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Focus on Exam 18

1 X fx

f =

=

=

Σ

Σ

1250

500

2 5 [ ]Shown

X ~ Po(25)

P( )

X x e

x x

x

= = =minus2 5 2 5

0 1 2 hellip

x P( X = x ) Oi

E i = np

i

( )O E

E

i i

i

minus

2

0 00821 50 41 19756

1 02052 98 102 02427

2 02565 120 128 05000

3 02138 113 107 03364

4 01336 61 67 05373

5 00668 29 33 04848

6 00278 19 14 17857

7 00099 10

0

10

5

2

7 128578 00031

Σ 500 500 71482

H 0 Y is a Poisson random variable

H 1 Y is not a Poisson random variable

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

c 2

= 715 lt χ 0 05 7

2

there4 Do not reject H 0 Y is a Poisson random variable

CHAPTER 18

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ACE AHEAD Mathematics (T) Third Term2

2 H 0 The playing time is normally distributed with mean 615 minutes and standard deviation

1 minute

H 1 The playing time is not normally distributed with mean 615 minutes and standard deviation

1 minute

X ~ N(615 12) z =minus

=

minus x xm

s

61 5

1

P P

P P

( ) ( )

( ) ( )

x

x

lt = lt minus

=

lt lt = minus lt lt minus

=

60 1 5

0 06681

60 61 1 5 0 5

0 24173

z

z

PP P

P P

( ) ( )

( ) ( )

61 62 0 5 0 5

0 38292

62 63 0 5 1 5

0

lt lt = minus lt lt

=

lt lt = lt lt

=

x z

x z

( ) ( )

( ) ( )

24173

63 64 1 5 2 5

0 06060

64 2 5

0

P P

P P

lt lt = lt lt

=

gt = gt

=

x

x

z

z

000621

Oi

E i= nP

( )O E

E

i i

i

minus

2

33 6681 171100

225 24173 11579

380 38292 00223

280 24173 60588

78 6060 49960

4 621 07865

301306

a = 005

v = 6 minus 1

= 5

χ

0 05 5

2

= 1107

c 2

= 301 gt χ 0 05 5

2

there4 Reject H 0 The playing time is normally distributed with mean 615 minutes and standard

deviation 1 minute

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Fully Worked Solutions 3

3 H 0 The number of defective screws per box follows a Poisson random variable with mean 1

H 1 The number of defective screws per box does not follows a Poisson random variable with

mean 1

X ~ Po(1) P( )

x x e

x

x

x

= = =minus1 1

0 1 2hellip

x P( X = x ) Oi

E i = np

i

( )O E

E

i i

i

minus

2

0 03679 35 3679 00871

1 03679 44 3679 14130

2 01839 14 1839 10480

3 00613

00153

00031

00006

1

803 013214 3

5 3

6 0

Σ 26802

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2

= 268 lt χ 0 05 3

2

there4 Do not reject H 0 The number of defective screws per box follows a Poisson random

variable with mean 1

4 (a) H 0 Age and sex are independent factors

H 1 Age and sex are dependent factors

00803

7

Oi

E i

( )O E

E

i i

i

minus

2

205 781 387

1495

( ) = 2021719 00396

182 714 387

1495

( ) = 18482280 00433

248 781 499

1495

( ) = 2606816 06169

Oi

E i

( )O E

E

i i

i

minus

2

251 714 499

1495

( ) = 2383183 06748

328 781 609

1495

( ) = 318146 03052

281 714 609

1495

( ) = 2908535 03338

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ACE AHEAD Mathematics (T) Third Term4

a = 005

v = 6 minus 1

= 5

χ

0 05 5

2

= 11070

c

2

= 201 lt χ 0 05 5

2

there4 Do not reject H 0 Age and sex are independent factors for insulin-dependent diabetes in

children under the age of 15

(b) Let p = probability of male in the proportion of diabetics

H 0 p = 05

H 1 p gt 05

Level of significant = 5

Let X be the number of diabetics who are males

X B~ ( ) since800 0 5 5np nqgt gt 5 hence N(400 200) X ~

there4

minus

=

minus

=

Z sim X 400

200

425 5 400

2001 80Test statistic

Reject H 0 if z gt 1645

Since 180 gt 1645 H 0 is rejected It is conclude that there are more diabetics who are

males than females

5 (a) X ~ B(4 021)

there4 P( X = x) = n

xC p q

x n xminus

P( X = 0) = 03895 there4 E 1 = 7790

P( X = 1) = 04142 there4 E 2 = 8283

P( X = 2) = 01651 there4 E 3 = 3303

P( X = 3) = 00293 there4 E 4 = 586

P( X = 4) = 00019 there4 E 5 = 038

Oi

E i

( )O E

E

i i

i

minus

2

79 7790 00155

81 8283 00404

34 3303 00285

6 624 00092

00936

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Fully Worked Solutions 5

χ 2 = 00942

v = 4 minus 1

= 3

χ

0 05 3

2

= 785

χ 2 = 00936 lt χ 0 05 32

there4 Do not reject H 0

(b) H 0 The data are random observations ~ Po(0 minus 9)

H 1 The data are not random observations ~ Po(0 minus 9)

P( X = 0) = 04066

P( X = 1) = 03659

P( X = 2) = 01647

P( X = 3) = 00494

P( X 4) = 1 ndash P( X 983100 3)

