CHAPTER 15netdrive.puiying.edu.hk/~ms/f6notes/MS-… · Web view · 2008-02-16EXERCISE 15.1...

CHAPTER 15SOME SPECIAL DISCRETE DISTRIBUTIONS

EXERCISE 15.1 Section 15.1 The Bernoulli distribution(page 332)

1. (a) Bernoulli, two possible outcomes: Male, Female

(b) Bernoulli, two possible outcomes: Fail, Pass

(c) not Bernoulli, more than two possible outcomes such as Christian, Catholic, Buddhist

(d) Bernoulli, two possible outcomes: Hit, Miss

2. (a) Bernoulli, two possible outcomes: Valid, Not valid

(b) not Bernoulli, there can be many makes, not just 2

(c) Bernoulli, two possible outcomes: Work, Fail

(d) not Bernoulli, more than two possible outcomes such as Chinese, English, German

3. An appropriate random variable for each of (a) – (d) is X such that

(a)

(b)

(c)

(d)

4. An appropriate random variable for each of (a) – (d) is X such that(a)

(b)

(c)

(d)

5. Mean =

= 0 0.8 + 1 0.2= 0.2

Variance =

= (0 – 0.2)2 0.8 + (1 – 0.2)2 0.2= 0.16

199

CHAPTER 15 SOME SPECIAL DISCRETE DISTRIBUTIONS

6. x 0 1

f (x)

mean = E(X)=

=

=

variance =

=

=

7. (a) p + 0.38 = 1 p = 0.62

(b) x 0 1

f (x) 0.38 0.62

mean = = 0 0.38 + 1 0.62= 0.62

variance = = (0 0.62)2 0.38 + (1 0.62)2 0.62= 0.235 6

standard deviation = = 0.485 4

8. (a)

200

SECTION 15.1 THE BERNOULLI DISTRIBUTION

(b) mean = p =

variance =

(c) E(3T + 2) = 3E(T) + 2 = + 2 = 4

Var (3T + 2) = 32 Var (T) = = 2

9. (a)

(b) mean = p = 0.63variance = p(1 p) = 0.63 0.27 = 0.170 1 standard deviation = = 0.412 4

(c) E(5 2W) = 5 2E(W) = 5 2 0.63 = 3.74Var (5 2W) = (2)2 Var (W) = 4 0.170 1 standard deviation of (5 2W) = 2 = 0.824 9

10. (a)

(b) mean = 0.92variance = 0.92 0.08 = 0.073 6

11. (a) The probability distribution of X is shown below.

x 0 1

P(X = x)

(b) Mean of X = 0 + 1 =

Variance of X =

=

(c) The mean of X is the proportion of times a white ball is obtained.The variance of X indicates the variability of the actual outcome.

12. (a) P(Y = 0) = P(no ace) =

P(Y = 1) = P(an ace) = =

201


(b)

(c) mean =

standard deviation =

13. (a)

(b) mean =

variance =

14. (a) Let P(female customer) = p.

Then P(male customer) = .

p + = 1

p =

The probability function of X is:

(b)

00 1 x

f(x)

13

23

Probability distribution of X

15. (a) Total number of ways in which the 8 books can be arranged = 8!When 2 particular books are adjacent to each other, they may be considered as a single object. Hence, the 7 objects may be arranged in 7! ways.But the 2 particular books may themselves be arranged in 2 ways. Number of ways the 8 books can be arranged with 2 particular books adjacent to each

other = 2 7!

Probability required =

202

SECTION 15.2 THE BINOMIAL DISTRIBUTION

(b) (i) From the result in (a), we obtain the probability distribution of X as follows.

x 0 1

P(X = x)

(ii) Mean of X =

Variance of X =

Standard deviation of X =

EXERCISE 15.2 Section 15.2 The binomial distribution(page 344)

1. (a) P(X = 2) =

(b) x 0 1 2 3

f (x)

mean =

=

=

=

=

variance =

=

=

2. (a) P(X = 4) = (0.2)4(0.8)= 0.006 4

(b) P(X 4) = P(X = 4) + P(X = 5)= 0.006 4 + (0.2)5(0.8)0

= 0.006 72

203


(c) Since X B(5, 0.2),mean of X = 5 0.2 = 1 = np

variance of X = 5 0.2 0.8 = 0.8

= npq

3. (a) P(X = 3) = = 0.254 1

(b) P(2 X 4) = = 0.296 5 + 0.254 1 + 0.136 1= 0.686 7

4. (a) P(Y = 13) = = 0.122 1

(b) P(Y 14) =

(c) P(Y 6) =

(d) mean = E(Y) = np = 20 0.55 = 11

standard deviation = = 2.225

5.

5(1 – p) =

p =

n = = 20

6. (a) P(2 X 6) =

(b) p = np 2 = npq

(i) (ii)(iii)

0.30.50.6

2.13.54.2

1.471.751.68

(c) The variance is largest when p = 0.5.[Note: In fact, for any B(n, p), the variance is largest when p = 0.5 as p(1 p) is maximum

when p = 0.5.]

204


(d) (i) When p = 0.3,

x 0 1 2 3 4 5 6 7

f (x) 0.082 0.247 0.318 0.227 0.097 0.025 0.004 0.000

(ii) When p = 0.5,

x 0 1 2 3 4 5 6 7

f (x) 0.008 0.055 0.164 0.273 0.273 0.164 0.055 0.008

(iii) When p = 0.6,

x 0 1 2 3 4 5 6 7

f (x) 0.002 0.017 0.077 0.194 0.290 0.261 0.131 0.028

7. (a) P(tongue roller) = p =

Let X be the number of tongue rollers in the sample.

X ~ B(5, ).

P(no tongue rollers) = P(X = 0) =

(b) P(3 tongue rollers) = P(X = 3) =

205


8. (a) Let X be the number of applicants having no marketing experience in the sample.Then X ~ B(6, 0.7).The required probability = P(X = 4) = = 0.324 1

(b) The required probability = P(X = 5) = = 0.302 5

9. (a) Let Y be the number of deals made.Then X ~ B(7, 0.1).P(2 deals) = P(X = 2) = = 0.124 0

(b) P(less than 2 deals) = P(X = 0) + P(X = 1)= = 0.478 3 + 0.372 0= 0.850 3

10. (a) Let X be the number of smokers in the sample.Then X ~ B(8, 0.4).The required probability = P(X = 7) = = 0.007 864

= 0.007 9 (4 d.p.)

(b) P(more than 6) = P(X 6)= P(X = 7) + P(X = 8)= 0.007 864 + (0.4)8

= 0.008 5 (4 d.p.)

11. (a) Let X be the number of germinating seeds.

Then X ~ B(9, ).

