2002 Mathematics

7/27/2019 2002 Mathematics

1/28

! ! !Hong Kong Examinations Authority

All Rights Reserved 2002

2002-AS-M & S1

HONG KONG EXAMINATIONS AUTHORITY

HONG KONG ADVANCED LEVEL EXAMINATION 2002

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(E) answer book.

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

4. Unless otherwise specified, all working must be clearly shown.

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

2002-ASL

M & S


2/28

2002-AS-M & S2 1 All Rights Reserved 2002

SECTION A (40 marks)

Answer ALL questions in this section.

Write your answers in the AL(E) answer book.

1. Let tet

x3

22

5 and tet

y2

2

10 ( 0zt ) . If 2

d

d

x

y, find the value

of t.(5 marks)

2. An adventurer estimates the volume of his hot air balloon by

SS 53

4)(V 3 rr , where r is measured in metres and V is measured in

cubic metres. When the balloon is being inflated, r will increase with time

0)(tt in such a way that

te

t

23

18)(r ,

where t is measured in hours.

(a) Find the rate of change of volume of the balloon at t 2 . Give youranswer correct to 2 decimal places.

(b) If the balloon is being inflated over a long period of time, what will thevolume of the balloon be? Give your answer correct to 2 decimalplaces.

(5 marks)

3. A researcher modelled the number of bacteria )N(t in a sample t hours afterthe beginning of his observation by ktat 900)(N , where )0(!a and k are

constants. He observed and recorded the following data:

t (in hours) 0.5 1.0 2.0 3.0)(N t 1100 1630 2010 2980

The researcher made one mistake when writing down the data for N(t).

Express ln N(t) as a linear function of t and use the graph paper on Page 2 todetermine which one of the data was incorrect, and estimate the value of N(2.5)correct to 3 significant figures.

(4 marks)


3/28

2002-AS-M & S

3

2

All Rights Reserved 2002

Go on to the next page

Page Total

3. (Continued) Fill in the details in the first three boxes above and tie this

sheet INSIDE your answer book.

Candidate Number Centre Number

Seat Number


4/28


This is a blank page.


5/28



4. An engineer conducts a test for a certain brand of air-purifier in a smoke-filledroom. The percentage of smoke in the room being removed by the air-purifier

is given by S . The engineer models the rate of change of Sby

3)103(

8100

d

d

t

t

t

S,

where 0)(tt is measured in hours from the start of the test.

(a) Using the substitution 103 tu or otherwise, find the percentage ofsmoke removed from the room in the first 10 hours.

(b) If the air-purifier operates indefinitely, what will the percentage ofsmoke removed from the room be?

(5 marks)

5. Twelve boys and ten girls in a class are divided into 3 groups as shown in thetable below:

Group A Group B Group C

Number of boys 6 4 2

Number of girls 2 3 5

To choose a student as the class representative, a group is selected at random,then a student is chosen at random from the selected group.

(a) Find the probability that a boy is chosen as the class representative.

(b) Suppose that a boy is chosen as the class representative. Find theprobability that the boy is from Group A.

(5 marks)

6. Assume that the number of passengers arriving at a bus stop per hour follows aPoisson distribution with mean 5 . The probability that a passenger arriving atthe bus stop is male is 0.65 .

(a) Find the probability that 4 passengers arrive at the bus stop in an hour.

(b) Find the probability that 4 passengers arrive at the bus stop in an hourand exactly 2 of them are male.

(5 marks)


6/28


7. Twenty two students in a class attended an examination. The stem-and-leafdiagram below shows the distribution of the examination marks of these

students.

Stem (tens) Leaf (units)3 5 74 2 4 65 0 3 4 4 4 56 1 2 5 5 87 3 8 98 4 89 5

(a) Find the mean of the examination marks.

(b) Two students left the class after the examination and their marks are

deleted from the stem-and-leaf diagram. The mean of the remainingmarks is then increased by 1.2 and there are two modes. Find the twodeleted marks.

(c) Two students are randomly selected from the remaining 20 students.Find the probability that their marks are both higher than 75 .

(5 marks)

8. A flower shop has 13 roses of which 2 are red, 5 are white and 6 are yellow.Mary selects 3 roses randomly and the colours are recorded.

(a) Denote the red rose selected by R , the white rose by W and the yellowrose by Y .

List the sample space (i.e. the set of all possible combinations of thecolours of roses selected, for example, 1R 2W denotes that 1 red roseand 2 white roses are selected).

(b) Find the probability that Mary selects exactly one red rose.

