ACJC 2013 JC2 H2 Mathematics REVISION SET G

COMPLETE SOLUTIONS: STATISTICS Part I

PERMUTATIONS AND COMBINATIONS

1.

(i) 420

!3!2

!7 (ii) 120

!2

!43

5 C (iii) 4

23! P 72

2.

3.

(i) (4−1)! ×5! = 720 (ii) (5 – 1)! × 5

3C × 3! = 1440 (iii) 8! = 40320

No. of different selections = 102 1 =1023

4.

(i) Arrangement in a row: 9C4 x 4! = 3,024

Arrangement in a circle: 5C5 x (51)! = 24

No. of ways = 3,024 24 = 72,576

(ii) Case : John & Mary beside each other in a row

No. of ways = (7C2 3! 2!) (

5C5 (51)!) = 6,048

Case : John & Mary beside each other in a circle

No. of ways = ((41)! 7C3 2!) (4!) = 10,080

Total no. of ways = 16,128

(iii) Arrangement in a circle: 6C2 (31)!

3C2 2! = 180

Arrangement in a row: 4! = 24

Total no. of ways = 180 24 = 4,320

5. (a) Case 1: 4 and 5 in the second and fourth positions 2 3! 12

Case 2: 3 and 5 in the second and fourth positions 2 2 4

Total number 12 4 16

(b) Number of ways 4 3 4 3 4 3

. . .3! 2041 3 2 2 3 1

Alternative: 7

4.3! 3! 204C

6.

(i) Number of ways required = 104

= 100 00

(ii) Number of ways for last digit = 5

Number of ways required = 9 8 7 6 5 5 = 75600

(iii) Number of ways for 3 odd digits = 3

5C = 10

Number of ways for 3 even digits = 3

5C = 10

Number of ways required = 10 10 6! = 72000

7. (a)

(b) (i)

(ii)

(iii)

55!

Number of ways 2! 10 76805

4Number of ways 6 1296

6

4Number of ways 4! 360C

Alternative solution: Number of ways 6 5 4 3 360

6 2

2 1

4! 4!Number of ways 210

3! 2!2!C C

8.

(i) Number of ways

15 15 150 12 C C 32768 1 15 32752

OR

1515 15 15 15 15

2 3 4 15

2

32752r

r

C C C C C

(ii)(a) Number of ways 9! 362880 OR 39 6 3

3 3 3 3! 362880C C C

(ii)(b) Number of ways 6

12! 7 6048! 0C

OR 26 7 6 3

1 1 3 32! 63! 0480C C C C

(iii) Number of ways = (6 1)! 6 5 4 = 14400

OR Number of ways 3 62 29 1 ! 7 1 ! 3! 6 1 4! 14 00P P

PROBABILITY

1 (a)

(b)

(i) ( ) 1P A B and ( ) ( ) ( ) ( )P A B P A P B P A B 0.7 0.8 1 = 0.5

(ii) ( ' ) ( ) ( ) 1P A B P A P A B (Use Venn Diagram)

( ) 0.85 0.7 1 0.55P A B

( ) 0.55 11

( ) 0.6875( ) 0.8 16

P A BP A B

P B

A & B are not independent as P(AB) P(A) (or ( ) ( ). ( )P A B P A P B )

(iii) If ( ) 0.5P A B , then ( ) 1P A B , event A & event B are exhaustive.

[Note: Exhaustive events are events whose total probabilities sum up to 1.]

Total no. of points = 1 2 3 ... 9 10 55 OR 11

2 55C

Prob. = 11

1

10 9 8 80.0503

55 54 53 159C OR

1011 3

1 55

3

8

159

CC

C

2 (i)

1 1 1P Abbey is first and Betty is sixth in the queue =

8 7 56

6! 1 1 6 5 4 3 1 1

Or = Or = 8! 56 8 7 6 5 4 3 56

(ii) Required probability

=P(Abbey is first) + P(Betty is second) – P(Abbey is first and Betty second)

1 7 1 1 1 13 7! 7! 6! 13

= + = or + = 8 8 7 8 7 56 8! 8! 8! 56

Alternatively,

Required probability

= P(Abbey is first but Betty is not second) + P(Abbey is not first but Betty is second)

+ P(Abbey is first and Betty second) 6 6! 6 6! 6! 13

8! 8! 8! 56

3 P( | ) 0.4B A'

P( )0.4

P( )

A' B

A'

P( ) 0.4 P( ) 0.4 0.3 0.12A' B A'

P( ) P( ) P( )B A' B A B

1 0.6 0.12 P( )A B P( ) 0.28A B

P( ) P( )A B 0.7 0.4 0.28

P( )A B

A and B are independent events.

Let x be P( )B' C . C does not overlap A. Therefore C is drawn as shown below.

Probability is not negative, 0x .

Since total probability cannot exceed 1,

0.82 1x 0.18x

Combining the inequalities, we have 0 0.18x . 0.1 P( ) 0.28C

4 (i)

2 2 2 3! 2 2 2

P club wins more than one match = 13 3 3 2! 3 3 3

20 or 0.741

27

Alternative solution

Let X be the number of matches, out of 3, that Champion Football Club wins.

X ~ B(3, 2/3)

( 1) 1 ( 1) 0.741 (3 s.f.)P X P X

A B

C

0.42 0.28 0.1

(ii)

(iii)

2 1 2

P club draws =13 5 15

3

P club obtains exactly 3 points =P(1 win and 2 losses) P(3 draws)

2 1 1 3! 2 278= or 0.0824

3 5 5 2! 15 3375

rd

1 1 2

5 5 3P club wins only 3 match | obtains exactly 3 points =278

3375

45 or 0.324

139

5 60.036.025.054.017.072.063.091.020.075.0 a

605.005445.009.0 aa

Prob required er)Prob(Offic

Officer) andNPFA Prob(Fail

0.60

36.0605.025.0 0908.0 (3 s.f.)

Prob required 909.00908.01 (3 s.f.)

Prob required officer)-Prob(Non

officer)-Non andNPFA Prob(Fail

0.601

64.0605.025.0395.025.0

489.0 (3 s.f.)