The required probability = P(X = 7) = = 0.306 1

(b) P(at least 2 seeds will germinate) = 1 P(X = 0) P(X = 1)

=

= 0.999 96

12. (a) Let X be the number of cardholders in the sample.Then X ~ B(10, 0.85).P(exactly 8 cardholders) = P(X = 8) = = 0.275 9

(b) P(not less than 8 cardholders) = P(X 8)= P(X = 8) + P(X = 9) + P(X = 10)= 0.275 9 + = 0.820 2

13. Let X be the number of defective chips in the sample of 11.Then X B(11, 0.9),i.e. P(X = x) = ,for x = 0, 1, 2, … , 11.(a) P(4 to 6 chips are defective)

= P(X = 4) + P(X = 5) + P(X = 6)=

206


= 0.002 7

(b) P(at least 9 chips are defective)= P(X = 9) + P(X = 10) + P(X = 11)= = 0.910 4

14. Let X be the number of packets in the sample of 12 which weigh more than 1 kg.Then X B(12, 0.93),i.e. P(X = x) = , for x = 0, 1, 2, … , 12.

(a) P(X > 9) = P(X = 10) + P(X = 11) + P(X = 12)= = 0.953 2

(b) P(at least one packet will weigh less than 1 kg)= 1 – P(all the 12 packets will weigh more than 1 kg)= 1 – P(X = 12)= 1 – (0.93)12

= 0.581 4

15. (a) Let X be the number of students going to school by MTR.Then X ~ B(15, 0.6).The required probability = P(X 14)

=

= 0.005 2(b) The required probability = P(X 5)

=

16. (a) Let Y be the number of residents voting for Mr. X in the sample.Then Y ~ B(80, 0.55)P(more than 60) = P(Y 60)

=

(b) P(between 45 and 60 inclusive) = P(45 Y 60)

=

17. X B(20, 0.25)i.e. P(X = x) = , for x = 0, 1, 2, … , 20.

(a) E(X) = 20 0.25 = 5 E(X) = np

Expected score of Jane = 5 5 = 25 marks

207


(b) x P(X = x)* P(X x)*

0 0.003 17 0.003 171 0.021 14 0.024 312 0.066 95 0.091 263 0.133 90 0.225 164 0.189 69 0.414 855 0.202 33 0.617 186 0.168 61 0.785 797 0.112 41 0.898 208 0.060 89 0.959 09

* Correct to 5 decimal places so that answers in probabilities derived from the table may be given correct to 4 decimal places.

(c) From the table in (b),(i) P(not more than 25 marks)

= P(not more than 5 correct answers)= P(X 5)= 0.617 2 (4 d.p.)

(ii) P(not more than 30 marks)= P(X 6)= 0.785 8 (4 d.p.)

(d) P(X 8) = 1 – P(X 7)= 1 – 0.898 20, from the table in (b)= 0.101 8 (4 d.p.)

Hence, if a student attempts the test by wild guesses, he/she still has a quite high probability (> 0.10) of obtaining 8 or more correct answers (i.e. 40 or more marks).On the other hand,P(X 9) = 1 – P(X 8)

= 1 – 0.959 09= 0.040 9 (4 d.p.),

which is less than 0.05 and is low enough.Hence, we would require a student to obtain at least 9 correct answers, i.e. to score at least 45 marks before he/she could be considered as not attempting by wild guesses.

(e) Suppose Jane obtains X correct answers and hence (20 – X) incorrect answers. Then the score she obtains is

M = 5X – (20 – X)= (6X – 20) marks

(i) Jane’s expected score isE(M) = 6E(X) – 20

= 6 5 – 20 E(X) = 5, from (a)= 10 marks

208


(ii) P(Jane will pass)= P(M 40 marks)= P(6X – 20 40)= P(X 10)= 1 – P(X 9)= 1 – P(X 8) – P(X = 9)= 1 – 0.959 09 –

P(X 8) from (b)

= 1 – 0.959 09 – 0.027 06= 0.013 9

18. (a) Let n be the number of missiles required.P(at least 1 hit) 0.95

i.e. 1 P(no hit) 0.95 1 (1 0.8)n 0.95

0.05 0.2n

log 0.05 n log 0.2 n 1.86Hence, the least number of missiles required = 2.

(b) Let X be the number of missiles hitting the target.Then X ~ B(6, 0.8).P(at least 2) = 1 P(X = 0) P(X = 1)

= = 0.998 4

19. (a) Let X be the number of persons having hypertension in the group.Then X ~ B(10, 0.15).The required probability = P(X 2)

= 1 [P(X = 0) + P(X = 1)]= 1 = 1 [0.196 9 + 0.347 4]= 0.455 7

(b) Let Y be the number of persons having hypertension in the second group.Then Y ~ B(10, 0.15).The required probability = P(X + Y = 2)

= P(X = 0 and Y = 2) + P(X = 1 and Y = 1) + P(X = 2 and Y = 0)= 0.196 9 0.275 9 + 0.347 4 0.347 4 + 0.275 9 0.196 9= 0.229 3

20. (a) Let X be the number of short-in-length piles in the sample of 12. Then X B(12, p)i.e. P(X = x | company) = , for x = 0, 1, 2, … , 12, where p = the probability of a short-in-length pile and the value depends on the supplying company.

(i) If the piles were supplied by company A, then p = 0.10. P(X = 2 | A) =

= 0.230 1

209


(ii) If the piles were supplied by company B, then p = 0.05. P(X = 2 | B) =

= 0.098 8

(iii) If the piles were supplied by company C, then p = 0.02. P(X = 2 | C) =

= 0.021 6

(b) The given information and the results in (a) are displayed in the following tree diagram.

Referring to the above tree diagram and with obvious notations,

(i) P(supplier was A | X = 2)=

=

=

=

= 0.532 3

(ii) P(supplier was B | X = 2)

=

= 0.342 8

(iii) P(supplier was C | X = 2)

=

= 0.124 9

210

Bayes’ Theorem


21. (a) Assuming the claim is true,X B(10, 0.70)

i.e. P(X = x) = , for x = 0, 1, 2, … , 10.

(i) Mean of X = 10 0.70 = 7Standard deviation of X = = 1.449 1

(ii) P(X = 5) = = 0.102 9

P(X 5) =

= 0.150 3

(b) From (a)(ii), if the claim is true,P(X = 5) = 0.102 9 and P(X 5) = 0.150 3

Both these probabilities are quite high (exceeding 0.10). Hence, the finding that only 5 out of 10 interviewees favour the new regulation is not sufficient evidence to reject the claim. That is, we should regard the claim is true.