(c) Given that Mary has selected exactly one red rose, find the probabilitythat only one of the other two roses is white.

(6 marks)


7/28



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.

9. Lactic acid in large amounts is usually formed during vigorous physicalexercise, which leads to fatigue. The amount of lactic acid, M, in muscles ismeasured in m mol/L. A student modelled the rate of change of the amount oflactic acid in his muscles during vigorous physical exercise by

t

e

t

Mt

3

12

d

d 32

( 40 dd t ) ,

where t is the time measured in minutes from the start of the exercise.

(a) The student used the trapezoidal rule with 5 sub-intervals to estimatethe amount of lactic acid formedafter the first 2.5 minutes of exercise.

(i) Find his estimate.

(ii) Find

t

e

t

t

3

12

d

d 32

2

2

and hence determine whether his estimate

is an over-estimate or an under-estimate.(5 marks)

(b) The student re-estimated the amount of lactic acid formedby expanding

t

et

312 3

2

as a series in ascending powers of t.

(i) Expandt3

1and hence find the expansion of

t

et

312 3

2

in

ascending powers of t as far as the term in 3t .

(ii) By integrating the expansion oft

et

312 3

2

in (i), re-estimate the

amount of lactic acid formed after the first 2.5 minutes ofexercise.

(7 marks)

(c) The student wanted to predict the amount of lactic acid formed in hismuscles after the first 4 minutes of exercise. He decided to use themethod in (b) to estimate the amount of lactic acid formed. Brieflyexplain whether his method was valid.

(3 marks)


8/28


10. Let

1

)(f

cx

baxx and 3)1()3()(g xxx , where a , b and c are

constants. It is known that )0(g)0(f , )3(g)3(f and )2(g)2(f .

(a) (i) Find the values of a , b and c .

(ii) Find the horizontal and vertical asymptotes of the graph of

)(fx .

(iii) Sketch the graph of )(fx and its asymptotes. Indicate the

point(s) where the curve cuts the y-axis.(5 marks)

(b) (i) Find all relative extreme point(s) and point(s) of inflexion of

)(g x .

(ii) On the diagram drawn in (a)(iii), sketch the graph of )(g x .

Indicate all the relative extreme point(s) and the point(s) ofinflexion, the point(s) where the graph cuts the coordinates axes

and where it cuts the graph of )(fx .

(6 marks)

(c) Find the area enclosed by the graphs of )(fx and )(g x .

(4 marks)


9/28



11. A food store manager notices that the weekly sale is declining, so he starts apromotion plan to boost the weekly sale. He models the rate of change of

weekly sale G by

208

82

d

d2

tt

t

t

G( 0tt ) ,

where G is the weekly sale recordedat the end of the week in thousands ofdollars and t is the number of weeks elapsed since the start of the plan.Suppose that at the start of the plan ( i.e. t = 0 ) , the weekly sale is50 thousand dollars.

(a) (i) Express G in terms of t.

(ii) At the end of which week after the start of the plan will the

weekly sale be the same as at the start of the plan? (5 marks)

(b) (i) At the end of which week after the start of the plan will theweekly sale drop to the least?

(ii) Find the increase between the weekly sales of the 5th and the6th weeks after the start of the plan.

(iii) The store manager decides that once such increase of weeklysales between two consecutive weeks is less than 0.2 thousanddollars, he will terminate the promotion plan. At the end ofwhich week after the start of the plan will the plan be terminated?

(6 marks)

(c) Let 1t and 2t be the roots of 0d

d2

2

t

G, where 21 tt . Find 2t .

Briefly describe the behaviour of G andt

G

d

dimmediately before and

after 2t .

(4 marks)


10/28


12. Two researchers want to study the distribution of the number of car accidents ata certain road junction in a month. They have collected the data over 40

months as shown in the following table. They suggest that the distribution canbe modelled by a Poisson distribution.

Expected number of months *Number of caraccidents

Observed numberof months Researcher A Researcher B

0 12 12.99 12.05

1 15 14.61 14.46

2 9 8.22 b

3 4 a 3.47

* Correct to 2 decimal places.

(a) Researcher A uses the sample mean of the distribution as the mean ofthe Poisson distribution. Find the value of a in the table correct to 2decimal places.

(3 marks)

(b) Researcher B tries to fit the data by using a Poisson distribution withanother mean.

(i) Find the mean used by researcher B.

(ii) Find the value of b in the table correct to 2 decimal places.

(2 marks)

(c) The absolute values of the differences between observed and expectednumbers are regarded as errors. The distribution with a smaller totalsum of errors will fit the data better. Which Poisson distribution will fitthe data better ?