= 00134

Oi

E i

( )O E

E

i i

i

minus

2

79 8132 00662

81 7318 04627

34 3294 00341

5

16

9 88

2 6812 56

34262

39891

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2 = 399 lt χ

0 05 3

2

there4 Do not reject H 0 The data are random observations of a random variable Po(09)

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ACE AHEAD Mathematics (T) Third Term6

6 H 0 The data is normally distributed with mean 65 kg and standard deviation 15 kg

H 1 The data is not normally distributed with mean 65 kg and standard deviation 15 kg

Weight Oi

E i = np

i

( )O E

E

i i

i

minus

2

P i ( x )

x lt 30

30 983100 x lt 40

40 983100 x lt 50

0

3

15

18

0 982

3 79

11 09

1586 02888

000982

00379

011090

50 983100 x lt 60 25 2109 07249 021090

60 983100 x lt 70 23 2609 03660 026090

70 983100 x lt 80 18 2109 04527 011090

80 983100 x lt 90 12

4

0

16

11 09

3 79

0 982

1586 00012

011090

00379

000982

90 983100 x lt 100

100 983100 x

1834

a = 0025

v = 5 minus 1

= 4

χ

0 025 4

2

= 1114

χ 2

= 183 lt χ 0 025 4

2

there4 Do not reject H

0 The data is normally distributed with mean 65 kg and standard

deviation 15 kg

7 E( X

) = x x xf d ( )0

infin

int = 0 20

0 2

x

x

xinfin

minus

int e

d = ( )

minus

minus infin

x x

e 0 2

0 + e d

minusinfin

int 0 2

0

x

x

= 00 2

0 2

0

minus

minus infin

e

x

= 0 0 1

0 2minus minus

= 5 [Shown]

P(0 lt X 983100 3) = 0 20 2

0

3

e d minus

int x

x = 04512

P(3 lt X 983100 5) = 0 20 2

3

5

e d minus x

x = 01809

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term2

2 H 0 The playing time is normally distributed with mean 615 minutes and standard deviation

1 minute

H 1 The playing time is not normally distributed with mean 615 minutes and standard deviation

1 minute

X ~ N(615 12) z =minus

=

minus x xm

s

61 5

1

P P

P P

( ) ( )

( ) ( )

x

x

lt = lt minus

=

lt lt = minus lt lt minus

=

60 1 5

0 06681

60 61 1 5 0 5

0 24173

z

z

PP P

P P

( ) ( )

( ) ( )

61 62 0 5 0 5

0 38292

62 63 0 5 1 5

0

lt lt = minus lt lt

=

lt lt = lt lt

=

x z

x z

( ) ( )

( ) ( )

24173

63 64 1 5 2 5

0 06060

64 2 5

0

P P

P P

lt lt = lt lt

=

gt = gt

=

x

x

z

z

000621

Oi

E i= nP

( )O E

E

i i

i

minus

2

33 6681 171100

225 24173 11579

380 38292 00223

280 24173 60588

78 6060 49960

4 621 07865

301306

a = 005

v = 6 minus 1

= 5

χ

0 05 5

2

= 1107

c 2

= 301 gt χ 0 05 5

2

there4 Reject H 0 The playing time is normally distributed with mean 615 minutes and standard

deviation 1 minute

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Fully Worked Solutions 3

3 H 0 The number of defective screws per box follows a Poisson random variable with mean 1

H 1 The number of defective screws per box does not follows a Poisson random variable with

mean 1

X ~ Po(1) P( )

x x e

x

x

x

= = =minus1 1

0 1 2hellip

x P( X = x ) Oi

E i = np

i

( )O E

E

i i

i

minus

2

0 03679 35 3679 00871

1 03679 44 3679 14130

2 01839 14 1839 10480

3 00613

00153

00031

00006

1

803 013214 3

5 3

6 0

Σ 26802

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2

= 268 lt χ 0 05 3

2

there4 Do not reject H 0 The number of defective screws per box follows a Poisson random

variable with mean 1

4 (a) H 0 Age and sex are independent factors

H 1 Age and sex are dependent factors

00803

7

Oi

E i

( )O E

E

i i

i

minus

2

205 781 387

1495

( ) = 2021719 00396

182 714 387

1495

( ) = 18482280 00433

248 781 499

1495

( ) = 2606816 06169

Oi

E i

( )O E

E

i i

i

minus

2

251 714 499

1495

( ) = 2383183 06748

328 781 609

1495

( ) = 318146 03052

281 714 609

1495

( ) = 2908535 03338

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ACE AHEAD Mathematics (T) Third Term4

a = 005

v = 6 minus 1

= 5

χ

0 05 5

2

= 11070

c

2

= 201 lt χ 0 05 5

2

there4 Do not reject H 0 Age and sex are independent factors for insulin-dependent diabetes in

children under the age of 15

(b) Let p = probability of male in the proportion of diabetics

H 0 p = 05

H 1 p gt 05

Level of significant = 5

Let X be the number of diabetics who are males

X B~ ( ) since800 0 5 5np nqgt gt 5 hence N(400 200) X ~

there4

minus

=

minus

=

Z sim X 400

200

425 5 400

2001 80Test statistic

Reject H 0 if z gt 1645

Since 180 gt 1645 H 0 is rejected It is conclude that there are more diabetics who are

males than females

5 (a) X ~ B(4 021)

there4 P( X = x) = n

xC p q

x n xminus

P( X = 0) = 03895 there4 E 1 = 7790

P( X = 1) = 04142 there4 E 2 = 8283

P( X = 2) = 01651 there4 E 3 = 3303

P( X = 3) = 00293 there4 E 4 = 586

P( X = 4) = 00019 there4 E 5 = 038

Oi

E i

( )O E

E

i i

i

minus

2

79 7790 00155

81 8283 00404

34 3303 00285

6 624 00092

00936

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Fully Worked Solutions 5

χ 2 = 00942

v = 4 minus 1

= 3

χ

0 05 3

2

= 785

χ 2 = 00936 lt χ 0 05 32

there4 Do not reject H 0

(b) H 0 The data are random observations ~ Po(0 minus 9)