22. X B(16, 0.07),i.e. P(X = x) = , for x = 0, 1, 2, … , 16.

(a) (i) P(X = 0) = (0.93)16 = 0.313 1

(ii) P(X = 1) = = 0.377 1

(iii) P(X = 2) = = 0.212 9

(b) P(the purchaser would accept the batch)= P(X 1)= P(X = 0) + P(X = 1)= 0.313 1 + 0.377 1, from the results in (a)(i) and (ii)= 0.690 2

(c) Let Xi = number of defective scanners in the ith sample (i = 1, 2).Then Xi B(8, 0.07).

(i) (1) P(batch accepted on the first sample)= P(X1 = 0)= (0.93)8

= 0.559 6

(2) P(second sample required)= P(X1 = 1)= (0.07)(0.93)7

= 0.337 0

(3) P(batch accepted)= P(batch accepted on the first sample)

+ P(second sample required) P(batch accepted on the second sample)= P(X1 = 0) + P(X1 = 1) P(X2 = 0)= 0.559 6 + 0.337 0 0.559 6= 0.748 2

211


(ii) No. The two-stage sampling has a higher probability (0.748 2 against 0.690 2) of accepting the batch, and hence has a higher risk of accepting defective scanners.

(iii) Expected number of scanners inspected= 8 + 8 P(second sample required)= 8 + 8 0.337 0= 10.70 Expected cost = $50 10.70

= $535 (nearest dollar)

23. (a) Let X be the number of defective items in the sample.Then X ~ B(4, 0.05).P(reject) = 1 P(accept)

= 1 P(X 1)= 1 (0.95)4 4(0.95)3 (0.05)= 0.014 0

(b) Let Y be the number of batches rejected.Then Y ~ B(10, 0.014 0).(i) P(none of the batches will be rejected) = P(Y = 0)

= (1 0.014 0)10

= 0.868 5

(ii) P(at least one batch will be rejected) = 1 P(Y = 0)= 1 0.868 5= 0.131 5

(iii) P(exactly 3 batches will be rejected) = P(Y = 3)= = 0.000 3

24. (a) Let Y be the number of patients recovered out of 20 patients.Then Y ~ B(20, p).When p = 0.1,P(Y = 13) =

= 3.708 109

= 0.000 0 (4 d.p.)Similarly, we can obtain the required probabilities for the other values of p as shown in the following table.

p 0.1 0.2 0.3 0.4 0.5

P(Y = 13) 0.000 0 0.000 0 0.001 0 0.014 6 0.073 9

p 0. 6 0.7 0.8 0.9

P(Y = 13) 0.165 9 0.164 3 0.054 5 0.002 0

212

SECTION 15.3 THE GEOMETRIC DISTRIBUTION

(b)

0.05

0.10

00.1 0.2 0.90.3 0.4 0.5 0.6 0.7 0.8 p

P(Y=13)

0.15

0.20

(c) From the graph, P(Y = 13) is maximum when p = 0.65.

EXERCISE 15.3 Section 15.3 The geometric distribution(page 353)

1. (a) P(X = 3) =

(b) mean =

variance =

(c) x 1 2 3 4 5 6 7 8

P(X = x) 0.286 0.204 0.146 0.104 0.074 0.053 0.038 0.027

0.1

0.2

01 2 3 4 5 6 7 8 x

f(x)

0.3

Probability distribution for X

2. (a) P(Y 2) = P(Y = 1) + P(Y = 2) = 0.4 + 0.6 0.4 = 0.64

(b) mean = = 2.5

213


standard deviation = = 1.936

(c) y 1 2 3 4 5 6 7

P(Y = y) 0.4 0.24 0.144 0.086 0.052 0.031 0.019

0.1

0.2

01 2 3 4 5 6 7 y

f(y)

0.3

0.4

Probability distribution for Y

3. (a) mean = = 5

p = 0.2P(2 Z 5) = P(Z = 3) + P(Z = 4) + P(Z = 5)

= (0.8)2 (0.2) + (0.8)3 (0.2) + (0.8)4 (0.2)= 0.312 3

(b) standard deviation = = 4.472

4. (a) variance = = 6

6p2 + p 1 = 0(3p 1) (2p + 1) = 0

p = (rejected)

p =

and mean =

= 3

(b) P(W 8) = P(W = 8) + P(W = 9) + P(W = 10) + . . .

=

=

214


=

=

5. p =

The required probability = = 0.040 7

6. (a) p = 0.05P(the first defective is the 6th vase) = (0.95)5 (0.05) = 0.038 7

(b) P(the first defective is the 8th vase) = (0.95)7 (0.05) = 0.034 9

7. Let X be the number of tosses required until the drawing pin lands “point up”. Then X has a

geometric distribution with parameter p = , i.e.

P(X = x) = , for x = 1, 2, …

(a) P(X = 4) = = 0.080 0

(b) P(X > 4) = P(a “point up” will not occur in the first 4 tosses)

=

= 0.106 6

8. X has a geometric distribution with probability p = for a success (i.e. for the line to be

connected):

P(X = x) = , for x = 1, 2, …

(a) P(X = 3) = = 0.146 5

(b) P(X 4) = 1 – P(X > 4)= 1 – P(not connected in the first 4 dials)

= 1 –

= 0.847 4

(c) Mean of X = = = 2.67 (2 d.p.)

Standard deviation of X =

215


=

=

= 2.11 (2 d.p.) 9. Let X be the number of the persons passing until the first one enters the shop. Then X has a

geometric distribution with probability p = 0.1 of entering the shop, i.e.P(X = x) = (0.9)x–1(0.1), for x = 1, 2, …

(a) P(the first person entering the shop is the 6th who pass) = P(X = 6)= (0.9)5(0.1)= 0.059 0

(b) P(the first person entering the shop is one of the first 10 persons who pass)= P(X = 1) + P(X = 2) + + P(X = 10)= 0.1 + 0.9 0.1 + + (0.9)9 0.1= 0.651 3Alternatively,P(the first person entering the shop is not one of the first 10 persons who pass) = (0.9)10

The required probability = 1 – (0.9)10

= 0.651 3

10. (a) P(hitting not until the 3rd shot) = (0.3)2 (0.7) = 0.063

(b) Let Y be the number of shots till the target is hit.P(hitting before the 4th shot) = P(Y = 1) + P(Y = 2) + P(Y = 3)

= 0.7 + (0.3) (0.7) + (0.3)2 (0.7)= 0.973

(c) P(hitting after the 4th shot) = P(Y = 5) + P(Y = 6) + P(Y = 7) + . . .= (0.3)4 (0.7) + (0.3)5 (0.7) + (0.3)6 (0.7) + . . .

=

= 0.008 1

11. (a) Let X be the number of calls before getting a helper.P(getting a helper at the 3rd call) = P(X = 3)

(b) P(getting a helper within 4 calls)= P(X 4)= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

=

= (or 0.974 4)

216


12. Let X be the number of throws the boy requires to knock the bottle off the shelf. Then X has a

geometric distribution with probability of success p = , i.e.