6 (i) P(A M ) =

200 1

400 2

(ii) P(M ' C ') 9 1

,20 2

P A P A M P A

A and M are not independent.

(i) No. of immigrants in the sample

= 0.2 200 250 0.3 130 300 0.05 120 225

P(voter supports Party A given voter is an immigrant) 0.2 450

0.4225

(ii) Number of immigrants supporting Party C = 0.05 120 = 6

P(exactly one immigrant voter supporting Party C or exactly one female voter

supporting Party A (or both))

exactly 1 immigrant voted for C exactly 1 female voted for A bothP P P

6 994 250 750 250 6 744

1 2 1 2 1 1 1

1000

3

0.434C C C C C C C

C

Alternatively, Required Probability

=6 994 993 250 750 749 6 250 744

3 3 3!1000 999 998 1000 999 998 1000 999 998

= 0.434

7 P(all are greater than 3) =

2 3 2. .

6 4 3 =

1

6

Let X be the event : “each of the three numbers is greater than 3”

Y be the event : “sum of the three numbers is equal to 13”

P(Y) = P((2,7,4), (5,4,4)) = 3 1 1 1 2 1

. . . .6 4 3 6 4 3

= 5

72

Required probability = P( )X Y = P( ) P( ) P( )X Y X Y

= 1 5 1 2 1

. .6 72 6 4 3 =

5

24

P(a player wins a particular game) =

22 4 2 4 2

. . ......13 13 13 13 13

=

22 4 4

1 ......13 13 13

= 4

13

2 1.

13 1

= 2

9

8 (a) P(Chris is late in a day) =

4

5+

2

5 (

4

5)(

2

5) =

22

25

Let X be the number of days Chris is late out of 4 days. X~B(4, 22

25).

P(Chris was late exactly thrice in a week | Chris was late on Mon)

= P(Chris was late twice in the remaining four days of the week)

= P(X = 2) 0.0669

(b) P(Chris was delayed at A | Chris was late)

= P(Chris was delayed at A)/P(Chris was late)

4

10522 11

25

(c) P(Chris was delayed at B | he was delayed at exactly one stop)

= P(Chris was delayed at stop B only)/P(delayed at exactly one stop)

=

1 2

15 54 3 1 2 7

5 5 5 5

( )( )

( )( ) ( )( )

9

7 sopranos, 6 altos, 3 tenors, 4 basses need 3 members

(i) P(all 3 are tenors) = 3

320

3

1

1140

C

C = 0.000 877 (3 sf)

(ii) P(exactly 1 bass) = 4 16

1 220

3

8

19

C C

C

= 0.421 (3 sf)

9 (iii) P(2 women | exactly 1 bass)

= P(2 women and exactly 1 bass) ÷ P(exactly 1 bass)

=

13 4 202 1 3

4 16 201 2 3

312 130.65

480 20

C C C

C C C

exactly

(iv) P(exactly 1 tenor or exactly 1 bass or both)

= P(exactly 1 tenor) + P(exactly 1 bass) P(1 tenor and 1 bass)

= 3 17 4 16 3 4 13

1 2 1 2 1 1 120

3

732 61

1140 95

C C C C C C C

C

= 0.642 (3 sf)

10 (i) P(Amanda and Beryl first) =

1 1 1

7 6 42

(ii) P(Amanda first or Beryl last, or both)

= P(Amanda first) + P(Beryl last) P(Amanda first and Beryl last)

= 1 1 1 1 11

7 7 7 6 42

(iii) P(no two girls stand next to each other) = P(girls and boys alternate)

= 4 3 3 2 2 1 1 1

7 6 5 4 3 2 1 35 G B G B G B G

(iv) P(all four girls stand next to each other) = 4 3 2 1 3 2 1 4

47 6 5 4 3 2 1 35

[ GGGG ] BBB or B [ GGGG ] BB or … 4 cases

(v) P(all 4 girls stand next to each other | ≥ 2 girls stand next to each other)

P(all 4 girls stand next to each other)

P( 2 girls stand next to each other)

P(all 4 girls stand next to each other)

1 P(no 2 girls stand next to each other)

4

4 2351 34 17

135

11 (i) P(4 different pictures) =

10 9 8 7 63

10 10 10 10 125 or 0.504 exactly

(ii) P(exactly 3 different pictures) = 10 9 8 1 54

610 10 10 10 125

or 0.432 exactly

Note: 6 ways of arranging pictures A A - -

(iii) P(picture has X or Y or both) = 1 P(no X and no Y at all)

= 8 8 8 8 369

110 10 10 10 625

= 0.5904 exactly = 0.590 (3 sf)

11(iv) Given P(at most n cards are needed to complete the set) > 0.99, we have

P(at most n cards are needed to complete the set)

= P(need 1 more card) + P(need 2 more) + P(need 3 more) + … + P(need n more)

=

2 3 11 9 1 9 1 9 1 9 1

...10 10 10 10 10 10 10 10 10

n

=

1 91

10 10

91

10

n

= 9

110

n

> 0.99 0.9n < 0.01

n > (lg0.01) ÷ (lg0.9) n > 43.7

Therefore, least n is 44.

BINOMIAL and POISSON DISTRIBUTIONS

1 Let X be the number of students out of n who use AIKON brand mobile phones.

When n = 30, 1

~ 30,5

X B

Expected number of students who use AIKON brand phone = 1

30 65

P 5 0.172X

P 1 0.99 1 P 0 0.99

4P 0 0.01 0.01

5

lg 0.0120.6

4lg

5

n

X X

X

n

[alternatively, use GC Y1= binompdf()]

Hence least n = 21.

2 (i)

(ii)

Required probability =1 P( all seeds selected do not germinate)

16 15 14 13 12 2321

20 19 18 17 16 323

Let X be the number of seeds that germinate, out of 5. X ~ B(5, 0.10)

Required probability P( 1) 1 P( 1)X X = 1 0.91854 = 0.0815

Binomial distribution is used in part (ii) based on the assumption that the probability

of success is a constant when the sample size is small as compared to a large

population.