(5 marks)


11/28



12. (Continued)

(d) Assume the Poisson distribution that fits the data better in (c) is adoptedand 30% of car accidents involve a bus.

(i) Find the probability that the number of car accidents at the roadjunction in a month is 3 and only one of which involves a bus.

(ii) Find the probability that the number of car accidents at the roadjunction in a month is 3 and only the third car accident involvesa bus.

(iii) Given that the number of car accidents at the road junction in amonth is 3 and only one of which involves a bus, find theprobability that the third car accident involves a bus.

(5 marks)


12/28


13. The weight of each bag of self-raising flour in a batch produced by a factoryfollows a normal distribution with mean 400 g and standard deviation 10 g .

A bag of flour with weight less than 376 g is underweight, and more than424 g is overweight.

(a) Find the probability that a randomly selected bag of flour

(i) is underweight;

(ii) is overweight.(3 marks)

(b) If a bag of flour is either underweight or overweight, it will beclassified as a substandard bag by the director of the factory. Thedirector randomly selects 50 bags as a sample from the batch.

(i) Find the probability that there is no substandard bag of flour inthe sample.

(ii) Find the probability that there are no more than 2 substandardbags of flour in the sample.

(5 marks)

(c) A wholesaler is only concerned about the number of bags of flour whichare underweight. The wholesaler re-analyses the sample of 50 bags offlour in (b).

(i) Find the probability that in the sample there is only 1substandard bag and it is not underweight.

(ii) Find the probability that there are no more than 2 substandardbags in the sample and no underweight bag of flour in thesample.

(iii) Given that in the sample there are no more than 2 substandardbags, find the probability that there is no underweight bag in thesample.

(7 marks)


13/28


14. Suppose the number of customers visiting a supermarket per minute follows aPoisson distribution with mean 6 .

(a) Find the probability that the number of customers visiting thesupermarket in one minute is more than 2 .

(3 marks)

(b) Suppose the amount $X spent by a customer in the supermarket follows

a normal distribution N(P, 2V ) .

Probability distribution of the amount spent by a customer

Amount spent($X)

Probability *

X< 100 0.063

100 dX< 200 0.364200 dX< 300 1a

300 dX< 400 2a

Xt 400 0.006

* Correct to 3 decimal places.

(i) Using the probabilities provided in the above table, find the

values of P and V correct to 1 decimal place.

Hence find the values of 1a and 2a correct to 3 decimal

places.

(ii) What is the median of the normal distribution?

(iii) Given that a customer spends less than $200 , find theprobability that the customer spends more than $50 .

(iv) Find the probability that there are 5 customers visiting thesupermarket in a minute and exactly 2 of them each spends lessthan $200 .

(12 marks)

END OF PAPER


14/28


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

3.03.1

3.23.33.4

.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 ofz. Areas for negative values ofz are obtained by symmetry.

z0

A(z)

xezAz

x

d2

1)(

0

2

2

S


15/28


2002

Section A

1. 6ln5

1t

2. (a) The rate of change 173.35 m3/h

(b) The volume of the balloon will be 920.49 m3

.

3. 900ln)ln()(Nln takt

2440)5.2(N |

4. (a) The percentage of smoke removed is 25.3125% .

(b) 45% of smoke will be removed.

5. (a) The required probability 28

15

(b) The required probability 15

7

6. (a) The required probability is 0.1755 .

(b) The required probability is 0.0545 .

7. (a) Mean = 61

(b) The deleted marks are 54 and 44 .

(c) The required probability is191 .

8. (a) The sample space is

{ 1R 2W,1R 1W1Y, 1R 2Y, 2R 1W, 2R 1Y, 1W2Y, 2W1Y, 3W, 3Y} .

(b) The required probability is13

5.

(c) The required probability is11

6.


16/28


2002

Section B

9. (a) (i) t 0 0.5 1.0 1.5 2 2.5

t

M

d

d 4 4.78496 5.84320 7.24875 9.10480 11.55161

M = 55161.114[2

5.0d

3

125.2

0

3

2

| tte

t

)]1048.924875.78432.578496.4(2

17.3788 (m mol/L)

(ii) t

e

t

Mt

3

12

d

d 32

,

2

3

2

2

3

2

3

2

3

2

)3(

)23(4

)3(33

212

3

12

d

d

t

et

t

e

t

e

t

e

t

tttt

andt

t

et

tt

t

e

t

3

2

3

23

2

2

2

)3(3

)269(8

3

12

d

d

? 0d

d

d

d2

2

!

t

M

t(for 5.20 dd t )

So,t

M

d

dis concave upward on [0, 2.5] .