P(X = x) = , for x = 1, 2, …

(a) P(knocking the bottle off within 3 throws) = P(X = 1) + P(X = 2) + P(X = 3)

=

= 0.488

(b) P(knocking the bottle off eventually)= P(knocking the bottle off in one of the 5 chances)= 1 – P(not knocking the bottle off in one of the 5 chances)

= 1 –

= 0.672 3

13. Let X be the number of the interviewees required until one is selected.Then X has a geometric distribution with probability p = 0.15 of a success, i.e.

P(X = x) = (0.85)x–1(0.15), for x = 1, 2, …

(a) P(X = 5) = (0.85)4(0.15)= 0.078 3

(b) P(the person selected is the 7th one interviewed or later)= P(none of the first 6 persons interviewed is selected)= (0.85)6

= 0.377 1

14. (a) The required probability = (0.25)4 (0.75) = 0.002 9

(b) The required probability= P(there are 2 out of the first 6 cases with family problem)

P(the 7th case is with family problem)= = 0.024 7

15. (a) P(the first test he passes is the 4th one) = = 0.104 6

(b) P(winning the award after 8 tests)= P(he passes 1 out of the first 7 tests) P(he passes the 8th test)

=

= 0.076 5

16. (a) Let S denote the event that a question is answered correctly and F denote the event that a question is answered wrongly.P(boy A wins at his 3rd attempt) = P(FFFFS)

= (0.8)4 (0.2)= 0.081 92

(b) P(boy A wins the prize) = P(S) + P(FFS) + P(FFFFS) + . . .= 0.2 + (0.8)2 (0.2) + (0.8)4 (0.2) + . . .

217


=

=

The ratio of the chances of winning for A and B = = 5 : 4

218


17. (a) The required probability =

(b) The required probability =

(c) The required probability =

18. (a) E(X) =

= 1 0.5 + 2 0.2 + 3 0.15 + 4 0.1 + 5 0.05= 2

Var(X) =

= (–1)2 0.5 + 02 0.2 + 12 0.15 + 22 0.1 + 32 0.05= 1.5

(b) The toll, in dollars, for a car with X passengers isT = 3X + 20

E(T) = 3E(X) + 20= 3 2 + 20, from the result in (a)= 26

i.e. the expected toll charge for a car is $26.

Var(T) = 32 Var(X)= 9 1.5, from the result in (a)= 13.5

Standard deviation of the toll charge for a car= $ = $3.67

(c) P(X 3) = P(X = 1) + P(X = 2) + P(X = 3)= 0.5 + 0.2 + 0.15, from the given distribution= 0.85

(d) Let U be the number of cars, among the 7 selected cars, that have not more than 3 passengers.

Then U B(7, 0.85), using the result in (c) P(U = 5) =

= 0.209 7

(e) For the toll charge of a car to exceed $30, there must be k passengers in it, where k is given by

3k + 20 > 30

i.e. k >

Hence, the car must carry 4 or 5 passengers.P(4 or 5 passengers) = 0.1 + 0.05 = 0.15

Then Y has a geometric distribution with probability of success p = 0.15.

(i) P(Y = 4) = (0.85)4–1(0.15) = 0.092 1

219


(ii) P(Y > 4) = P(first 4 cars pay toll charge not exceeding $30 each)= (0.85)4

= 0.522 0

220

SECTION 15.4 THE POISSON DISTRIBUTION

EXERCISE 15.4 Section 15.4 The Poisson distribution(page 366)

1. (a) (i) P(X = 2) = = 0.146 5

(ii) P(X = 3) = = 0.195 4

(b) variance = = 4

(c) x 0 1 2 3 4 5 6 7 8 9

f (x) 0.018 0.073 0.147 0.195 0.195 0.156 0.104 0.060 0.030 0.013

2. (a) (i) P(Y = 4) = = 0.133 6

(ii) P(2 Y 4) = P(Y = 2) + P(Y = 3) + P(Y = 4)

= + 0.133 6

= 0.256 5 + 0.213 8 + 0.133 6= 0.603 9

(b) standard deviation = = 1.581 1

(c) y 0 1 2 3 4 5 6 7 8

f (y) 0.082 0.205 0.257 0.214 0.134 0.067 0.028 0.010 0.003

221


3. (a) For Po(1.5),

P(X = x) = , for x = 0, 1, 2, …

The numerical values for x = 0, 1, 2, … , 9 are shown below.

x 0 1 2 3 4 5 6 7 8 9

P(X = x) 0.223 1 0.334 7 0.251 0 0.125 5 0.047 1 0.014 1 0.003 5 0.000 8 0.000 1 0.000 0

For Po(5),

P(X = x) = , for x = 0, 1, 2, …

The numerical values for x = 0, 1, 2, … , 9 are shown below.

x 0 1 2 3 4 5 6 7 8 9

P(X = x) 0.006 7 0.033 7 0.084 2 0.140 4 0.175 5 0.175 5 0.146 2 0.104 4 0.065 3 0.036 3

(b) Bar chart for Po(1.5):

Bar chart for Po(5):

(c) The bar chart for Po(1.5) shows that the distribution is positively skewed. The probability drops rapidly as x increases and so the mean and the standard deviation remain quite small.The bar chart for Po(5) shows that the distribution is fairly symmetrical about the mean 5. The dispersion is much larger than that of Po(1.5).

4. (a) Let X be the number of defective ones in the sample of 20.Then X B(20, 0.025).i.e. P(X = x) = , for x = 0, 1, 2, … , 20.

(i) P(X = 0) = (0.975)20 = 0.602 7

(ii) P(X = 3) = = 0.011 6

222


(b) The mean of X = 20 0.025

= np for the B(n, p)

= 0.5Using the Po(0.5) as an approximation to the B(20, 0.025), we have

P(X = x) , for x = 0, 1, 2, …

(i) P(X = 0) e–0.5 = 0.606 5

(ii) P(X = 3) = 0.012 6

5. (a) Let X be the number of bulbs producing black tulips.Then X B(50, 0.03)

(i) P(X = 1) = = 0.337 2

(ii) P(X = 2) = = 0.255 5

(b) The mean of X = 50 0.03= 1.5

Using the Po(1.5) as an approximation to the B(50, 0.03), we have

(i) P(X = 1) e–1.5(1.5) = 0.334 7

(ii) P(X = 2) = 0.251 0

6. (a) For B(10, ), P(X = x) = , for x = 0, 1, 2, … , 10

For B(25, ), P(X = x) = , for x = 0, 1, 2, … , 25

For B(200, 0.025), P(X = x) = , for x = 0, 1, 2, … , 200

For Po(5), P(X = x) = , for x = 0, 1, 2, …

The required results as shown below.

x

B(n, p)

Po(5)n = 10, p = n = 25, p = n = 200, p =

0.025

0 0.001 0 0.003 8 0.006 3 0.006 7

1 0.009 8 0.023 6 0.032 4 0.033 7

2 0.043 9 0.070 8 0.082 7 0.084 2

3 0.117 2 0.135 8 0.140 0 0.140 4

(b) Such a binomial distribution may be approximated by the Po(np).