H 1 The data are not random observations ~ Po(0 minus 9)

P( X = 0) = 04066

P( X = 1) = 03659

P( X = 2) = 01647

P( X = 3) = 00494

P( X 4) = 1 ndash P( X 983100 3)

= 00134

Oi

E i

( )O E

E

i i

i

minus

2

79 8132 00662

81 7318 04627

34 3294 00341

5

16

9 88

2 6812 56

34262

39891

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2 = 399 lt χ

0 05 3

2

there4 Do not reject H 0 The data are random observations of a random variable Po(09)

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ACE AHEAD Mathematics (T) Third Term6

6 H 0 The data is normally distributed with mean 65 kg and standard deviation 15 kg

H 1 The data is not normally distributed with mean 65 kg and standard deviation 15 kg

Weight Oi

E i = np

i

( )O E

E

i i

i

minus

2

P i ( x )

x lt 30

30 983100 x lt 40

40 983100 x lt 50

0

3

15

18

0 982

3 79

11 09

1586 02888

000982

00379

011090

50 983100 x lt 60 25 2109 07249 021090

60 983100 x lt 70 23 2609 03660 026090

70 983100 x lt 80 18 2109 04527 011090

80 983100 x lt 90 12

4

0

16

11 09

3 79

0 982

1586 00012

011090

00379

000982

90 983100 x lt 100

100 983100 x

1834

a = 0025

v = 5 minus 1

= 4

χ

0 025 4

2

= 1114

χ 2

= 183 lt χ 0 025 4

2

there4 Do not reject H

0 The data is normally distributed with mean 65 kg and standard

deviation 15 kg

7 E( X

) = x x xf d ( )0

infin

int = 0 20

0 2

x

x

xinfin

minus

int e

d = ( )

minus

minus infin

x x

e 0 2

0 + e d

minusinfin

int 0 2

0

x

x

= 00 2

0 2

0

minus

minus infin

e

x

= 0 0 1

0 2minus minus

= 5 [Shown]

P(0 lt X 983100 3) = 0 20 2

0

3

e d minus

int x

x = 04512

P(3 lt X 983100 5) = 0 20 2

3

5

e d minus x

x = 01809

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

892019 Answer Maths TChapter 18

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

892019 Answer Maths TChapter 18

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Fully Worked Solutions 3

3 H 0 The number of defective screws per box follows a Poisson random variable with mean 1

H 1 The number of defective screws per box does not follows a Poisson random variable with

mean 1

X ~ Po(1) P( )

x x e

x

x

x

= = =minus1 1

0 1 2hellip

x P( X = x ) Oi

E i = np

i

( )O E

E

i i

i

minus

2

0 03679 35 3679 00871

1 03679 44 3679 14130

2 01839 14 1839 10480

3 00613

00153

00031

00006

1

803 013214 3

5 3

6 0

Σ 26802

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2

= 268 lt χ 0 05 3

2

there4 Do not reject H 0 The number of defective screws per box follows a Poisson random

variable with mean 1

4 (a) H 0 Age and sex are independent factors

H 1 Age and sex are dependent factors

00803

7

Oi

E i

( )O E

E

i i

i

minus

2

205 781 387

1495

( ) = 2021719 00396

182 714 387

1495

( ) = 18482280 00433

248 781 499

1495

( ) = 2606816 06169

Oi

E i

( )O E

E

i i

i

minus

2

251 714 499

1495

( ) = 2383183 06748

328 781 609

1495

( ) = 318146 03052

281 714 609

1495

( ) = 2908535 03338

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ACE AHEAD Mathematics (T) Third Term4

a = 005

v = 6 minus 1

= 5

χ

0 05 5

2

= 11070

c

2

= 201 lt χ 0 05 5

2

there4 Do not reject H 0 Age and sex are independent factors for insulin-dependent diabetes in

children under the age of 15

(b) Let p = probability of male in the proportion of diabetics

H 0 p = 05

H 1 p gt 05

Level of significant = 5

Let X be the number of diabetics who are males

X B~ ( ) since800 0 5 5np nqgt gt 5 hence N(400 200) X ~

there4

minus

=

minus

=

Z sim X 400

200

425 5 400

2001 80Test statistic

Reject H 0 if z gt 1645

Since 180 gt 1645 H 0 is rejected It is conclude that there are more diabetics who are

males than females

5 (a) X ~ B(4 021)

there4 P( X = x) = n

xC p q

x n xminus

P( X = 0) = 03895 there4 E 1 = 7790

P( X = 1) = 04142 there4 E 2 = 8283

P( X = 2) = 01651 there4 E 3 = 3303

P( X = 3) = 00293 there4 E 4 = 586

P( X = 4) = 00019 there4 E 5 = 038

Oi

E i

( )O E

E

i i

i

minus

2

79 7790 00155

81 8283 00404

34 3303 00285

6 624 00092

00936

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Fully Worked Solutions 5

χ 2 = 00942

v = 4 minus 1

= 3

χ

0 05 3

2

= 785

χ 2 = 00936 lt χ 0 05 32

there4 Do not reject H 0

(b) H 0 The data are random observations ~ Po(0 minus 9)

