1

ACJC 2015 JC2 H2 Mathematics REVISION SET G

Solutions to Binomial and Poisson Distributions

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

When n = 30, 1

~ B 30,5

X

Expected number of students who use AIKON brand phone = 1

30 65

P 5 0.172X . 1(ii)

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

16 15 14 13 12 2321

20 19 18 17 16 323

.

2(ii) 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. 2(last

part)

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(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( ) (independence)

( 0) 0.01P X Y

(1.2 ) 0.01e

1.2 4.605170186 3.41 . 3(ii) For a day , X + Y ~ Po(4.61)

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

3(last

part)

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 (3s.f.) .

4(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).

1.2

2

4(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.

4(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.

4(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.

5(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 . 5(ii) 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

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

in each two-hour period or there arent. (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.

5(iv) 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

3

P 10V 0.0060881936 = 0.00609. 6(i) (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 . 6(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 .

6(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 .

7(i) 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.

7(ii) 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 . 7(iii)

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!

.

7(iv) ~ 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

7(v) 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).

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8(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 .

8(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).

9(a) 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.

5

9(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.).

9(b)(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. 9(last

part)

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.).

10(i) Let X denote the demand for GreatRun tyres in a randomly chosen month.

X ~ Po(4)

Required prob. = P 5X = 0.7851303874 = 0.785 (3 s.f.). 10(ii) Using GC,

No. of tyres sold, x Prob.

0 P(X = 0) = 0.01832

1 P(X = 1) = 0.07326

2 P(X = 2) = 0.14653

3 P(X = 3) = 0.19537

4 P(X = 4) = 0.19537

5 P 5X = 0.37116 Hence, the most probable number of tyres sold is 5.

10(iii) Let the number of tyres the garage should keep in stock be N.

P 0.001

1 P 0.001

X N

X N

Using GC,

N P X N 10 0.00284 > 0.001

11 0.000915 < 0.001

Hence the least number of tyres the garage should keep in stock at the beginning of

the month is 11.

10(iv) Let Y denote the demand for GreatRun tyres in a randomly chosen week.

Y ~ Po(1)

P 1 1 P 1Y Y = 0.2642411176 Let W denote the number of weeks where more than 1 GreatRun tyre is sold out of

4 weeks.

W ~ B(4, 0.26424)

Required prob. = P 2 1 P 2W W = 0.059174 = 0.0592 (3 s.f.) 10(v) Let A denote the number of months the garage was not able to meet monthly

6

demands out of 120 months.

A ~ B(120, 1 0.78513) i.e. A ~ B(120, 0.21487) Since n = 120 is large, np = 25.7844 > 5, nq = 20.24410597 > 5,

A ~ N(25.7844, 20.24410597) i.e. A ~ N(25.784, 20.244) approximately

Required prob. = P 36 | 12A A

P 36 12

P 12

P 36

P 12

P 35.5by continuity correction

P 12.5

0.015432

0.0154 (3 s.f.).

A A

A

A

A

A

A

11(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 11(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).

11(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.

11(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,

7.27.2, approx

52X N

.

7 0.2954667 0.295P X Alternative method:

7

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

11(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. 11(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.

12(a)(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 . 12(a)(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 .

12(b)(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).

12(b)(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.

13(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.).

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

0.15N 0.15,

50X

approx.

ie. N 0.15, 0.003X approximately.

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

14(a) ~ 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

8

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.

14(b) 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).

15(i) 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.

15(ii) 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 .

15(iii) 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)

.

15(iv) 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.

15(last

part) Since sample size = 50 is large,

6.3~ (9, )

50X N approx by Central Limit Theorem

P( 10) 0.00242X .

16(i) 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

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 .

16(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

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

smaller.

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

Limit Theorem, 6

~ N 6, approx.Tn

9

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.

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

(out of 25).

.

17(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

.

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

enquiries at the Police Force's booth.

Since large, , approximately.

Required Prob .

17 (last

part)

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.

1(25, )

6X B

( 10) ( 9) 0.99526P X P X

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

(60,0.0047426)Y B

60n 0.28455 5np 0(0.28455)Y P

( 2)P Y 0.30459 0.305