223


7. (a) Let X be the number of faulty invoices in the sample.Then X ~ B(50, 0.02).

n = 50, p = 0.02 = np = 50 0.02 = 1X can be approximated by Po(1).P(no faulty invoice) = P(X = 0) = e1 = 0.367 9

(b) P(less than 3 faulty invoices)= P(X = 0) + P(X = 1) + P(X = 2)

= 0.367 9 +

= 0.919 7

8. (a) Let X be the number of claims in a year.

Then X can be approximated by Po , i.e. Po(2.5).

P(1 claim) = P(X = 1) = e2.5 2.5 = 0.205 2

(b) P(more than 2 claims) = 1 P(X = 0) P(X = 1) P(X = 2)

=

= 0.456 2

9. (a) Let X be the number of bulbs burned out within a month.X ~ Po(80 0.015), i.e. Po(1.2)

The required probability = P(X = 3)

=

= 0.086 7

(b) The required probability = P(X = 0) + P(X = 1) + P(X = 2)

=

= 0.879 5

10. (a) Let X be the number of employees taking sick leave on a day.Then X ~ B(400, 0.03).It is approximated by Po() with = 400 0.03 = 12.P(at most 2 employees) = P(X = 0) + P(X = 1) + P(X = 2)

=

= 0.000 5

(b) P(at least 5 employees) = 1 P(X 4)= 1 P(X 2) P(X = 3) P(X = 4)

= 1 0.000 522

= 0.992 4

224


11. (a) P(X = 5) = = 0.160 6

(b) P(X 3)= P(X = 0) + P(X = 1) + P(X = 2)

=

= 0.002 5 + 0.014 9 + 0.044 6 = 0.062 0

(c) P(X 3) = 1 P(X 3)= 1 0.062 0= 0.938 0

12. (a) Let X be the number of errors in an English dictation.Then X ~ Po(5).P(no error) = P(X = 0) = e5 = 0.006 7

(b) P(less than 2 errors) = P(X = 0) + P(X = 1)

= 0.006 7 +

= 0.040 4

13. (a) Let X be the number of accidents in a month.Then X ~ Po(4).

P(2 accidents) = P(X = 2) = = 0.146 5

(b) P(more than 2 accidents) = P(X 2)= 1 P(X = 0) P(X = 1) P(X = 2)

=

= 0.761 9

14. (a) Let X be the number of typhoons in a year.Then X ~ Po(5.5).

P(4 typhoons) = P(X = 4) = = 0.155 8

(b) P(5 to 7 typhoons) = P(X = 5) + P(X = 6) + P(X = 7)

=

= 0.452 0

15. (a) Let X be the demand for chickens on a day.Then X ~ Po(20).

P(15 chickens) = P(X = 15) = = 0.051 7

(b) P(19 to 22 chickens) = P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22)

=

= 0.339 2

225


16. (a) Let X be the number of customers served in one hour.Then X Po(6).

P(X = 8) = = 0.103 3

(b) For 5-minute intervals, the mean number of customers served = 6 = 0.5

The number Y of customers served in a 5-minute interval has the Po(0.5).P(Y > 2) = 1 – P(Y 2)

= 1 – P(Y = 0) – P(Y = 1) – P(Y = 2)

= 1 –

= 0.014 4

(c) For 2-hour periods, the mean number of customers served = 2 6 = 12 The number Z of customers served in a 2-hour period has the Po(12).

P(Z = 10) = = 0.104 8

17. (a) Let X be the number of snails found in a 1 m2 vegetable bed.Then X Po(15).P(X > 4) = 1 – P(X 4)

= 1 – P(X = 0) – P(X = 1) – P(X = 2) – P(X = 3) – P(X = 4)

=

= 0.999 1

(b) For a vegetable bed of area 1.5 m 2 m (i.e. 3 m2), the mean number of snails = 3 15 = 45

The number Y of snails found in this area of vegetable bed has the Po(45).P(Y < 5) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y= 4)

=

= 5.356 10–15

18. (a) Let X be the number of calls in a year.Then X ~ Po(3).P(more than 4 calls)= 1 P(X 4)= 1 P(X = 0) P(X = 1) P(X = 2) P(X = 3) P(X = 4)

=

= 1 0.049 8 0.149 4 0.224 0 0.224 0 0.168 0= 0.184 8

(b) The expected cost= $500 P(X = 1) + $300 P(X = 2) + $300 P(X = 3) + $300 P(X = 4) + $300 P(X = 5)= $500 0.149 4 + $300 (0.224 0 + 0.224 0 + 0.168 0 + 0.100 8)= $289.74

226


19. (a) Let X be the number of bacteria in a sample.Then X ~ Po(3). P(3 or more bacteria) = 1 P(X = 0) P(X = 1) P(X = 2)

=

= 0.576 8

(b) Let Y be the number of samples containing 3 or more bacteria.Then Y ~ B(5, 0.576 8). The required probability = P(Y = 2)

= = 0.252 2

20. (a) Let X and Y be the numbers of containers ordered by the two distributors in a week respectively.Then X ~ Po(1) and Y ~ Po(1.5).P(the total order is exactly 2 containers)= P(X + Y = 2)= P(X = 0) P(Y = 2) + P(X = 1) P(Y = 1) + P(X = 2) P(Y = 0)

=

= 0.256 5

(b) P(X + Y = 0) = P(X = 0) P(Y = 0) = = 0.082 1P(X + Y = 1) = P(X = 0) P(Y = 1) + P(X = 1) P(Y = 0)

=

= 0.205 2P(the total order is more than 2 containers)= P(X + Y 2)= 1 P(X + Y 2)= 1 (0.082 1 + 0.205 2 + 0.256 5)= 0.456 2

21. (a) Let X be the number of patients diagnosed by doctor A in a 10 minute interval.

mean of X = 12 = 2

and X ~ Po(2).

