1999 Maths

Hong Kong Examinations AuthorityAll Rights Reserved 1999

99-AS-M&S1

HONG KONG EXAMINATIONS AUTHORITY

HONG KONG ADVANCED LEVEL EXAMINATION 1999

MATHEMATICS AND STATISTICS AS-LEVEL

8.30 am 11.30 am (3 hours)This paper must be answered in English

1. This paper consists of Section A and Section B.

2. Answer ALL questions in Section A, using the AL(C)1 answer book.

3. Answer any FOUR questions in Section B, using the AL(C)2 answer book.

4. Unless otherwise specified, numerical answers should either be exact or givento 4 decimal places.

99-ASLM & S

99-AS-M&S2 1

All Rights Reserved 1999

SECTION A (40 marks)Answer ALL questions in this section.Write your answers in the AL(C)1 answer book.

1. It is given that 1)1(

2

3

+

+=

x

xxe

xy , where x > 0 .

(a) Find the value of y when 1=x .

(b) Find the value of x

ydd

when 1=x .

(5 marks)

2. (a) Expand xe 2 in ascending powers of x as far as the term in 3x .

(b) Using (a), expand xe

x

2

21

)1( + in ascending powers of x as far as the term

in 3x .

State the range of values of x for which the expansion is valid.(6 marks)

3. A test was carried out to see how quickly a class of students reacted to a visualinstruction to press a particular key when they played a computer game. Theirreaction times, measured in tenths of a second, are recorded and the statisticsfor the whole class are summarized below.

Lower quartile Upper quartile Median Minimum MaximumBoys 8 14 11 5 17Girls 9 16 11 7 21

(a) Draw two box-and-whisker diagrams in your answer book comparingthe reaction times of boys and girls.

(b) Suppose a boy and a girl are randomly selected from the class. Whichone will have a bigger chance of having a reaction time shorter than1.1 seconds? Explain.

(5 marks)

99-AS-M&S3 2

All Rights Reserved 1999Go on to the next page

4. The total number of visits N to a web site increases at a rate of

)118(dd 2

131

ttt

N+= (0 t 100) ,

where t is the time in weeks since January 1, 1999. It is known that N = 100when t = 1 .

(a) Express N in terms of t .

(b) Find the total number of visits to the web site when t = 64 .(5 marks)

5. 60% of passengers who travel by train use Octopus. A certain train has 12compartments and there are 10 passengers in each compartment.

(a) What is the probability that exactly 5 of the passengers in acompartment use Octopus?

(b) What is the mean number of passengers using Octopus in acompartment?

(c) What is the probability that the third compartment is the first one to haveexactly 5 passengers using Octopus?

(6 marks)

6. At a school sports day, the timekeeping group for running events consists of 1chief judge, 1 referee and 10 timekeepers. The chief judge and the referee arechosen from 5 teachers while the 10 timekeepers are selected from 16students.

(a) How many different timekeeping groups can be formed?

(b) If it is possible to have a timekeeping group with all the timekeepersbeing boys, what are the possible numbers of boys among the 16students?

(c) If the probability of having a timekeeping group with all thetimekeepers being boys is

3643

, find the number of boys among the

16 students.(6 marks)

99-AS-M&S4 3


7. Three control towers A , B and C are intelecommunication contact by means ofthree cables X , Y and Z as shown inFigure 1. A and B remain in contact onlyif Z is operative or if both cables X and Yare operative. Cables X , Y and Z aresubject to failure in any one day withprobabilities 0.015 , 0.025 and 0.030respectively. Such failures occurindependently.

(a) Find, to 4 significant figures, the probability that, on a particular day,(i) both cables X and Z fail to operate,(ii) all cables X , Y and Z fail to operate,(iii) A and B will not be able to make contact.

(b) Given that cable X fails to operate on a particular day, what is theprobability that A and B are not able to make contact?

(c) Given that A and B are not able to make contact on a particular day,what is the probability that cable X has failed?

(7 marks)

A

B CX

YZ

Figure 1

99-AS-M&S5 4


SECTION B (60 marks)Answer any FOUR questions in this section. Each question carries 15 marks.Write your answers in the AL(C)2 answer book.