H 1 The data are not random observations ~ Po(0 minus 9)

P( X = 0) = 04066

P( X = 1) = 03659

P( X = 2) = 01647

P( X = 3) = 00494

P( X 4) = 1 ndash P( X 983100 3)

= 00134

Oi

E i

( )O E

E

i i

i

minus

2

79 8132 00662

81 7318 04627

34 3294 00341

5

16

9 88

2 6812 56

34262

39891

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2 = 399 lt χ

0 05 3

2

there4 Do not reject H 0 The data are random observations of a random variable Po(09)

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ACE AHEAD Mathematics (T) Third Term6

6 H 0 The data is normally distributed with mean 65 kg and standard deviation 15 kg

H 1 The data is not normally distributed with mean 65 kg and standard deviation 15 kg

Weight Oi

E i = np

i

( )O E

E

i i

i

minus

2

P i ( x )

x lt 30

30 983100 x lt 40

40 983100 x lt 50

0

3

15

18

0 982

3 79

11 09

1586 02888

000982

00379

011090

50 983100 x lt 60 25 2109 07249 021090

60 983100 x lt 70 23 2609 03660 026090

70 983100 x lt 80 18 2109 04527 011090

80 983100 x lt 90 12

4

0

16

11 09

3 79

0 982

1586 00012

011090

00379

000982

90 983100 x lt 100

100 983100 x

1834

a = 0025

v = 5 minus 1

= 4

χ

0 025 4

2

= 1114

χ 2

= 183 lt χ 0 025 4

2

there4 Do not reject H

0 The data is normally distributed with mean 65 kg and standard

deviation 15 kg

7 E( X

) = x x xf d ( )0

infin

int = 0 20

0 2

x

x

xinfin

minus

int e

d = ( )

minus

minus infin

x x

e 0 2

0 + e d

minusinfin

int 0 2

0

x

x

= 00 2

0 2

0

minus

minus infin

e

x

= 0 0 1

0 2minus minus

= 5 [Shown]

P(0 lt X 983100 3) = 0 20 2

0

3

e d minus

int x

x = 04512

P(3 lt X 983100 5) = 0 20 2

3

5

e d minus x

x = 01809

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term4

a = 005

v = 6 minus 1

= 5

χ

0 05 5

2

= 11070

c

2

= 201 lt χ 0 05 5

2

there4 Do not reject H 0 Age and sex are independent factors for insulin-dependent diabetes in

children under the age of 15

(b) Let p = probability of male in the proportion of diabetics

H 0 p = 05

H 1 p gt 05

Level of significant = 5

Let X be the number of diabetics who are males

X B~ ( ) since800 0 5 5np nqgt gt 5 hence N(400 200) X ~

there4

minus

=

minus

=

Z sim X 400

200

425 5 400

2001 80Test statistic

Reject H 0 if z gt 1645

Since 180 gt 1645 H 0 is rejected It is conclude that there are more diabetics who are

males than females

5 (a) X ~ B(4 021)

there4 P( X = x) = n

xC p q

x n xminus

P( X = 0) = 03895 there4 E 1 = 7790

P( X = 1) = 04142 there4 E 2 = 8283

P( X = 2) = 01651 there4 E 3 = 3303

P( X = 3) = 00293 there4 E 4 = 586

P( X = 4) = 00019 there4 E 5 = 038

Oi

E i

( )O E

E

i i

i

minus

2

79 7790 00155

81 8283 00404

34 3303 00285

6 624 00092

00936

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Fully Worked Solutions 5

χ 2 = 00942

v = 4 minus 1

= 3

χ

0 05 3

2

= 785

χ 2 = 00936 lt χ 0 05 32

there4 Do not reject H 0

(b) H 0 The data are random observations ~ Po(0 minus 9)

H 1 The data are not random observations ~ Po(0 minus 9)

P( X = 0) = 04066

P( X = 1) = 03659

P( X = 2) = 01647

P( X = 3) = 00494

P( X 4) = 1 ndash P( X 983100 3)

= 00134

Oi

E i

( )O E

E

i i

i

minus

2

79 8132 00662

81 7318 04627

34 3294 00341

5

16

9 88

2 6812 56

34262

39891

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2 = 399 lt χ

0 05 3

2

there4 Do not reject H 0 The data are random observations of a random variable Po(09)

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ACE AHEAD Mathematics (T) Third Term6

6 H 0 The data is normally distributed with mean 65 kg and standard deviation 15 kg

H 1 The data is not normally distributed with mean 65 kg and standard deviation 15 kg