3 (a)

(b)

(i) Let X denote the no of patients out of 20 who will not recover. (20,0.02)X B

( 2) 1 ( 2) 0.00707P X P X

(ii) Let Y denote the no of sample out of 50 that has more than half patient who

will not recover. 0.392

N(0.4, )50

Y approximately by CLT

( 0.5) 0.129P Y

(i) Let Y denote the no of patients out of 50 who will recover. (50, )Y B p

( 48) 0.64P Y

( 48) 0.36P Y

From GC, p = 0.97481 = 0.975 (3 sig fig)

(ii) Let Y denote the no of patients out of 50 who will recover. (50,0.97481)Y B

Let Y’ denote the no of patients out of 50 who will not recover.

' (50,0.02519)Y B

Since n > 50, p < 0.1, np = 1.2595 < 5, ' (1.2595)oY P approximately

So ( 46) ( ' 5)P Y P Y 1 ( ' 4)P Y = 0.00940

4 (i) Let X be the number of flaws in a roll of a particular design of wallpaper.

Then Po(0.15)X .

Required probability 2P( 1)P( 1)X X 2P( 1) 1 P( 1)X X

0.00263 (3 s.f.)

(ii) Since 50n is large, by Central Limit Theorem, 0.15

N 0.15,50

X

approx.

ie. N 0.15, 0.003X approximately.

P( 0.3) 0.00309X (3 s.f.)

5 (i)

(ii)

(iii)

(iv)

1. Each of the students is equally likely to answer the question correctly (i.e.

constant p throughout all trials)

2. Whether a student answers the question correctly or not is independent of the

other students doing so.

Let X be the random variable “no of students out of 30 students who could do the

Differential Equation question”. X ~ (30,0.3)B

P(X 6) =1 P(X 5) =0.92341 0.923

Let S be the r.v. “no of students out of 8 who could do that question”. S~B(8, 0.3)

Let T be the r.v “no of students out of 22 who could do that question”. T~B(22, 0.3)

P(S=2)P(T 4)P(only 2 among first 8 could do that question| X 6) = 0.299

P(X 6)

Let Y be the r.v. “no of students out of n who could do that question”. ~ ( ,0.3)Y B n

P(Y 5) >0.9

From G.C,

Therefore the largest possible value of n is 11.

5 Since sample size = 50 is large,

6.3~ (9, )

50X N approx by Central Limit Theorem

P( 10) 0.00242X

6 ~ B( , )X n p

45

npq np 45

q and 15

p

P( 1) 0.92X

1 P( 0) 0.92X

P( 0) 0.08X

45

0.08n OR

11

12

0.8 0.0859 0.08

0.8 0.0687 0.08

11.3n

Least value of n is 12

13

~ B(8, )X so 83

( )E X and 169

( )Var X

By CLT, 8 163 9

~ ( 60, 60)S N or 3203

~ (160, )S N

( 162) 0.423P S (3 s.f)

7 Let X be the number of requests for cars on a particular day. ~ Po(4)X

Let Y be the number of requests for vans on a particular day. ~ Po(2)Y

(i) Let T be the number of requests for vehicles on a particular day. ~ Po(6)T

prob.req d ( 11) 1 ( 11) 0.0201P T P T

(ii) Either demand for a car or a van is not met. Thus

prob.req d ( 7 or 4) 1 ( 7 and 4)

1 ( 7) ( 4) 0.101

P X Y P X Y

P X P Y

(iii) The event in (i) is a subset of the event in (ii). Thus the value obtained in (i)

will be smaller.

(iv) Let n be the number of days needed. Assume that n is large. By Central Limit

Theorem, 6

~ N 6, approx.Tn

6

7) 0.001

7 60.001

0.9996

(

n

T

P Z

P

P

nZ

Thus 3.09023 57.36

nn

Thus least number of days required is 58.

8 i Let X = “no of print jobs sent to the colour printer” and Y = “no of print jobs sent to

the laser printer”

For a day , X + Y ~ Po(1.2 ) (independence)

( 0) 0.01P X Y

(1.2 ) 0.01e

1.2 4.605170186 3.41

8 ii For a day , X + Y ~ Po(4.61)

( 3) 0.163 (3 . .)P X Y s f

8 iii In 1 hr, let X ~ Po(0.15) , Y ~ Po(0.42625) , X + Y ~Po (0.57625)

In 7 hrs, let A ~ Po(1.05) , B ~ Po(2.98375) , A + B ~ Po (4.03375)

Prob req =

(Prob of 2 jobs in 1st hr, 1 job in next 7 hrs) +(Prob of 3 jobs in 1st hr, 0 job in next 7 hrs)

Prob of 3 jobs in a day

( 2) ( 1) ( 3) ( 0)

0.16250022

P X Y P A B P X Y P A B

0.09331032 0.07142884 0.01792335 0.0177078

0.16250022

0.0430 (3 . .)s f

9 (i) Let W be the random variable no. of air bubbles in 1 randomly chosen plastic

sheet. Then W ~ Po(1

2 )

P(W 3) = 1 − P(W 2) = 0.0143877 ≈ 0.0144 (to 3 s.f.) (shown)

(ii) Let V be the random variable no. of air bubbles in 5 randomly chosen plastic

sheets. Then V ~ Po(1

2 5) = Po(2.5)

Using GC, P(V = 1) = 0.2052

P(V = 2) = 0.2565 (highest probability)

P(V = 3) = 0.2138

Hence the most likely number of air bubbles is 2.

(iii) Let X be the random variable no. of plastic sheets with at least 3 air bubbles out

of 15 plastic sheets. Then X ~ B(15, 0.0144)

P(X 2) = 1 − P(X 1) = 0.0192244 ≈ 0.0192

(iv) Let Y be the random variable no. of rejected crates out of 100 crates.

Then Y ~ B(100, 0.0192244)

Since n = 100 > 50 and np < 5, Y ~ Po(1.92244) approximately

P(Y < 2) = P(Y 1) = 0.42741≈ 0.427

10 (i) Let X be the number of arrivals at the airport in a two-hour period. ~ Po 8X

P 13 1 P 12 0.063797 0.0638X X

10 (ii)

(iii)

(iv)

Let W be the number of arrivals in a one-hour period.