P(doctor A will diagnose 3 patients) = P(X = 3) = = 0.180 4

(b) Let Y be the number of patients diagnosed by doctor B in a 10 minute interval.

mean of Y = 18 = 3

and Y ~ Po(3).P(the 2 doctors will diagnose a total of 3 patients)= P(X + Y = 3)= P(X = 0) P(Y = 3) + P(Y = 1) P(Y = 2) + P(X = 2) P(Y = 1) + P(X = 3) P(Y = 0)

227


=

= 0.135 3 0.224 0 + 0.270 7 0.224 0 + 0.270 7 0.149 4 + 0.180 4 0.049 8= 0.140 4

(c) P(X + Y = 0) = P(X = 0) P(Y = 0)= 0.135 3 0.049 8= 0.006 7

P(X + Y = 1) = P(X = 0) P(Y = 1) + P(X = 1) P(Y = 0)= 0.135 3 0.149 4 + 0.270 7 0.049 8= 0.033 7

P(X + Y = 2) = P(X = 0) P(Y = 2) + P(X = 1) P(Y = 1) + P(X = 2) P(Y = 0)= 0.135 3 0.224 0 + 0.270 7 0.149 4 + 0.270 7 0.049 8= 0.084 2

P(the 2 doctors will diagnose more than 3 patients together)= P(X + Y 3)= 1 P(X + Y 3)= 1 (0.006 7 + 0.033 7 + 0.084 2 + 0.140 4)= 0.735 0

22. (a) Let X i be the number of unhealthy pigs in lot i, i = 1 and 2.Then X i ~ B(300, 0.01),which is approximated by Po(300 0.01), i.e. Po(3).P(no unhealthy pigs between two lots)= P(X1 = 0) P(X2 = 0)= = 0.002 5

(b) P(2 unhealthy pigs between two lots)= P(X1 = 0) P(X2 = 2) + P(X1 = 1) P(X2 = 1) + P(X1 = 2) P(X2 = 0)

=

= 0.044 6

23. (a) Let X be the number of flaws in each roll.Then mean of X = 0.5 10 = 5and X ~ Po(5).P(no flaw in the first row) = e5 = 0.006 738 = 0.006 7 (4 d.p.)

(b) The required probability= P(the first 5 rolls have no flaw) P(the 6th roll contains one or more flaws)= (0.006 738)5 (1 0.006 738)= 1.379 1011

(c) The required probability = = 0.000 7

24. (a) Let X be the number of defective articles in a daily sample.Then X ~ B(300, 0.006),which is approximated by Po(300 0.006), i.e. Po(1.8).P(no defective articles in a sample) = P(X = 0) = = 0.165 3

(b) P(at least 1 defective in a sample) = 1 0.165 3 = 0.834 7The required probability = 0.834 72 = 0.696 7

228


(c) The required probability= 1 P(0 or 1 defective in the 2 samples)= 1 P(X = 0) P(X = 0) P(X = 0) P(X = 1) P(X = 1) P(X = 0)

=

= 0.874 325. (a) By calculator, the sample mean = 3

(b) X may be expected to follow a Poisson distribution because, approximately,(1) the aeroplanes may be assumed to arrive randomly in a specified time interval;(2) the arrivals are independent;(3) no more than one aeroplane can arrive simultaneously;(4) the mean number of arrivals in a given time interval is proportional to the length of the

interval.

(c) (i) X Po(3)P(3 X 5) = P(X = 3) + P(X = 4) + P(X =5)

=

= 0.492 9

(ii) For 1 hour, the mean number of arrivals = 4 3 = 12Hence, the number Y of arrivals in an hour has Po(12).P(Y 5) = P(Y = 0) + P(Y = 1) + + P(Y = 5)

=

= 0.020 3

26. (a) By calculator, = 1.96

(b) The Poisson distribution is appropriate for describing the data because, approximately,(1) the bacterial colonies grow randomly and independently in a continuous region (the

agar plate);(2) no two or more colonies can grow simultaneously at the same point on the plate;(3) the mean number of colonies growing in a given region is proportional to the area of

the region.

(c) Assume X Po(1.96).

(i) P(X 1) = 1 – P(X = 0)= 1 – = 0.859 1

(ii) On a 200-cm2 agar plate, the mean number of colonies growing = 2 1.96 = 3.92Let Y be the number of bacterial colonies on a 200-cm2 agar plate. Then Y Po(3.92).P(Y 2) = 1 – P(Y = 0) – P(Y = 1)

= 1 – = 0.902 4

(iii) Let Z be the number of the 100-cm2 agar plates inspected until the one has bacterial colonies. Then Z has a geometric distribution with probability of success p = 0.859 1, from the result of (i).P(Z = 3) = (1 – 0.859 1)2(0.859 1)

= 0.017 1

229


(iv) Suppose n 100-cm2 agar plates should be inspected. Let U be the number of agar plates, in the sample of n, which have at least one bacterial colony.Then U B(n, 0.859 1), using the result in (i).

P(U 1) = 1 – (0.140 9)n

230

REVISION EXERCISE 15

For this probability to be greater than 0.95,1 – (0.140 9)n > 0.95

(0.140 9)n < 0.05n log 0.140 9 < log 0.05n(–0.851 09) < –1.301 03

n > = 1.529

At least 2 agar plates should be inspected.

REVISION EXERCISE 15 Revision exercise 15(page 372)

1. From the given box-and-whisker diagram, 60 is the upper quartile of the distribution of scores. Hence, the proportion of students having scores higher than 60 is 0.25.Then X has a Bernoulli distribution with probability of success p = 0.25.

(a) The probability distribution of X is

x 0 1

P(X = x) 0.75 0.25

(b) Mean of X = p = 0.25Variance of X = p(1 – p) = 0.187 5

2. X B(5, 0.3)

(a) The probability distribution of X is given byP(X = x) = , for x = 0, 1, 2, … , 5.

The numerical values are shown below.

x 0 1 2 3 4 5

P(X = x) 0.168 1 0.360 2 0.308 7 0.132 3 0.028 4 0.002 4

The distribution is also represented graphically by the following bar chart.

(b) P(X 3) = P(X = 3) + P(X = 4) + P(X = 5)= 0.132 3 + 0.028 4 + 0.002 4, from the table in (a)= 0.163 1

231


(c) E(X) = 5 0.3 = 1.5

E(X) = np

Var(X) = 5 0.3 0.7 = 1.05 Var(X) = npq

3. P(working properly)= 1 P(4 defective components) P(3 defective components)= 1 (0.15)4 (0.15)3 (0.85)= 0.988 0

4. The number of male monkey in a sample ~ B(10, 0.4). The expected value = np = 10 0.4 = 4

The standard deviation = = 1.549

5. (a) Let X be the number of shots A has to fire in order to hit B.Then X has a geometric distribution with probability of success p = 0.6.

P(X = 3) = (0.4)2(0.6) = 0.096

(b) By an argument similar to that in (a),P(the first shot that B will hit A is the 3rd shot) = (0.35)2(0.65)

= 0.079 625= 0.079 6 (4 d.p.)