Weight Oi

E i = np

i

( )O E

E

i i

i

minus

2

P i ( x )

x lt 30

30 983100 x lt 40

40 983100 x lt 50

0

3

15

18

0 982

3 79

11 09

1586 02888

000982

00379

011090

50 983100 x lt 60 25 2109 07249 021090

60 983100 x lt 70 23 2609 03660 026090

70 983100 x lt 80 18 2109 04527 011090

80 983100 x lt 90 12

4

0

16

11 09

3 79

0 982

1586 00012

011090

00379

000982

90 983100 x lt 100

100 983100 x

1834

a = 0025

v = 5 minus 1

= 4

χ

0 025 4

2

= 1114

χ 2

= 183 lt χ 0 025 4

2

there4 Do not reject H

0 The data is normally distributed with mean 65 kg and standard

deviation 15 kg

7 E( X

) = x x xf d ( )0

infin

int = 0 20

0 2

x

x

xinfin

minus

int e

d = ( )

minus

minus infin

x x

e 0 2

0 + e d

minusinfin

int 0 2

0

x

x

= 00 2

0 2

0

minus

minus infin

e

x

= 0 0 1

0 2minus minus

= 5 [Shown]

P(0 lt X 983100 3) = 0 20 2

0

3

e d minus

int x

x = 04512

P(3 lt X 983100 5) = 0 20 2

3

5

e d minus x

x = 01809

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 5

χ 2 = 00942

v = 4 minus 1

= 3

χ

0 05 3

2

= 785

χ 2 = 00936 lt χ 0 05 32

there4 Do not reject H 0

(b) H 0 The data are random observations ~ Po(0 minus 9)

H 1 The data are not random observations ~ Po(0 minus 9)

P( X = 0) = 04066

P( X = 1) = 03659

P( X = 2) = 01647

P( X = 3) = 00494

P( X 4) = 1 ndash P( X 983100 3)

= 00134

Oi

E i

( )O E

E

i i

i

minus

2

79 8132 00662

81 7318 04627

34 3294 00341

5

16

9 88

2 6812 56

34262

39891

a = 005

v = 4 minus 1

= 3

χ 0 05 3

2

= 7815

c 2 = 399 lt χ

0 05 3

2

there4 Do not reject H 0 The data are random observations of a random variable Po(09)

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ACE AHEAD Mathematics (T) Third Term6

6 H 0 The data is normally distributed with mean 65 kg and standard deviation 15 kg

H 1 The data is not normally distributed with mean 65 kg and standard deviation 15 kg

Weight Oi

E i = np

i

( )O E

E

i i

i

minus

2

P i ( x )

x lt 30

30 983100 x lt 40

40 983100 x lt 50

0

3

15

18

0 982

3 79

11 09

1586 02888

000982

00379

011090

50 983100 x lt 60 25 2109 07249 021090

60 983100 x lt 70 23 2609 03660 026090

70 983100 x lt 80 18 2109 04527 011090

80 983100 x lt 90 12

4

0

16

11 09

3 79

0 982

1586 00012

011090

00379

000982

90 983100 x lt 100

100 983100 x

1834

a = 0025

v = 5 minus 1

= 4

χ

0 025 4

2

= 1114

χ 2

= 183 lt χ 0 025 4

2

there4 Do not reject H

0 The data is normally distributed with mean 65 kg and standard

deviation 15 kg

7 E( X

) = x x xf d ( )0

infin

int = 0 20

0 2

x

x

xinfin

minus

int e

d = ( )

minus

minus infin

x x

e 0 2

0 + e d

minusinfin

int 0 2

0

x

x

= 00 2

0 2

0

minus

minus infin

e

x

= 0 0 1

0 2minus minus

= 5 [Shown]

P(0 lt X 983100 3) = 0 20 2

0

3

e d minus

int x

x = 04512

P(3 lt X 983100 5) = 0 20 2

3

5

e d minus x

x = 01809

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term6

6 H 0 The data is normally distributed with mean 65 kg and standard deviation 15 kg

H 1 The data is not normally distributed with mean 65 kg and standard deviation 15 kg

Weight Oi

E i = np

i

( )O E

E

i i

i

minus

2

P i ( x )

x lt 30

30 983100 x lt 40

40 983100 x lt 50

0

3

15

18

0 982

3 79

11 09

1586 02888

000982

00379

011090

50 983100 x lt 60 25 2109 07249 021090

60 983100 x lt 70 23 2609 03660 026090

70 983100 x lt 80 18 2109 04527 011090

80 983100 x lt 90 12

4

0

16

11 09

3 79

0 982

1586 00012

011090

00379

000982

90 983100 x lt 100

100 983100 x

1834

a = 0025

v = 5 minus 1

= 4

χ

0 025 4

2

= 1114

χ 2

= 183 lt χ 0 025 4

2

there4 Do not reject H

0 The data is normally distributed with mean 65 kg and standard

deviation 15 kg

7 E( X

) = x x xf d ( )0

infin

int = 0 20

0 2

x

x

xinfin

minus

int e

d = ( )

minus

minus infin

x x

e 0 2

0 + e d

minusinfin

int 0 2

0

x

x

= 00 2

0 2

0

minus

minus infin

e

x

= 0 0 1

0 2minus minus

= 5 [Shown]

P(0 lt X 983100 3) = 0 20 2

0

3

e d minus

int x

x = 04512

P(3 lt X 983100 5) = 0 20 2

3

5

e d minus x

x = 01809

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 7

P(5 lt X 983100 7) = 0 20 2

3

7

e d minus x

x = 01213

P(7 lt X 983100 9) = 0 20 2

7

9

e d minus x

x = 00813

P(9 lt X 983100 11) = 0 20 2

9

11

e d minus x

x = 00545

P(11 lt X 983100 13) = 0 20 2

11

13

e d minusint x

x = 00365

P( X gt 13) = 0 213

0 2

infin

minus

int e d x

x

= 00743

H 0 The data is from the same probability density function

H 1 The data is not from the same probability density function

Time ( seconds) Oi

E i= np

( )O E i i

i

minusminus

2

E P( X

)