Let Y be the number of departures in a one-hour period.

~ Po 4W ; ~ Po 3Y ; ~ Po 7W Y

P 2 9P 2 | 9

P 9

P 0, 9 P 1, 8 P 0 P 9 P 1 P 8

P 9 P 9

0.0051052524 0.0503453384 0.0503 (ans)

0.1014046695

Y W YY W Y

W Y

Y W Y W Y W Y W

W Y W Y

(1) There are two mutually exclusive outcomes, either there are at least 13 arrivals

in each two-hour period or there aren’t.

(2) The probability of having at least 13 arrivals for each two hour period remains

constant for each of the 60 two-hour periods.

(3) There is a fixed number of 60 two-hour periods independently selected under

consideration.

Let V be the number of two-hour periods, out of 60, with at least 13 arrivals each.

~ B 60,P 13V X

Since n = 60 ( > 50, large) and P 13 0.0638p X ( < 0.1, small), such that np

= 3.828 ( < 5), we have ~ Po 3.828V approximately.

At most 50 two-hour periods with less than 13 arrivals each means the same as at

least 10 two-hour periods with at least 13 arrivals each.

P 10 1 P 9V V 0.0060899731 = 0.00609

Note:

Students may choose to use the more accurate value for P 13 0.0637971966X .

If they do so, the following values will be obtained:

np = 3.827831796 and ~ Po 3.827831796V

P 10V 0.0060881936 = 0.00609.

11 (i)

(ii)

(iii)

The average number of incoming calls received per hour is constant throughout the

opening hours of the mall.

OR

The probability of 2 or more incoming calls received in a very short interval of time

is negligible.

Let X be the number of incoming calls received in an hour. ~ Po 6.75X .

P 8 1 P 7 0.364 (3sf)X X

Required probability is 2 4!

P 6 P 7 P 82!

X X X

2 4!

0.48759 0.14832 0.36409 0.115 (3sf)2!

11(iv)

(v)

~ Po 81Y . So 81 , 9 .

Since 81 10 , 2~ N 81, 9Y approximately.

P 81 9 81 9 P 72 90

P 72.5 89.5 by continuity correction

0.6551 (4dp)

Y Y

Y

Let W be the number of busy days in 14 days.

~ B 14, P 90 ,that is, ~ B 14, 0.14593 .W Y W

Required probability is

P 2 P 90 0.29191 0.14593W Y 0.0426 (3sf)

12 (i) Let X be the number of employees, out of n , with good performance.

, 0.26X B n

( 1) 0.96P X 1 ( 0) 0.96P X

( 0) 0.04P X

0 0

0.26 1 0.26 0.040

nn

0.74 0.04n

ln 0.04

ln 0.74n

10.6902n Thus greatest 10n

(ii) Let Y be the number of employees, out of 120, with excellent performance.

120, 0.04 4.8Y B Po since n is large and np < 5.

( 7) 1 ( 7) 0.113P Y P Y

(iii) Let W be the number of employees, out of 20, with average or poor

performance. 20, 0.7W B

12 17P W W

12 17

17

P W W

P W

12 16

16

P W

P W

16 11

16

P W P W

P W

0.8929131955 0.1133314628

0.8929131955

0.873

13 (i) Let X be the number of calls made in a one-hour period. )9(~ PoX

( 9) 1 ( 9) 0.413P X P X (to 3 sf) (Shown)

Let Y be the number of periods where more than 9 calls are made out of k periods.

)413.0 ,(~ kBY

~ (0.413 , 0.2424 )Y N k k approximately

Given that )20( YP >0.9

1.0)5.19(

9.0)5.19(

YP

YP

19.5 0.4130.1

0.2424

19.5 0.4131.282

0.2424

19.5 0.413 1.282 0.2424

kP Z

k

k

k

k k

47.2 1.53k k

13(ii) Let T be the total number of calls made in one day. )90(~ PoT

Since 10 , )90,90(~ NT approximately.

)5.995.89()10090( TPTP after continuity correction.

= 0.363 (to 3 sf)

14 Let X be the number of students who make enquiries at the Police Force's booth (out

of 25). 1

(25, )6

X B

( 10) ( 9) 0.99526P X P X

(i) Method 1

Since 60n is large, by Central Limit Theorem,

1 525

1 6 6N 25 ,

6 60X

approximately.

i.e. 25 125

N ,6 2160

X

4 6 0.756P X (3s.f.)

Method 2

Let 1 2 60

1... (1500, )

6T X X X B

(4 6) (240 360)P X P T ( 360) ( 239)P T P T

0.76541 0.765

(ii) Let Y be the number of classes (out of 60) having 10 or more students

making enquiries at the Police Force's booth.

(60,0.0047426)Y B

Since 60n large, 0.28455 5np , 0(0.28455)Y P approximately

Required Prob ( 2)P Y 0.30459 0.305

Quota sampling.

Disadvantage:

[1] Sample may be biased as the interviewers are allowed to select students who are

more approachable to fulfill the quota required.

[2] Sample may not be representative of the student cohort as the male to female ratio

may not be 2:3 as stated in the sample.

[3] Quota Sampling method is not random and as a result the sample may be biased

as interviewers are allowed to select the students in any manner to fulfill the quota.

15(a)

(i)

Let X denote the random variable for the number of demands per hour for a court in

this sports hall on a weekend. Then Po(7.2)X

P(courts are fully booked on a particular time slot on a Saturday)

P( 6) 1 P( 5) 0.7241025 0.724(shown)X X

15

(a)

(ii)

Let Y denote the random variable for the number of hours on an entire weekend for

which the courts are fully booked. Then B(30, 0.7241025).Y

Since 30n is large, 21.723 5,np (1 ) 8.2769 5,n p

N(21.723, 5.9933)Y approximately.