8. In a 100 m race, the speeds, AS m/s and BS m/s , of two athletes A and Brespectively can be modelled by the functions

+= tttS A 1204

4731

9625256 23

and ktB teS

=

50183

,

where k is a positive constant and t is the time measured from the start inseconds.

It is known that A finishes the race in 12.5 seconds and during the race, A andB attain their respective top speeds at the same time.

(a) Find the top speed of A during the race. (3 marks)

(b) Find the value of k . (3 marks)

(c) Suppose the model for B is valid for 0 t 12.5 . Use the trapezoidalrule with 5 sub-intervals to estimate the distance covered by B in 12.5seconds. (3 marks)

(d) Find 2

2

dd

t

S B . Hence or otherwise, state with reasons whether B

finishes the race ahead of A or not. (3 marks)

(e) In the same race, the speed, CS m/s , of another athlete C is modelledby

2]2ln)2[ln(50

+

+=

t

tSC .

Determine whether or not C is the last one to finish the race among thethree athletes. (3 marks)

Go on to the next page

99-AS-M&S6 5


9. An ecologist studies the birds at Mai Po Nature Reserve. Only 21% of thebirds are residents, i.e. found throughout the year. The remaining birds aremigrants. The ecologist suggests that the number N(t) of a certain species ofmigrants can be modelled by the function

btae

t+

=

10003)N( ,

where a , b are positive constants and t is the number of days elapsed sincethe first one of that species of migrants was found at Mai Po in that year.

(a) This year, the ecologist obtained the following data:t 5 10 15 20

N(t) 250 870 1 940 2 670

(i) Express

1)N(0003ln

t as a linear function of t .

(ii) Use the graph paper on Page 6 to estimate graphically the valuesof a and b correct to 1 decimal place.

(5 marks)

(b) Basing on previous observations, the migrants of that species start toleave Mai Po when the rate of change of N(t) is equal to one hundredthof N(t). Once they start to leave, the original model will not be validand no more migrants will arrive. It is known that the migrants willleave at the rate r(s) per day where ss 60)r( = and s is the numberof days elapsed since they started to leave Mai Po. Using the values ofa and b obtained in (a)(ii),

(i) find )(N t , and show that N( t) is increasing;

(ii) find the greatest number of the migrants which can be found atMai Po this year;

(iii) find the number of days in which the migrants can be found atMai Po this year.

(10 marks)

99-AS-M&S7 6


Page Total

9. (Contd) If you attempt Question 9, fill in the details in the first three boxesabove and tie this sheet INSIDE your answer book.

Candidate Number Centre Number Seat Number

1)N(0003ln

t

t0

1

5 10 15 20

1

2

3

4

2


99-AS-M&S8 7


This is a blank page.

99-AS-M&S9 8


10. A criminologist has developed a questionnaire for predicting whether ateenager will become a delinquent. Scores on the questionnaire can range from0 to 100 , with higher values indicating a greater criminal tendency. Thecriminologist sets a critical level at 75 , i.e., a teenager scores more than 75will be classified as a potential delinquent (PD). Extensive studies have shownthat the scores of those considered non-PDs follow a normal distribution with amean of 65 and standard deviation of 5 . The scores of those considered PDsfollow a normal distribution with a mean of 80 and standard deviation of 5 .

(a) Find the probability that(i) a PD will be misclassified,(ii) a non-PD will be misclassified.

(4 marks)

(b) What is the probability that out of 10 PDs, not more than 2 will bemisclassified? (3 marks)

(c) If a sociologist wants to ensure that only 1 in 100 PDs should bemisclassified, what critical level of score should be used?

(3 marks)

(d) It is known that 10% of all teenagers are PDs. Will the probability ofteenagers misclassified by the sociologist in (c) be greater than thatmisclassified by the criminologist? Explain.

(5 marks)


99-AS-M&S10 9


11. Letx

xx

=

246)f( ( 2x ) and

be

eax

x

x

+

=

+ 1)g(2

, where a and b are constants.

Let 1C and 2C be the curves )f(xy = and )g(xy = respectively.Figure 2 shows 2C for 83 x .