0 lt x 983100 3 26 4512 81023 04512

3 lt x 983100 5 19 1809 00458 01809

5 lt x 983100 7 16 1213 12347 01213

7 lt x 983100 9 15 813 58053 00813

9 lt x 983100 11 10 545 37986 00545

11 lt x 983100 13

x gt 13

14

0

14

3 65

7 43

11 08

0769500365

00743

197565

a = 005

v = 5 minus 1

= 4

χ

0 05 4

2

= 9488

χ 2

= 198 gt χ 0 05 4

2

there4 Reject H 0 The data is not a good fit model

8 H 0 The data is normally distributed with mean 487 and standard deviation 149

H 1 The data is not normally distributed with mean 487 and standard deviation 149

X ~ N(487 1492)

there4 P(a lt x lt b) = Pa

Z bminus

lt lt minus

48 7

14 9

48 7

14 9

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

892019 Answer Maths TChapter 18

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

892019 Answer Maths TChapter 18

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

892019 Answer Maths TChapter 18

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

892019 Answer Maths TChapter 18

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term8

Marks Oi

E i

( )O E

E

i i

i

minusminus

2

0 minus 9

10 minus 19

0

1919

2 15

10 3512 50

3380

20 minus 29 50 3685 4693

30 minus 39 47 8495 16954

40 minus 49 125 12645 0017

50 minus 59 132 12210 0803

60 minus 69 100 7645 7254

70 minus 79 27 3100 0516

80 minus 89 0 970 9700

43317

a = 005

v = 8 minus 1

= 7

χ

0 05 7

2

= 1407

χ 2

= 433 gt χ 0 05 7

2

there4 Reject H 0 The data is not normally distributed with mean 487 and standard deviation 149

9 H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

H 1 The data is not normally distributed with mean 180 cm and standard deviation 3 cm

X ~ N(180 32)

Height Oi

E i = P

i

( )O E

E

i i

i

minusminus

2

P( X )

x lt 175 2

15

478

110900805

000956

175 x lt 177 002218

177 x lt 179 29 2109 29667 004218

179 x lt 181 25 2609 00455 005218

181

x lt 183 12 2109 39179 004218183 x lt 185 10

7

1109

47800805

002218

x 185 000956

70911

a = 005

v = 5 minus 1

= 4

χ 0 05 4

2

= 9488

17

1587

17

1587

892019 Answer Maths TChapter 18

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1119

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

892019 Answer Maths TChapter 18

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 9

χ 2

= 709 lt χ 0 05 4

2

there4 Do not reject H 0 The data is normally distributed with mean 180 cm and standard deviation 3 cm

10 H 0 The number of babies born in I II III and IV quadrant are in the ratio 4 2 1 3

H 1 The number of babies born in I II III and IV quadrant are not in the ratio 4 2 1 3

Quadrant P( X ) Oi

E i

( )O E

E

i i

i

minusminus

2

I 04 142 120 40333

II 02 52 60 10667

III 01 34 30 05333

IV 03 72 90 36000

92333

a = 001

v = 4 minus 1

= 3

χ 0 01 3

2

= 1135

c 2

= 923 lt χ 0 01 3

2

there4 Do not reject H 0 The number of babies born in I II III and IV quadrant are in the ratio

4 2 1 3

11 X ~ B(5 05)

P i

O2

E i = mp

( )O E

E

i i

i

minusminus

2

00313 2 31301651

01563 15 1563

03125 S 3125( )

S minus 31 25

31 25

2

03125 69 minus S 3125( )

37 75

31 25

2 minus S

01563 12 1563

00313 2 313 12078

there4 sum

c 2 =

sum

( )O E

E

i i

i

minus

2

= 13729 +

( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

a = 01

v = 4 minus 1

= 3

17

1876

14

1876

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term10

there4 0 1 3

2

= 6251

We reject H 0 if c

2 c

0 1 3

2

there4 H 0not being rejected if c

2 lt 6251

there4 13729 + ( )

S minus 31 25

31 25

2

+ ( )

37 75

31 25

2 minus S

lt 6251

2S

2 minus 138S + 395 lt 0

there4 S = 2639 lt S lt 4260

there4 The smallest value of S is 27

12 H 0 The colour of a lost button is independent of boyrsquos form

H 1 The colour of a lost button is dependent of boyrsquos form

24(2100) 9(1125) 27(2775) 60

26(2450) 13(1313) 31(3238) 70

2(525) 5(281) 8(694) 15

4(525) 3(281) 8(694) 15

160

Oi

E i

( )O E

E

i i

i

minusminus

2

24 21000 04286

9 11250 04500

27 27750 00203

26 24500 00918

13 13130 00013

31 30380 00588

6 10500 19286

8 5620 10079

16 13880 03238

43111

v = (3 minus 1)(2 minus 1)

= 2

there4 c

0 05 2

2

= 5991

c 2

= 431 lt 5991

there4 Do not reject H 0 The colour of a lost button is not associate with the schoolboyrsquos form