P(the courts are fully booked for at least 20 randomly chosen hours on both Saturday

and Sunday of a particular week)

P( 20) P( 19.5) (by Continuity correction)

0.81807 0.818 (to 3 significant figur

Y Y

es)

15(b)

By plotting 1Y poissonpdf (21.6, )x in GC and using the “Table” function (as below)

From the GC, most probable value is 21.

15(c) Let X denote the average number of demands for a court between 0700 and 0800

per Sunday for 52 randomly chosen Sundays.

Since n = 52 is large, by Central Limit Theorem, 7.2

7.2, approx52

X N

.

7 0.2954667 0.295P X

Alternative method:

1 2 52 374.4X X X Po

Since 374.4 10 , 1 2 52 374.4, 374.4 approxX X X N

1 2 52

1 2 52

1 2 52

P 7 P 752

P 364

P 364.5 0.304450 0.304

X X XX

X X X

X X X

15(d)

(i)

The probability that one of the courts is booked will be affected by the event that

another court is booked. Hence the trials do not occur independently (the probability

of a court being booked is not consistent) and so a binomial model would probably

not be valid.

15(d)

(ii)

As people have to work during weekdays, the average number of demands will be

fewer on the weekdays. Hence the mean on the weekday is different from that on a

weekend.

16 pnX ,B~

11

1

11

, 0, 1, 2, ..., 1.

1

!

1 ! 1 !

!(1 )

! !

!( )! ( )(shown)

( 1)!( 1)! 1 ( 1)(1 )

n kk

k

n kkk

np p

kpk n

npp p

k

np

k n k

np

k n k

k n k p n k p

k n k p k p

When n = 10 and 1

3p , if 1k kp p , then 1 1k

k

p

p

,

110

83 1 10 2( 1) 3 82 3

13

k

k k k k

k

Thus k = 3, 4, …., 9. So 3 4 10...p p p

Conversely, if 1k kp p , then 8

3k .

Thus k = 0, 1, 2 . So, 0 1 2 3p p p p .

Since 3p is the greatest, therefore the most probable number of successes is 3.

(b)

(i)

Let X be the no. of adults, out of 8, having some knowledge of a foreign language.

B(8,0.3)X

P( 2) 0.552X (3 s.f.)

(ii) Let Y be the no. of adults, out of 400, having some knowledge of a foreign language.

B(400, 0.3)Y

Since n = 400 is large, 120 5np and 280 5nq ,

~ N(120 ,280 0.3) i.e ~ N(120, 84) approximately.Y Y

P( ) 0.9Y n P( 0.5) 0.9Y n using continuity correction

From GC, when 132, P( 0.5) 0.89522 0.9n Y n

when 133, P( 0.5) 0.91369 0.9n Y n

Least value of n =133

Let T be the no. of adults, out of 400, having some knowledge of the particular

foreign language. B(400, 0.01)T

Since n = 400 is large and 4 5np , ~ Po(4) approximately.T

P ( 4)T = 0.629 (3 s.f.)

NORMAL DISTRIBUTION

1 2~ (45000,2000 )A N 2~ (30000,1850 )B N

(i) 1 22 ~ (30000,22845000)A B B N

1 22 25000 0.85224 0.852P A B B (3 sig fig)

Assumption: The distributions of the lifespans of all televisions are independent of each

other.

(ii) Let W denote the number of plasma televisions out of 50 with a life span of

more than 30000 hours. ~ (50,0.5)W B

(14 22) (15 21)P W P W ( 21) ( 14)P W P W 0.160

(iii) 22000

~ 45000,X Nn

( 46500) 0.99P X 46500 45000

0.992000

P Z

n

3

0.994

nP Z

3

2.326354

n

9.6212n the least value of n is 10

(iv) (I) will be greater as A > 25000 and B > 25000 is a subset of A+B > 50000.

2

2

P( 3) 0.2 P( 3) 0.8

2.56~N(2.56, ),

0.44 0.44P 0.8 0.8416212335 0.5228 0.523

X X

XX Z

Z

)0.54663968 N(5.12,~21 XXW

P( 4 6X ) = 0.818

20.5228~ N 2.56,

5

P( 3) 0.029923

X

X

Expected No. = 0.029923 100 = 2.99

972.0)2.4P(

72)0.03656558 N(4.564,~5

)N(5.9,0.35~

32121

2

S

YYYXXS

Y

3 C ~ N(260, 102) G ~ N( ,

2)

C + G ~ N(260 + , 102 +

2) and C – G ~ N(260 – , 10

2 +

2)

075396.0)290(P GC 924604.0)290(P GC

924604.0100

260290P

2

Z 4367.1

100

30

2

301004367.1 2 --- (1)

99168.0)270(P GC

99168.0100

260270P

2

Z 3946.2

100

10

2

101003946.2 2 --- (2)

Solving (1) & (2), =15 and 2100 = 10.440 3

Let B be the random variable denoting the weight of the contents of a packet of biscuit

type B. B ~ N(180, 152)

~354321 CBBBBB N( 5(180) – 3(260) , 5(152) + 9(10

2) ) = N(120, 2025)

103P 54321 CBBBBB = 1 – 103B10P 54321 CBBBB = 0.995

B ~ N(180, 152)

120

15,180N~

2

B

072064.0)182(P_

B 0721.0

Expected number of samples to get sample mean less than 182g

= )072064.01(100 = 92.7936 93 (nearest integer) or 92.8 (3 sf)

4 2 2 2~ N(68, ), ~ N(65,8 ), ~ N(70,10 )C T S

85 68P( 85) 0.05 P 0.05

85 681.6449 10.335 10.3 (3 s.f.)

C Z

2

2 1 2 5 10~ N(68,10 ) ~ N 68,

5 5

C C CC C

P( 75) 0.058762 0.0588 (3 s.f.)C

Let 0.2 0.2 0.6

E( ) 0.2(68) 0.2(65) 0.6(70) 68.6

X C T S

X

2

2 2 2 2 210Var( ) 0.2 0.2 (8 ) 0.6 (10 ) 39.36 ~ N(68.6,39.36)

5X X

P( 80) 0.034601 0.0346 (3 s.f.)X

or , and are independent.C C T S

5 Let T be the r.v. denoting “the mass of a randomly chosen jar of Tasty jam”, and

Y be the r.v. denoting “the mass of a randomly chosen jar of Yummy jam”.