(c) P(A and B will first hit each other at the 3rd shot)= P(A will first hit B at the 3rd shot) P(B will first hit A at the 3rd shot)= 0.096 0.079 625= 0.007 6 (4 d.p.)

6. (a) Let the first red car observed be the Xth car.Then X has a geometric distribution with probability of success p = 0.2.

P(X = 4) = (0.8)3(0.2) = 0.102 4

(b) Let Y be the number of red cars among the first 8 cars observed.Then Y B(8, 0.2).P(Y 2) = 1 – P(Y = 0) – P(Y = 1)

= = 0.496 7

7. (a) Let X be the number of accidents per day.Then X ~ Po(1.3)

and P(X = x) =

P(exactly one accident) = P(X = 1) = = 0.354 3

(b) P(at least two accidents) = 1 P(X = 0) P(X = 1)

= 1 0.354 3

= 0.373 2

232


8. For the geometric distribution of X,

mean = = 2

p = 0.5

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

=

= 0.112 0

9. (a) The Poisson distribution model is the most appropriate.The required assumptions are:(1) the numbers of calls occurring in non-overlapping time intervals are independent;(2) the probability of a single call occurring in a short time interval is proportional to the

length of the time interval;(3) the probability of more than one call in such a short time interval is negligible.

(b) Let X be the number of calls in a minute.X ~ Po(3)

P(more than 4 calls)= P(X 4)= 1 P(X = 0) P(X = 1) P(X = 2) P(X = 3) P(X = 4)

=

= 0.184 7

10. P(a box contains 3 defective articles and is delivered)

=

= 0.030 0

11. Suppose that in the sample of 12 bus drivers, X of them have stomach problem. Then X B(12, 0.05).

(a) P(X 2) = P(X = 0) + P(X = 1) + P(X = 2)= = 0.980 4

(b) The mean of the binomial distribution = 12 0.05 = 0.6Hence, the Poisson approximation is Po(0.6).

P(X 2)

= 0.976 9

12. [Public Examination question](a) 0.384 0 (b) 0.139 5

13. [Public Examination question](a) 0.036 (b) 0.881 7 (c) 0.025 9

14. [Public Examination question](a) 0.066 8 (b) 0.431 1 (c) 0.135 9

233


15. [Public Examination question](a) 161.5 cm (b) (i) 0.104 7 (ii) 0.068 4

16. [Public Examination question](a) 0.200 7 (b) 6 (c) 0.128 2

17. [Public Examination question](a) 160 160 (b) 10, 11, 12, 13, 14, 15, 16 (c) 12

18. [Public Examination question]0.925 5

19. (a) (i) Let X be the number of sales in a day.

Then X ~ .

P(X = 4) =

(ii) P(X = 3) =

(iii) P(X = 2) =

P(X 2) =

(b) The expected value of daily bonus

= $

= $10.56

20. (a) (i) Let X be the number of breakdowns in a month.Then X ~ Po(5).

P(X = 3) = = 0.140 4

(ii) P(3 breakdowns in each of 4 successive months)

=

= 0.000 4(iii) For a 2-month period, the average rate of breakdown = 2 5 = 10

P(4 breakdowns in a 2-month period)

=

= 0.018 9

(b) The assumption that the number of breakdowns follows a Poisson distribution is reasonable. This is because we can regard breakdowns occur independently and the probability of a single breakdown occurring in a short time interval is proportional to the length of the time interval.

234


21. (a) (i) Let X be the number of faults in a blanket.Then X ~ Po(3).P(at least 1 fault) = 1 P(X = 0)

= 1

= 0.950 2

(ii) P(at most 2 faults) = P(X = 0) + P(X = 1) + P(X = 2)

=

= 0.423 2

(b) P(X = 3) = = 0.224 042

The required probability = (0.224 042)2 (0.775 958)2

= 0.306 8

22. (a) Let X be the number of people in the sample who are opposed to the project.Then X ~ B(30, 0.8)The required probability = P(X 26)

= P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30)= + = 0.122 7

(b) The required probability = (0.8)2 (0.2)= 0.128

(c) The required probability = = 0.078 6

23. (a) It is reasonable to assume that the probability for the pointer to fall onto a particular sector is proportional to the angle of the sector. Then the probability distribution of X is shown in the following table.

x 1 2 3 4 5

P(X = x)

(b) P(even number) = P(X = 2) + P(X = 4)

=

=

(c) P(product is even) = P(at least one of the numbers is even)= 1 – P(both numbers are odd)

= , using the result in (b)

=

235


(d) Suppose that out of the 6 times, Florence gets an even number Y times.

Then Y B(6, ), using the result p = in (b).

P(Y = 3) =

= 0.301 1

(e) P(X = 3) = =

Let Z be the number of spins required to get the first “3”.

Then Z has a geometric distribution with probability of success p = .

P(Z = 5) =

= 0.081 3

24. (a) (i) Area of region A = 52 cm2 = 25 cm2

Area of region B = (102 – 52) cm2 = 75 cm2

Area of region C = (152 – 102) cm2 = 125 cm2

Total area of the target = 225 cm2

P(hitting region A) =

P(hitting region B) =

P(hitting region C) =

The probability distribution of X is shown below.

x 1 5 10

P(X = x)

(ii) Expectation of Joe’s score in one throw, E(X)=

=

=

(b) (i) P(scoring 5 points both times)= [P(X = 5)]2

=

236


=

(ii) P(total score > 10 points)= P(at least one of the scores = 10)= 1 – P(none of the 2 scores is 10)

= 1 –

=

(c) Let Y be the number of times Joe hits region A.

Then Y B(18, ).

(i) P(Y = 3) =

= 0.191 3

(ii) Expected number of times Joe hits region A= E(Y)

=

= 2

(d) (i) P(hitting region A at least once)= 1 – P(not hitting region A in any of the n throws)

= 1 –

(ii) For the above probability to be greater than 0.5,

> 0.5

< 0.5

< log 0.5

n > 5.88 The least value of n is 6.

25. (a) The bar chart is shown below.

237

E(Y) = np


(b) The distribution of X shows a slightly positive skewness. The probabilities P(X = 2) and P(X = 3) are equal; this is a characteristic of the Po(3).

(c) By calculator, = 3.003 = 1.735 8

(d)= P(3.003 – 1.735 8 < X < 3.003 + 1.735 8)= P(1.267 < X < 4.739)= P(2 X 4), since X takes integer values in the table= 0.224 + 0.224 + 0.168, from the given table= 0.616

(e) P(X > 5) = 0.050 + 0.023 + 0.008 + 0.003 + 0.001, from the given table= 0.085

Let Y be the number of observed values of X which are greater than 5.Then Y B(6, 0.085).