Combine

892019 Answer Maths TChapter 18

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

892019 Answer Maths TChapter 18

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

892019 Answer Maths TChapter 18

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

892019 Answer Maths TChapter 18

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 11

13 H 0 The sample is from the Po(15) distribution

H 1 The sample is not from the Po(15) distribution

x P( X = x) Oi

E i = 100P

( )O E

E

i i

i

minusminus

2

0 02231 25 2231 03243

1 03347 31 3347 01823

2 02510 23 2510 01757

3 01255 13 1255 00161

4 00471 5 47103112

5 00186 3 186

10096

a = 001 v = 5 minus 1

= 4

there4 c

0 01 4

2

= 1328

c 2 = 101 lt

c 0 01 4

2

there4 Do not reject H 0 The sample is from the Po(15) distribution

14 H 0 The number of cars of each colour of that model is the same

H 1 The number of cars of each colour of that model is not the same

Oi

E i

( )O E

E

i i

i

minusminus

2

13 15 02667

14 15 00667

16 15 00667

17 15 02667

06668

a = 005

v = 4 minus 1

= 3

there4 c 2

0053 = 782

c 2 = 0667 lt c

2

0053

there4 Do not reject H

0 The number of cars of each colour is the same

8

657

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

892019 Answer Maths TChapter 18

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

892019 Answer Maths TChapter 18

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copy O xford Fajar Sdn Bhd (008974-T) 2014

ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1519

copy O xford Fajar Sdn Bhd (008974-T) 2014

Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

892019 Answer Maths TChapter 18

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term12

15 H 0 The level of education and the opinion on the social issues are independent

H 1 The level of education and the opinion on the social issues are dependent

Level of EducationOpinion on Social issue

Total Agree Disagree

University468(1227)

1300 = 44172

468 73

130026 28

( )= 468

College577(1227)

1300 = 54460

577 73

130032 40

( )= 577

High School255(1227)

1300 = 24068

255 75

130014 32

( )= 255

Total 1227 73 1300

Oi

E i

( )O E

E

i i

i

minusminus

2

450 44172 01552

547 54460 00106

230 24068 04739

18 2628 26088

30 3240 01778

25 1432 79653

113916

a = 001

v = (3 minus 1)(2 minus 1)

= 2

c 2001 2

= 9210

c

2 = 11392 gt c

2

001 2

there4 Reject H

0 The level of education and the opinion on the social issues are dependent

16 In hypothesis testing the significance level is the criterion used for rejecting the null

hypothesis Test statistic c

2 = 1938

v = 7

there4 Reject H 0 if c 2 c

2

a 7

From table c

2001 7

= 18475 and c

20005 7

= 20278 gt c

2 001 7

Hence the smallest value of a

is estimated as 005

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 13

17 (a) P( X = x) = eminus092 092 x

x

X P( X ) E = np

0 03985 23830

1 03666 21923

2 01687 10088

3 00517 3092

4 00119 712

5 00026 155

(b)

Oi

E i

( )O E

E

i i

i

minusminus

2

241 23830 00306

211 21923 03090

104 10088 00965

35 3092

712

3959

155

05384

00020

01467

15500

7

0

(i) c 2

= ( )O E

E

i i

ii

minus

=sum

2

1

6

= 00306 + 03090 + 00965 + 05384 + 00020 + 155

= 25265 asymp 253

(ii) c 2

=( )O E

E

i i

ii

minus

=

sum2

1

4

= 00306 + 03090 + 00965 + 01467

= 05828 asymp 0583

The value of test statistic obtained from the combined frequencies is very much lower then

the value of test statistic without combining any frequency

(c) H 0 The data fits the Poisson distribution with mean 092

H 1 The data does not fit the Poisson distribution with mean 092 a

= 005

v = 4 minus 1 = 3

c

2005

= 782

c

2 = 0583 lt c

2005 2

there4 Do not reject H 0 The data fits the Poisson distribution with mean 092

18 H 0 The educational level is related to the opinion on caning

H 1 The educational level is not related to the opinion on caning

42

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term14

Education Level Yes No Not Sure

High school (323)(290)

8201142317=

213 290

82075 3293

( )=

284 290

820100 4390

( )=

College 323 230

820

90 5976( )

=

213 230

820

59 7439( )

=

284 230

820

79 6585( )

=

University 323(300)

8201181707=

213 300

82077 9268

( )=

284 300

820103 9024

( )=

Oi

E i

( )O E

E

i i

i

minusminus

2

125 11423 10154

98 9060 06044

100 11817 27938

65 7533 14166

68 5974 11420

80 7793 00550

100 10044 00019

64 7966 30785

120 10390 24948

126024

a = 001

v = (3 minus 1)(3 minus 1)

= 4

c

2

001 4 = 1328

c

2002 4

= 1114

c 2 = 126 lt c 2001 4

c 2 = 126 gt c 2002 4

there4 Do not reject H 0 at 1 level of significant But reject H

0 at 2 level of significant

19 (a)