T ~ N(300, 42), Y ~ N(350, 5

2).

T1 – T2 ~ N( 0, 32)

P( | T1 – T2 | > 10 ) = 2P( T1 – T2 > 10 ) = 0.0771

3 4!355 345 0.00253

3!P Y P Y

Let A = 1 10 1 5

15

T T Y Y

E(A) = 1 4750 950

300 10 350 515 15 3

Var(A) = 2 2

2

1 194 10 5 5

1515

P(317 < A < 322 ) = 0.384

6 Let T be the time taken to sign an autograph. Then 2N(3, 0.8 )T .

P( ) 0.12T k P( ) 0.88T k

Using GC, 3.94k (3 s.f.).

1 2 N(0, 1.28)T T

Required probability 1 2 1 2P( 2) 1 P( 2 2)T T T T = 0.0771

Required probability 2 3 5![P( 4)] [P( 3)]

2!3!T T

2 3(0.10565) (0.5) 10 0.0140 (3 s.f.).

7

8 Let X and Y be the weights of girls and boys respectively. X~ N (47, 52), Y ~ N (53, 4

2)

1 2Y Y ~ N (53 – 53, 42

+ 42) or N (0, 32)

1 2( 1.5)P Y Y = 1 2( 1.5 1.5)P Y Y = 0.209 (3 s.f.)

1 2 2Y Y X ~ N ( 2(53) – 2(47), 42

+ 42 + 2

2(5

2) ) or N (12, 132)

1 2( 2 0)P Y Y X = 0.852

1 2 3 1 2 7

1( ... )

10M X X X Y Y Y

2 2

2

3 47 7 53 3 5 7 4~ ,

10 10M N

or N(51.2, 1.87)

( 51) 0.442P M

9 Let A be the mass of an avocado. 2115 9A N ,

P 110 115 0 21074X .

Required prob= 2

3 0 21074 0 5 . . = 0.158 (3sf)

Let K be the mass of a kiwi. 282K N ,

P 90 0 1055 P 90 0 8945X X . .

90 82P 0 8945Z

.

81 2508

. 6 40 .

1 2 1 2 3

2

1 1509 324 92

5 5 5

K K A A AN

, . or 101 8 12 9968N . , .

1 2 1 2 3 100 0 6915

K K A A AP

.

Let 1 2 1 2 30 012 0 015T K K A A A . .

2 2 2 20 012 164 0 015 345 0 012 6 40 0 015 9T N . . , . . .

7 143 0 024123T N . , .

0 99P T a . 7 5403a .

Least value of a = 8

10 Let S denote the delivery time of a standard parcel via Super Express service.

36 36

P 36 0.36944 P 0.36944 0.33334 ---- (1)S Z

P 48 0.25249

48 48P 48 0.74751 P 0.74751 0.66667 ---- (2)

S

S Z

From (1), we have 36 0.33334 ---- (3)

Subst. (3) into (2), we get

48 36 0.333340.66667

12 0.33334 0.66667

1.00001 12 12 (correct to nearest whole number)

From (3), we have 36 0.33334 11.99988 40 (correct to nearest whole number)

10 Let R denote the delivery time of a standard parcel via Regular service.

Let E denote the delivery time of a standard parcel via Express service.

R ~ N(100, 242), E ~ N(72, 18

2), R – E ~ N(28, 900)

P 20 P 20 P 20 0.660 (3s.f.)R E R E R E

1 2

1 148~ , 644

3 3R E S S N

1 2 1 2

1 1P P 0 0.974 (3 s.f.)

3 3R E S S R E S S

11 (i) X~ time taken by route A. X~ N(14, 2.52)

P(X > 12) = 0.788

(ii) Y ~ time taken by route B. Y~ N(12, 7.3)

X – Y ~N(2, 2.52 + 7.3) so X – Y ~N(2, 13.55)

P(|X – Y| 3) = 1 P(3 X – Y 3) = 0.480

Let T = X1 + X2 + X3 + Y1 + Y2 ~ N(42 + 24, 3(2.52)+2(7.3)) or N(66,33.35)

P( T < 70) = 0.756

12 Let X be mass of an abalone in grams. 2~ 180,X N

200 0.08P X 200 180

0.08P Z

200.92P Z

From G.C.,

201.4051 14.234 14.2 (3 s.f.) (shown)

165 0.14541 0.145 (3 s.f.)P X

214.2

~ 180, 15

X N

180 5 180 5 ( 180 5)

2 180 5 or 2 ( 180 5)

2 185 or 2 ( 175)

0.173 (3 s.f.)

P X P X P X

P X P X

P X P X

Alternative 1, 214.2

180 ~ 0, 15

X N

180 5 180 5 ( 180 5) 2 180 5 0.173 (3 s.f.)P X P X P X P X

Alternative 2, ~ 2700, 3024.6X N

180 5 2700 75 ( 2700 75) ( 2700 75)

2 2775 0.173 (3 s.f.)

P X P X P X P X

P X

12 Let C be cost of one abalone in dollars. 2450

~ 81, 6.391000

C X N

2

1 2 3 4 5 ~ 5 81, 5 6.39T C C C C C N ~ 405, 204.1605T N

420 0.147 (3 s.f.)P T

Alternatively,

Let 2

1 2 3 4 5 ~ 5 180, 5 14.2W X X X X X N ~ 900, 1008.2W N

Probability required 450

420 933.33 0.147 (3 s.f.)1000

P W P W

13 Let X be the r.v. for the lecture duration of a lesson. 2~ (0.8, )X N 1 0.8

0.2

( 1) 0.9 ( ) 0.9

1.28155 0.1560608291 0.156 (3 s.f.)

P X P Z

Let Y be the r.v. for the tutorial duration of a lesson. 2~ (1.1,0.195 )Y N

~ (1.9,0.062379982)X Y N

(2.3 2.5)( 2.5 | 2.3)

( 2.3)

0.04648286030.85088 0.851 (3 s.f.)