(a) Show that 0)(f > x for 2x . (1 mark)

(b) Find the equations of the horizontal and vertical asymptotes to 1C .(2 marks)

(c) It is given that )2g()2f( = and )1g()1f( = . Find the exact values ofa and b . (2 marks)

(d) Sketch 1C on Figure 2 and indicate the asymptotes, intercepts and thepoints of intersection of the two curves. (3 marks)

(e) Find )(g x . Hence explain briefly why there is no point of intersectionof the two curves beyond the range 83 x . (3 marks)

(f) Find the area of the region bounded by the two curves. (4 marks)

99-AS-M&S11 10


Page Total

11. (Contd) If you attempt Question 11, fill in the details in the first three boxesabove and tie this sheet INSIDE your answer book.

2C


x

y

0 22 4 6 8

5

10

15

5

10

Figure 2


99-AS-M&S12 11


This is a blank page.

99-AS-M&S13 12


12. A bus company finds that the number of complaints received per dayfollows a Poisson distribution with mean 10 . 40% of the complaintsinvolve the time schedule, 35% involve the manner of drivers, 13%involve the routes and 12% involve other things. These four kinds ofcomplaints are mutually exclusive and can be resolved to thepassengers satisfaction with probabilities 0.6 , 0.2 , 0.7 and 0.5respectively.

(a) If a complaint cannot be resolved to the passengers satisfaction, findthe probability that this complaint involves the manner of drivers.

(4 marks)

(b) Find the probability that on a given day,(i) there are 5 complaints,(ii) there are 5 complaints and 3 of them involve the time schedule.

(4 marks)

(c) Find the probability that on a given day, there are n complaints and 9of them involve the time schedule. (2 marks)

(d) (i) Show that xk

kex

kx 9

9 )!9( == .(ii) Find the probability that, on a given day, there are 9 complaints

involving the time schedule.(5 marks)


99-AS-M&S14 13


13. A herbal tea for curing a certain disease is prepared and sold by manydifferent shops. A researcher has collected the herbal tea from 100randomly selected shops and counted the number of different kinds ofmedicinal herbs in the tea. The results are shown in the first twocolumns of Table 1.

(a) Find the sample mean number of different kinds of medicinal herbs inthe herbal tea. (1 mark)

(b) The researcher tried to fit the data by a binomial ( 7=n ), a Poisson anda normal distribution. The sample mean was used as the mean of thedistributions. Class intervals (0.5, 0.5] , (0.5, 1.5] , , (7.5, 8.5]were used to calculate the expected frequencies under the normaldistribution. Fill in the missing values in Table 1. (7 marks)

(c) Suppose the absolute values of the differences between observed andexpected frequencies are regarded as errors. The distribution with thesmallest maximum error is considered as the best fit. Which distributionis the best? (1 mark)

(d) A man drank a cup of this herbal tea bought from a randomly selectedshop each day starting from the first day he felt sick. Find theprobability, under the best distribution found in (c), that(i) the fourth day is the first day that he actually got medicinal herbs

in the tea;

(ii) in ten consecutive days he drank, at least two cups of this teawhich contained exactly three kinds of medicinal herbs.

(6 marks)

99-AS-M&S15 14


Pag

Page Total13. (Contd) If you attempt Question 13, fill in the details in the first three boxes

above and tie this sheet INSIDE your answer book.

Table 1 Observed and expected frequencies of the number ofmedicinal herbs in a particular herbal tea

Expected frequency *Number ofmedicinal herbs

Observedfrequency Poisson Binomial Normal

0 5 4.98 1.99 5.32

1 14 14.94 10.44 11.73

2 23 22.40 23.50 19.37

3 22 22.40 22.82

4 17 16.80 19.37

5 11 9.91

6 5 2.48

7 3 2.16 0.27 1.70

8 0 0.81 0.40

Total 100

* Correct to 2 decimal places.

