < .18):

z =

950

)81)(.19(.

19.14. =

n

QP

Pp

= -3.93

z =

950

)81)(.19(.

19.18. =

n

QP

Pp

= -0.79

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

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

Asignment Sum Solutionn q.A

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