0.0546288535

P X YP X Y X Y

P X Y

Let W be the r.v. for the number of overly long lessons, out of 100.

( 2.3) 0.0546288537P X Y so ~ (100,0.0546288537)W B

( 9) 0.9530987829 0.953 (3 s.f.)P W

Let L and T be the r.v. for the total duration of a lesson from the Lim and Tan centres

respectively. 2~ (1.9,0.062379982) ~ (2.2,0.4 )L N T N

20 19 ~ ( 3.8,82.711928)

(20 19 ) (20 19 0) 0.3380801501 0.338 (3 s.f.)

L T N

P L T P L T

Total charges from the Lim Centre is more than that from the Tan Centre only 33.8% of

the time. Thus pupils should choose the Lim Tuition Centre.

Alternative Method

Expected cost for Lim Centre = $20 x 1.9 = $38

Expected cost for Tan Centre = $19 x 2.2 = $41.80

Hence the pupil should sign up with Lim Centre.

14 (a) P P 1Z k p Z k p

Also P 1Z k p (By symmetrical property of normal curve about

P 1 2 1 2 1k Z k p p

(b)(i) Let X be the r.v. “mass of a randomly chosen pineapple (in kg)”. X ~ N 0.75,0.22( ) 2

21 2.02 ,0~ NXX

05.0P05.0P05.0 P 212121 XXXXXX 860.0

(ii) Let Y be the r.v. “mass of a randomly chosen papaya (in kg)”. Y ~ N 1.2,0.12( )

2222

71 1.042.075.2 ,2.1275.075.2~2...5.2 ssNsYXX

2

71 0.04s1.75 ,4.2125.13~2...5.2 sNsYXX

Given 82...5.2P 71 sYXX 0.765

765.0

04.075.1

4.2125.138P

2

s

sZ 235.0

04.075.1

125.54.2P

2

s

sZ

2.4s - 5.125

1.75+ 0.04s= -0.72248 72.1$s

(iii) Required proabability = 1 2 33!

P 2 P 2 P 22!

C C C

= 2 3!

0.40129 1 0.40129 0.2892!

15 (i) It is reasonable because P(Height of a boy < 0) 0 which is negligible.

(ii) 2~ N(155,5 )G , 2~ N(170,4 )B .

1 2E(3 ) 3 155 2 170 125G B B , 2 2 2

1 2Var(3 ) 3 5 2 4 257G B B

1 23 ~ N(125,257)G B B

1 2P(3 120)G B B 0.622

(iii) ~ (155,25)G N

P( 155 ) 0.85G k (OR) P( 155 ) 0.85G k

P( < 155 ) 0.85k G k P( < 155 ) 0.85k G k

P 0.855 5

k kZ

P( 155 ) 0.075G k

P 0.0755

kZ

155 147.802k

1.43955

k 7.20k

7.20k

(iv) Required probability = 2

P 165 P 165B B 0.00998

16

Let X kg and Y kg be the masses of a man

and a child respectively.

70

17 Let X be the random variable of the mass of an apple. 2~ ( ,30 )X N

70 150

2

g110 (by symmetry)

1 2 ~ (220,1800)X X N

407.040683185.023021 XXP

Let W be the random variable of the number of apples (out of 50) which are graded as ‘large’.

~ (50, ( 150))W B P X , ~ (50,0.09121121)W B

Since n=50 large and p=0.0912 small and np = 4.5605605 < 5

0~ P (4.5605605)W approximately

3 1 2 1 0.16688002P W P W = 0.83311998 0.833

Let Y be the random variable of the mass of an orange. 2~ (190, 24 )Y N

1 20.002 0.0015 0.0015

3

~ 0.2633333,0.000688

X Y YC

C N

( 0.25) 0.69438853 0.694P C

(iv) There is a good chance that at least

1 out of 9 hotel guests entering the

lift is a sumo wrestler, and thus the

probability of exceeding the safety

limit is greater than (iii).

OR

The mean and standard deviation

for the mass of a man in the hotel

would have changed when a large

number of sumo wrestlers checked

in.

x 150

SAMPLING SOLUIONS

1.

[JJC 2011 PRELIM P2 Q11]

The bags of sweets are ordered chronologically.

900/15=60

Randomly choose a bag of sweets from the first 60 and then choose every 60th

bag of

sweets thereafter.

A systematic sample is a better representation of all bags of sweets produced

throughout the day as a simple random sample may sample bags of sweets all

produced in the morning (or afternoon). 4

23! P 72

2.

[MJC 2011 PRELIM P2 Q10a]

Sampling frame is not available or a list of the concertgoers is not available.

First, obtain a list of the concertgoers by their ticket serial number and number them

from 1 to N.

Next, determine the sampling interval = 1000.01

Nk

N

Finally, use a graphing calculator to select a random number from 1 to 100 (e.g. 40)

and subsequently select every 100th

concertgoer from then (140th

, 240th

, 340th

, ...)

until 0.01N concertgoers are selected.

3.

[NYJC 2011 PRELIM P2 Q5]

Divide the student population into 2 mutually exclusive subgroups, for example,

males and females.

Interview any 100 males and 100 females to get a sample size of 200.

List the students in order.

By method of counting off, we divide the 1800 students into 200 intervals, each

interval with 9 students.

Within the 1st interval, choose a random start from the 1

st to 9

th student (inclusive),

then take the 9th

student thereafter.

4.

[RVH 2011 PRELIM P2 Q6]

Quota sampling was used.

Systematic sampling can be done by first obtaining a ordered list of all students in

Lee Hwa Junior College.

From the first 2000

4050

students in the list, a student is selected randomly, say the

5th

student.

The next 40th

student is selected and so on, i.e. the 45th

, 85th

, …

The sample is more evenly spread out over the population (esp. when students found

in canteen during break time usually come from the same classes or levels).

5. [NJC 2011 PRELIM P2 Q5]

(i)

Increase the number of students to be surveyed.

Survey representatives from different levels.