(i) P(Y = 2) = = 0.076 0

(ii) P(Y 2) = 1 – P(Y = 0) – P(Y = 1)= = 0.086 1

(f) Suppose we simulate 1 000 values of X.According to the given probability distribution, there should be 50 “0”s, 149 “1”s, … , 1 “10”. Without loss of generality, we can assume these values to be lined up consecutively so that the first 50 values are “0”s, the next 149 values are “1”s, … , the last value is “10”, as shown below:

…

Each value is given a 3-digit address, from 000 to 999. Now, using the random number generator of the calculator, generate a 3-place decimal number. Ignoring the decimal point, if the 3-digit number is in the interval [000, 049], consider it as a “0”; if the number is in the interval [050, 198], consider it as a “1”, and so on. Continue generating numbers in this way, we obtain the simulated values of X, e.g.

0, 2, 2, 1, 3, 4, 3, 5, …

26. Area of each of the 7 photos = 20 25 cm2 = 500 cm2

Average rate of bad spots per photo = 5 0.5 = 2.5Let X be the number of bad spots on a photo. Then X Po(2.5).

(a) (i) P(X = 0) = = 0.082 1

(ii) P(X = 1) = = 0.205 2

(iii) P(X = 2) = = 0.256 5

(iv) P(X 2) = 1 – P(X = 0) – P(X = 1)= 1 – 0.082 1 – 0.205 2= 0.712 7

238


239


(b) P(X = 2 | X 1)=

=

=

= , from the results in (a)(iii) and (a)(i)

= 0.279 4

(c) Let Y be the number of photos rejected, i.e. photos with 2 or more bad spots. Then Y B(7, 0.712 7), from the result in (a)(iv).

(i) P(Y = 0) = (0.287 3)7 = 0.000 2

(ii) P(Y = 2) = = 0.020 9

(d) For plant B, the average rate of bad spots on a finished photo of size 20 cm 25 cm = 5 P(no bad spots on a photo) = Using Po(5)

= 0.006 74The situation of the shop is represented by the tree diagram below.

With obvious notations,P(plant B | no bad spots)

=

=

= 0.034 0

27. For a glass pane of size 20 cm 20 cm (i.e. 400 cm2), the average number of flaws = 0.15 4 = 0.6

(a) Let X be the number of flaws in a glass pane. Then X Po(0.6).

(i) P(X = 0) = = 0.548 81 = 0.548 8 (4 d.p.)

(ii) P(X = 1) = = 0.329 29 = 0.329 3 (4 d.p.)

(iii) P(X 1) = 1 – P(X = 0)

240

Bayes’ Theorem


= 1 – 0.548 81= 0.451 19= 0.451 2 (4 d.p.)

(b) From (a)(iii), P(a flawed pane) = P(X 1) = 0.451 19Let Y be the number of flawed panes in a window.Then Y B(6, 0.451 19).

(i) P(Y = 0) = (0.548 81)6

= 0.027 32= 0.027 3 (4 d.p.)

(ii) P(Y = 1) = = 0.134 78= 0.134 8 (4 d.p.)

(iii) P(Y = 2) = = 0.277 01= 0.277 0 (4 d.p.)

(iv) P(Y > 2) = 1 – [P(Y = 0) + P(Y = 1) + P(Y = 2)]= 1 – (0.027 32 + 0.134 78 + 0.277 01)], from (i), (ii) and (iii)= 1 – 0.439 11= 0.560 9 (4 d.p.)

(c) P(a window is sold in the shop) = P(Y 2)= 0.439 11, from (b)(iv)

(i) P(window is flawless | window is sold in the shop)= P(Y = 0) | Y 2)

=

=

= , from (b)(i) and (iv)

= 0.062 22= 0.062 2 (4 d.p.)

(ii) Let Z be the number of windows the customer has to inspect to find a flawless one. Then Z has a geometric distribution with probability of success p = 0.062 22.P(Z = 3) = (1 – 0.062 22)2(0.062 22)

= 0.054 7

28. (a) The length of the chicken run = (20 – 2x) m A = (20 – 2x)x = 20x – 2x2

(b) (i) The probability distribution of A is shown below.

x 3.5 4 4.5 5 5.5

A = 20x – 2x2 45.5 48 49.5 50 49.5

P(X = x) 0.1 0.15 0.25 0.3 0.2

241


P(A > 49) = P(4.5 X 5.5)= 0.25 + 0.3 + 0.2, using the given probability distribution= 0.75

(ii) E(A) = = 45.5 0.1 + 48 0.15 + + 49.5 0.2= 49.025

i.e. The expected area of the chicken run = 49.025 m2

(c) A = 20x – 2x2

= 20 – 4x

For A to have a stationary value, = 0,

i.e. x = 5

= –4, which is negative.

Hence A is a maximum when x = 5. Maximum area of the chicken run

= (20 5 – 2 52) m2

= 50 m2

(d) (i) Let Y be the number of spots per lattice per day.Then Y Po(5).P(Y = 1) =

= 0.033 7

(ii) P(a lattice has no spots)= P(Y = 0)= = 0.006 74

P(2 out of 6 lattices have no spots)= = 0.000 7

29. [Public Examination question](a) The standard deviation of 4.7 is surprising.(b) 10.07; 1.834 (c) 0.100 3 (d) 10.57; 1.922

30. [Public Examination question](a)(i) 0.163 1 (ii) 0.486 8 (iii) 0.124 7; 0.191 0(b) 0.017 9

31. [Public Examination question](a)(i) 0.226 9 (ii) 0.429 (iii) 0.49(b)(i) 0.02 (ii) 0.83 (iii) 0.024 1

32. [Public Examination question]

(b) 0.052 8 (c) 0.418 4 (d) 0.678 6 (e)

33. [Public Examination question]

242


(a) 0.096 1 (b) 10.4 (c)(i) 0.644 5 (ii) 0.205 8(d) 1.020 6

243


34. [Public Examination question](a) 137 days (b) 0.99 (c) (i) 0.950 4 (ii) 0.925 9 (iii) 0.597 5

35. [Public Examination question](a) 0.323 3 (b) 2 (c) (i) 0.648 9 (ii) 0.166 9 (iii) 0.236 9

36. [Public Examination question](a) 0.519 5 (b) (i) 0.037 8 (ii) 0.008 7(c)

(d) (ii)

37. [Public Examination question](a) 1 300 (b) (i) 0.966 4 (ii) 0.033 2 (c) 0.012 0

38. [Public Examination question](a) (i) 0.040 8 (ii) 0.067 2 (iii) 0.997 3 (iv) 0.047 7(b) (ii) 0.018 3

39. [Public Examination question](a) 0.003 5 (b) (i) 0.106 0 (ii) 0.056 3 (c) 0.031 3(d) farmer (e) 2

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