END OF PAPER


99-AS-M&S16 15


Table:Area under the Standard Normal Curve

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.00.10.20.30.4

.0000

.0398

.0793

.1179

.1554

.0040

.0438

.0832

.1217

.1591

.0080

.0478

.0871

.1255

.1628

.0120

.0517

.0910

.1293

.1664

.0160

.0557

.0948

.1331

.1700

.0199

.0596

.0987

.1368

.1736

.0239

.0636

.1026

.1406

.1772

.0279

.0675

.1064

.1443

.1808

.0319

.0714

.1103

.1480

.1844

.0359

.0753

.1141

.1517

.1879

0.50.60.70.80.9

.1915

.2257

.2580

.2881

.3159

.1950

.2291

.2611

.2910

.3186

.1985

.2324

.2642

.2939

.3212

.2019

.2357

.2673

.2967

.3238

.2054

.2389

.2704

.2995

.3264

.2088

.2422

.2734

.3023

.3289

.2123

.2454

.2764

.3051

.3315

.2157

.2486

.2794

.3078

.3340

.2190

.2517

.2823

.3106

.3365

.2224

.2549

.2852

.3133

.3389

1.01.11.21.31.4

.3413

.3643

.3849

.4032

.4192

.3438

.3665

.3869

.4049

.4207

.3461

.3686

.3888

.4066

.4222

.3485

.3708

.3907

.4082

.4236

.3508

.3729

.3925

.4099

.4251

.3531

.3749

.3944

.4115

.4265

.3554

.3770

.3962

.4131

.4279

.3577

.3790

.3980

.4147

.4292

.3599

.3810

.3997

.4162

.4306

.3621

.3830

.4015

.4177

.4319

1.51.61.71.81.9

.4332

.4452

.4554

.4641

.4713

.4345

.4463

.4564

.4649

.4719

.4357

.4474

.4573

.4656

.4726

.4370

.4484

.4582

.4664

.4732

.4382

.4495

.4591

.4671

.4738

.4394

.4505

.4599

.4678

.4744

.4406

.4515

.4608

.4686

.4750

.4418

.4525

.4616

.4693

.4756

.4429

.4535

.4625

.4699

.4761

.4441

.4545

.4633

.4706

.4767

2.02.12.22.32.4

.4772

.4821

.4861

.4893

.4918

.4778

.4826

.4864

.4896

.4920

.4783

.4830

.4868

.4898

.4922

.4788

.4834

.4871

.4901

.4925

.4793

.4838

.4875

.4904

.4927

.4798

.4842

.4878

.4906

.4929

.4803

.4846

.4881

.4909

.4931

.4808

.4850

.4884

.4911

.4932

.4812

.4854

.4887

.4913

.4934

.4817

.4857

.4890

.4916

.4936

2.52.62.72.82.9

.4938

.4953

.4965

.4974

.4981

.4940

.4955

.4966

.4975

.4982

.4941

.4956

.4967

.4976

.4982

.4943

.4957

.4968

.4977

.4983

.4945

.4959

.4969

.4977

.4984

.4946

.4960

.4970

.4978

.4984

.4948

.4961

.4971

.4979

.4985

.4949

.4962

.4972

.4979

.4985

.4951

.4963

.4973

.4980

.4986

.4952

.4964

.4974

.4981

.4986

3.03.13.23.33.4

.4987

.4990

.4993

.4995

.4997

.4987

.4991

.4993

.4995

.4997

.4987

.4991

.4994

.4995

.4997

.4988

.4991

.4994

.4996

.4997

.4988

.4992

.4994

.4996

.4997

.4989

.4992

.4994

.4996

.4997

.4989

.4992

.4994

.4996

.4997

.4989

.4992

.4995

.4996

.4997

.4990

.4993

.4995

.4996

.4997

.4990

.4993

.4995

.4997

.4998

3.5 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998

Note: An entry in the table is the proportion of the area under the entire curve which is between z = 0and a positive value of z . Areas for negative values of z are obtained by symmetry.

z 0

A(z)xezA

zx

d21)(

02

2

= pi


4. (a) 8866 611

3

4

++= ttN

(b) 13912

5. (a) 0.2007

(b) 6

(c) 0.1282

6. (a) 160160

(b) 10, 11, 12, 13, 14, 15, 16.