P( X = x ) Oi

E i = 50P

i

( )O E

E

i i

i

minusminus

2

008

010

015

025

042

10

15

25

8

5

12

40

50

90

75

125

210

284444

00333

45000

38571

368348

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 15

(b) H 0 The data fits the probability distribution as suggested

H 1 The data does not fit the probability distribution as suggested

c 2

0053 = 782 Since c 2

= 368 gt c 2

0053 reject H 0 The data does not fit the suggested

probability function

20 (a) Cream A 30 45

100

+ times 100 = 75

Cream B25 65

100

+ times 100 = 90

(b) H 0 The condition of a patient is independent of the type of cream used

H 1 The condition of a patient is dependent of the type of cream used

Cream

Patients

No Recovery Partial Recovery Complete Recovery

A100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

B100 35

20017 50

( )=

100 55

20027 50

( )=

100 110

20055 00

( )= 100

Total 35 55 110 200

Oi

E i

( )O E

E

i i

i

minusminus

2

25 1750 32143

10 1750 32143

30 2750 02273

25 2750 02273

45 5500 11818

65 5500 11818

92468

a

= 005

v = (3 minus 1)(2 minus 1)

= 2

c 2005 2

= 5991

c 2 = 925 gt c 2005 2

there4 Reject H 0 The condition of the patient is dependent on the type of cream used

(c) The result in (a) agrees with the calculation in (b)

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

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Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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ACE AHEAD Mathematics (T) Third Term16

21 (a)

Smoker Long Ailment

Total Mild Severe

Light smoker 23 14 37

Heavy smoker 20 39 59Total 43 53 96

(b) H 0 The severity of lung ailment and the smoking habit are independent

H 1 The severity of lung ailment and the smoking habit are dependent

Smoker Lung Mild Ailment Severe

Light43 37

96

( )

= 165729 53 37

96

( )= 204271

Heavy43 59

96

( )

= 264271

53 59

96

( )= 325729

Oi

E i

O E

E

i i

i

minusminus minusminus(( ))0 52

23 165729 21198

14 204271 1719820 264271 13293

39 325729 10785

62474

a = 005

v = (2 minus 1)(2 minus 1)

= 1 (Yates correction)

c

0 05 1

2

= 3841

c

2 = 625 gt c 0 05 1

2

reject H

0 The severity of lung ailment and the smoking habit are dependent

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

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Fully Worked Solutions 17

22 H 0 The normal distribution with mean 65 kg and standard deviation 15 kg is an appropriate model

H 1 The normal distribution with mean 65 kg and standard deviation 15 kg is not an appropriate

model

X sim Ν(65 225)

Weight P( X = x ) Oi E i = 100 times P

( )O E

E

i i

i

minusminus

2

x 983100 30 00098 0

3

18

15

098

379

1586

1109

0288830 983100 x lt 40 00379

40 983100 x lt 50 01109

50 983100 x lt 60 02109 25 2109 07249

60 983100 x lt 70 02609 23 2609 03660

70 983100 x lt 80 02109 18 2109 04527

80 983100 x lt 90 01109 12

4

16

0

1109

379

1586

098

00012 90 983100 x lt 100 00379

x 100 00098

18336

a = 005

V = 5 minus 1

= 4

χ

0 05 4

2

= 9488

c 2 = 183 lt χ 0 05 4

2

there4 Do not reject H 0 The normal distribution with mean 65 kg and standard deviation 15 kg isan appropriate model

23

Oi

E i

( )O E

E

i i

i

minusminus

2

p minus q8

8

p

p

2

= p q

p

p + q8

8

p

p

= p q

p

2

p minus r 8

8

p

p

= p r

p

2

p + r 8

8

p

p

2

= p r

p

2

2 p + q + r 16

8

2 p

p

= 2 p ( )q r

p

+ 2

2

2 p minus q minus r 16

8

2

p

= 2 p ( )q r

p

+

2

2

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

892019 Answer Maths TChapter 18

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ACE AHEAD Mathematics (T) Third Term18

χ 2

2

1

6

= minus

=

sum( )O E

E

i i

ii

= 2 q

p

2

+ 2 r

p

2

+ 2 ( )q r

p

+ 2

2

= 3 3 2

2 2q r qr

p

+ +

=3 22 2( )q r qr

p

+ +

= 3 4

2( )q r qr

p

+ minus (Shown)

H

0 The rate of germination and the type of corn seeds are independent

H

1 The rate of germination and the type of cord seeds are dependent

Oi E i

( )O E

E

i i

i

minusminus

2

20 25 100

30 25 100

29 25 064

21 25 064

51 50 002

49 50 002

332

a

= 005

v = (2 minus 1)(3 minus 1)

= 2

χ 0 0

2

2 = 5991

χ 2

= 332 lt χ 0 05

2

2

there4 Do not reject H 0 The rate of germination does not depends on the types of corn seeds

24 H 0 The blood pressure follow a normal distribution with mean 12846 and standard deviation

1369

H 1 The blood pressure does not follow a normal distribution with mean 12846 and standard

deviation 1369

X ~ N(12846 13692)

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369

892019 Answer Maths TChapter 18

httpslidepdfcomreaderfullanswer-maths-tchapter-18 1919

Fully Worked Solutions 19

P( X = x ) Oi

E 2 = 250P

( )O E

E

i i

i

minusminus

2

00002 0

3

15

12

0050

4650

22200

1750000186 2335

00700

01795 52 44875 1131

02763 74 69075 0351

02557 67 63925 0148

01419 26 35475 2531

00472 12

4

16

0

11800

2625

14450

0025

00105 0166

00001

6662

a

= 005 v = 6 minus 1

= 5

χ

0 05

2

5 = 11070

c c 2 2

0 05 56 66= lt

there4 Do not reject H 0 The blood pressure follow a normal distribution with mean 12846 and

standard deviation 1369