Hence it is over-estimate.


17/28


(b) (i)

t3

1)

27

1

9

1

3

11(

3

1 32 ttt

3281

1

27

1

9

1

3

1ttt

t

e 32

32 )3

2(

!3

1)

3

2(

!2

1

3

21 ttt

3281

4

9

2

3

21 ttt

t

et

312 3

2

)81

1

27

1

9

1

3

1(12

32 ttt )

81

4

9

2

3

21(

32 ttt

3281

4

9

4

3

44 ttt

(ii) tt

et

d3

125.2

0

3

2

| 5.2

0

32 d)81

4

9

4

3

44( tttt

5.2

0

432

81

1

27

4

3

24

tttt

16.9637 (m mol/L)

(c) The expansion is valid only when1

31 t

33 t

Hence 30 d t ( as tt 0 )

? this method is not valid to estimate the amount of lactic acid for tt 3 .


18/28


10. (a) (i) )0(g)0(f

b 3 (1)

)3(g)3(f

013

3

c

ba

03 ba (2)

)2(g)2(f

512

2 c

ba

5102 cba (3)

Using (1) and (2) ,

a1

Using (3) , c 1

(ii)1

3)(f

x

xx

11

1

31

lim)(flim

rforfo

x

xxxx

The horizontal asymptote is 01y

The vertical asymptote is 01x


19/28


(iii)

(b) (i) gc(x) 23 )1()3(3)1( xxx

)2()1(4 2 xx

gs(x) 2)1(4)2()1(8 xxx

)1()1(12 xx

gc(x) 0 x1 or 2gs(2) 36 < 0

? g(x) is maximum when x 2 and g(2) 27? the maximum point is at (2, 27) .

gs(x) 0 x1 or 1

x x < 1 x1 1 1

x 1 0 x 1 0 gs(x) 0 0

g(1) 0 , g(1) 16? (1, 0) and (1, 16) are points of inflexion of g(x) .

x

y

O01y

01x

3

3

10

20

30

)(f xy 1 2


20/28


(ii)

(c) Using the graphs, the area is

3

0

3d

1

3)1()3( x

x

xxx

-

3

0

24d

1

41)386( x

xxxx

3

0

24d

1

4486 x

xxxx

3

0

235)1(ln4442

5

1

xxxxx

4ln45267

1y ( 1 ,0)

y

1 2O

10

30

20

2

3

x

(1, 16)

(2, 27)

(2, 5)

(0, 3) (3, 0)

)(g xy )(f xy

01x


21/28


11. (a) (i)

t

tt

tG d

208

822

Ctt )208(ln 2

When 0t , G = 50 .20ln50C

20ln50)208(ln2 ttG

(ii) ForG = 50,

5020ln50)208(ln2 tt

202082 tt

082 tt t= 0 or t= 8 .

At the end of the 8th week, the weekly sale is the same as

at the start of the promotion plan.

(b) (i)]4)4[(

)4(2

208

82

d

d22

t

t

tt

t

t

G

0d

d?

t

Gwhen t= 4

Since 0d

d

t

Gwhen t< 4

and 0d

d!

t

Gwhen t> 4 ,

G is least at t= 4 .At the end of the 4th week, the weekly sale is least.

(ii) )5()6( GG )20ln505(ln)20ln508(ln

4700.05

8ln | (thousand dollars)


22/28


(iii) 2.0)()1( tGtG

}20ln50]20)1(8)1{ln[(2

tt 2.0}20ln50)208{ln(

2 tt

2.0208

136ln

2

2

tt

tt

0)1320()68()1(2.02.022.0 ! etete

t < 3.94316 or t > 13.09015

0d

d t

Gwhen 0 < t< 4 , G is decreasing

? 94316.3t is rejected. t= 14 .

Thus the promotion plan will be terminated at the end of the 15thweek.

(c)208

82

d

d2

tt

t

t

G

22

2

2

2

)208(

)82)(82()208(2

d

d

tt

tttt

t

G

22 )208(

)6)(2(2

tt

tt

0d

d2

2

t

Gwhen t= 2 or t= 6 . ? 62 t

Although G keeps increasing,

tG

dd

increases immediately before t 6 ,

tG

dd

decreases immediately after t 6 .


23/28


12. (a) Let 1O be the sample mean of car accidents at the road junction in a month.

1O 40 4392151120 uuuu 1.125

Let X be the number of car accidents at the road junction in a month.