(c) 12

7. (a) (i) 0.00045

(ii) 0.00001125

(iii) 0.001189

(b) 0.030

(c) 0.3785


1999

Section B

8. (a)

+= tttS

A 1204

47

3

1

9625

256 23

t

SA

d

d=

+ 1202

47

9625

256 2tt

= )152)(16(9625

128 tt

t

SA

d

d

0

t

SB

d

d

> 0 for all t N(t) is increasing

(ii) If )(N t = )N(100

1t

( )213000

bt

bt

ae

abe

+ =

btae

+

1

0003

100

1

)1100(

1

=

bae

bt

)]1100(ln[3.0

1= bat

t 24.2242

N( )]1100(ln[3.0

1ba ) =

)]}1100(ln[3.0

1{

1

0003

+bab

ae

= 2900

The greatest number of migrants found at Mai Po is 2900.

(iii) Suppose all the migrants leave Mai Po in x days.

Then ssx

d600 = 2900

x

s

0

2

3

40

= 2900

x 17.3870

The number of days in which we can see the migrants is24.2242 + 17.3870 42


10. Let X be the score on the questionnaire.

(a) (i) P(classify as non-PD | PD)= P(X < 75 | X ~ N(80, 52))

= )5

8075(P

Z

= )2(P >Z

0.5 0.4772= 0.0228

(b) The probability that out of 10 PDs, not more than 2 will be misclassified

82102910

110 )1587.01()1587.0()1587.01)(1587.0()1587.01( ++ CC

0.7971

(c) Let 0x be the required critical level of score.

P(X < 0x | X ~ N(80, 52)) = 0.01

)5

80(P 0

68.3665 | X ~ N(65, 52))

= )6733.0(P >Z

0.5 0.2496

= 0.2504

P(misclassified) (0.01)(0.1) + (0.2504)(0.9) 0.2264

If a teenager is classified by the criminologist, then

P(misclassified) (0.1587)(0.1) + (0.0228)(0.9) 0.0364

0.2264 > 0.0364 The probability of teenagers miscalssified by the sociologist is greater

than that by the criminologist.


11. (a) )(f x = 2)2()46()2(6

x

xx

+ = 2)2(

8x

> 0 for 2x

(b) =

= x

xx

xx 246lim)f(lim

22 and =

=++ x

xx

xx 246lim)f(lim

22

2=x is a vertical asymptote to 1C .

612

46lim)f(lim =

=

x

xx

xx

6=y is a horizontal asymptote to 1C .

(c)

)2g()2f( = and )1g()1f( =

+

=

+

=

be

ea

be

ea

12

14

3

2

0

=

=

41

63

be

ea


(d)

1C

2C

1C

x

y

0 22 4 6 8

5

10

15

5

10

x = 2

y = 6

(2, 4)

(1, 2)

2

2/3


(e) 41

1

6)g(

2

3

=

+

x

x

e

e

e

ex

)(g x =

++

x

xxxx

e

eeee

e

e2

22

3

)1(

1

6 = xe

e

e

1

63

)(g x > 0 and hence g(x) is (strictly) increasing for all values of x .

For x < 3 , f(x) > 6 but g(x) < 6 .For x > 8 , f(x) < 6 but g(x) > 6 .Thus 1C and 2C has no point of intersection

beyond the range 83 x .

(f) Area of the region bounded by 1C and 2C

= xxx d))f()(g(1

2 = x

x

x

e

e

e

ex

x

d2

464

1

1

61

2

2

3

+

= ( ) xx

xxeee

e x d2

86d4d

1

6 1

2

1

2

1

2

2

3

+

= [ ] [ ] [ ]1 21 21 223 )2(ln8216 +++ xxexee e x 12.94312254 + 6 11.09035489 7.8528


12. Let N be the number of complaints received on a given day andX be the number of complaints involving the time schedule.

(a)time schedule resolved

not resolved

manner of drivers resolved not resolved

routes resolvednot resolved

other things resolvednot resolved

P(manner of drivers | not resolved)

= 5.012.03.013.08.035.04.04.0

8.035.0

+++

0.5195

(b) (i) P(N = 5) = !5

10 105 e

0.0378

(ii) P(N = 5 and X = 3) = ( )2353105 )6.0()4.0(!5

10C

e

0.0087

(c) P(N = n and X = 9) =

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1999 Maths