(ii) Since quality of program is to be considered, an appropriate stratum would be the

various levels of the school population.

Determine the number of students to be surveyed at each level which is proportional

to the relative size of the student population at that level i.e.

no. of students for the level100

total no. of students in the College .

In each stratum, randomly select the students based on the respective numbers

determined.

6. [TJC 2011 PRELIM P2 Q9a]

(a) (i) Simple Random Sampling method.

(ii) Simple Random Sampling is not appropriate as there is a possibility that the

sample is not a good representation of the student population. For example,

the sample could have an unevenly large number of one gender and less of

the other.

(iii) Stratified sampling could be used where the sample is obtained by dividing

the student population according to gender. The students are selected

randomly from each gender and the number of students selected from each

gender should be proportional to the sizes of each gender in the school.

The advantage of this method ensures a good representative sample of the student

population as each gender is proportionally represented.

Note for (iii): Quota Sampling is not acceptable as the sampling frame is available.

7.

[SAJC 2011 PRELIM P2 Q5]

(i) Quota Sampling

(ii) Possible answers :

(1)Information can be collected quickly as a sampling frame is not required.

(2)Cost is low.

(3)Ensures selection of adequate numbers of participants with appropriate

characteristics.

(iii) Stratified sampling method. This gives a more representative sample.

Or simple random sampling method. The data collected is free from bias as every

member of the population has an equal chance of being selected.

Or systematic sampling method. Sample is more evenly spread over the population of

working adults as the sample is taken throughout the ordered population.

8.

[TPJC 2011 PRELIM P2 Q9]

(i) Not good because not all employees in the office building may go to the canteen.

Use stratified sampling.

Put the employees into strata by gender i.e. male and female. Then select 640x150

1200

80 males and 560x150

1200 70 females randomly from 640 male employees and 560

female employees respectively.

Stratified sampling produces a more representative sample.

(ii) Let Y be the weight of female employees in kg. Then 2~ N(50, 2 )Y

Let X be the weight of male employees in kg. Then X 2N 68, 2

2

1 2 3 4 N 50 4, 2 4Y Y Y Y i.e. 1 2 3 4 N 200,16Y Y Y Y

2 23 ~ N 68 3, 2 3X i.e. 3 ~ N 204, 36X

Let 1 2 3 4 3T Y Y Y Y X . Then ~ N 200 204,16 36T i.e.

~ N 4, 52T

P 0 0.71045 0.710T

(iii) Let X be the weight of male employees in kg. ~X 2N 68, 2 and

22

~N 68,80

X

P | | 0.5 =P 68 0.5 68 0.5 =P 67.5 68.5X X X

0.97465 0.975

(iv) Let M be the total weight of k male employees in kg

2~ N 68 , 2M k k i.e. ~ N 68 , 4M k k

Let F be the total weight of 25 k female employees in kg

2~ N 50 (25 ), 2 (25 )F k k i.e. ~ N 1250 50 ,100 4F k k

Then ~ N 1250 18 ,100M F k

P 1500 0.987

250 18 250 18P 0.013 2.2262

100 100

M F

k kZ

Solving gives 15.125 15k

9.

[HCI 2011 PRELIM P2 Q5]

(a) Number the citizens from 1 to 500, 000.

Randomly select a citizen from the first 500 000

50001000

citizens.

Select every 5000th

citizen thereafter.

Stratified sampling with appropriate strata such as income levels is more appropriate

because the sample is more representative of the population, as citizens’ choice of

president may vary according to their income levels.

OR

Quota sampling is more appropriate as it is generally easier and faster to conduct than

systematic sampling. Hence it will require lesser amount of resources to conduct the

survey.

OR

Quota sampling is more appropriate as it can be conducted without knowledge of the

sampling frame since the editor may not have access to the sampling frame.

(b) Any of the following:

People may lie in the pre-election poll.

People may change their minds.

Sample may not be representative of the population even though the sample is

randomly chosen.

Sample size may not be large enough to reflect the votes of the citizens.

10.

(i) Let X be the number of lemon candies in a randomly selected packet of 20.

(20,0.24)X B

( ) 20 0.24 4.8E X ( ) 20 0.24 0.76 3.648Var X

Since 60 ( 50)n , by Central Limit Theorem, 3.648

4.8,60

X N

.

( 5) 0.20865208P X =0.209 (3sf)

(ii) The sample is biased, as only students are surveyed. Not everyone in the

population has an equal chance of being surveyed.

It will be difficult to get an exhaustive list of people of all age groups to do a proper

stratification. (no sampling frame)

Use Quota Sampling

11 (i)

(ii)

(iii)

A ‘random sample’ in this context refers to choosing groups of 100 students from the

population of 2000 in such a way that every group of 100 has the same chance of

being chosen and the choosing of the groups are independent of each other.

Randomly choose 22 students from the group who go by car, 38 from those who go

by bicycle and 40 from those who walk.

An advantage that stratified sampling would have compared to random sampling in

this context is that stratified sampling guarantees a representative sample from each

of the strata ‘car’, ‘bicycle’ and ‘walk’.

A better stratified sample of size 100 could have been achieved by considering the six

groups by year and method of travel.

12 72 ÷ 8 = 9

Randomly choose a claim from the first 9, then choose every subsequent 9th claim

thereafter, example the 5th claim was randomly chosen so choose the 14th, 23rd …

till 8 claims have been chosen.

The first 8 claims submitted could have been those which are simple cases or which

are claiming for less in terms of damages. Latter claims could be more complicated

cases which require more time for the claimants to handle. Hence using a systematic

sampling method is fairer as it spreads out the sample over the different types of

claims.

13 (i)

(ii)

(iii)

Systematic sampling.

Advantage:

It is easy to obtain a spread of students for the sample, e.g. students may come

according to classes, so this method allows the headteacher to choose students from

different classes.

Disadvantage:

Biased since this method leaves out students who do not eat lunch at that time.

An alternative sampling method which would be better in this case is the stratified

sampling method. The headteacher should have a list of all students in the school. She

could stratify the students by level and choose students randomly from each level

according to the proportion of students in each level.