For researcher A ,

)3(P40 Xa

125.13

!3

125.140 u e

08.3|

(b) For researcher B , let the mean be 2O . Then

(i) )0(P4005.12 X

24005.12O e

2O 4005.12ln

| 1.1998

(ii) )6732.8(!2

1998.140)2(40

1998.12

|u| eXPb | 8.67

(c) For the number of car accidents is 4 or more, the expected number of monthsobserved by researcher A is

10.1)08.322.861.1499.12(40 |

Let

TSE1 Total sum of errors for model fitted by researcher A

1ff0 E 10.1008.3422.8961.141599.1212

| 4.18For the number of car accidents is 4 or more, the expected number of monthsobserved by researcher B is

35.1)47.367.846.1405.12(40 |

TSE2 Total sum of errors for model fitted by researcher B

2ff0 E | 35.1047.3467.8946.141505.1212

2.8

As TSE2 < TSE1 , researcher B fits the data of car accidents better than

researcher A does.


24/28


(d) (i) P(the number of car accidents at the road junction in a month is 3

and one of which involves a bus)

P(X 3 and one of which involves a bus) P(one accident involves a bus |X 3) P(X 3)

2!3

)3.01(3.03223

1OO uu eC

1998.13

2

!3

1998.17.03.03 uuu e (| 0.038254)

| 0.0382

(ii) P(the number of car acidents at the road junction in a month is 3

and only the third car accident involves a bus)

P(X 3 and only the third car accident involves a bus)

=3

1P(X 3 and one of which involves a bus)

| 0.0127

(iii) P(X 3 and the third car accident involves a bus |X 3 andonly one of which involves a bus)

3

1.


25/28


13. Let Xg be the weight of a bag of self raising flour in the batch.

(a) (i) P(a bag of flour is underweight) P(X< 376)

)10

400376

10

400(P

X

)4.2(P Z

| 0.0082

(ii) P(a bag of flour is overweight) P(X> 424)

)10

400424

10

400(P

!X

)4.2(P !Z

| 0.0082

(b) (i) P(a bag of flour is substandard)

)424(P)376(P ! XX

| 0.0082 0.0082 0.0164

Let Y be the number of substandard bags in the sample.

P(there is no substandard bags in the sample) P(Y 0)

500500 )0164.01(0164.0 uC

509836.0 | 0.4374

(ii) P(Yd 2) P(Y 0) P(Y1) P(Y2)

4950150050

0 9836.00164.09836.00164.0 uu CC 48250

2 9836.00164.0 uC

| 0.43745 0.36469 0.14897| 0.9511


26/28


(c) Let W be the number of underweight bags in the sample.

(i) P(W 0, Y1) )1(P)1|0(P YYW

49501 )9836.0()0164.0(2

1Cu

| 0.1823

(ii) The required probability is P(W 0, Yd 2) )2,0(P)1,0(P)0,0(P YWYWYW

)2()2|0()1,0()0( YPYWPYWPYP

| 482502

2

)9836.0()0164.0(2

118235.043745.0 C

| 0.6570

(iii) The required probability is P(W 0 | Yd 2)

)2(P

)2,0(P

d

d

Y

YW

|95111.0

65704.0

| 0.6908


27/28


14. (a) Let N be the number of customers visiting the supermarket in one minute.

)2P( dN 62

0!

6

ek

k

k

62

66

!2

6

!1

6 eee

| 0.002479 0.01487 0.04462| 0.0620

? )2P( !N )2(P1 d N | 0.9380

(b) (i) X~ ),(N 2VP

P(X< 100) 0.063

063.0)100

(P

V

PZ

53.1100

|

V

P.. (1)

P(Xt 400) 0.006

006.0)400

(

tV

PZp

51.2400

|

V

P.. (2)

Solving (1) and (2), we get

P | 213.6

V | 74.26 | 74.3

)300200(P1 d Xa

)3.74

6.213200()

3.74

6.213300(

ZPZp

| 0.4484| 0.448

)400300(P2 d Xa

| 0.117


28/28


(ii) For normal distribution,

median mean 213.6

(iii) P(X> 50 |Xd 200)

)200100(P)100(P

)20050(P

dd

XX

X

364.0063.0

)18.020.2(P

d Z

|427.0

0714.04861.0

| 0.9712

(iv) The required probability

)5(P))200(P1()200(P 3252 NXXC

65

32

!56))364.0063.0(1()364.0063.0(10 e

| 10(0.1823)(0.1881) (0.1606)| 0.0551

Documents

2002 Mathematics