19[Anal Add Math CD]

11

Additional Mathematics SPM Chapter 19

© Penerbitan Pelangi Sdn. Bhd.

1. (a) X = {Number of heads} = {0, 1, 2, 3}

(b) X = {Number of blue marbles} = {0, 1, 2}

(c) X = {Number of trials that hit the target} = {0, 1, 2, 3, 4, 5, 6, 7, 8}

2. Let X represents the number of heads.p = P(H), q = P(T) = 1—

2 = 1—

2

(a) P(X = 3) = 3C3 p3q0

= 3C3 1 1—2 2

3

1 1—2 2

0

= 1—8

(b) P(X = 1) = 3C1 p1q2

= 3C1 1 1—2 2

1

1 1—2 2

2

= 3—8

(c) P(X > 1) = P(X = 1) + P(X = 2) + P(X = 3) = 3C1 p

1q2 + 3C2 p2q1 + 3C3 p

3q0

= 3C1 1 1—2 21 1—

2 22 + 3C2 1 1—

2 22

1 1—2 2

+ 3C3 1 1—2 2

3

1 1—2 2

0

= 3—8

+ 3—8

+ 1—8

= 7—8

Alternative P(X > 1) = 1 – P(X = 0) = 1 – 3C0 p

0q3

= 1 – 3C0 1 1—2 2

0

1 1—2 2

3

= 1 – 1—8

= 7—8

3. Let X represent the number of blue marbles obtained.

p = 8–––20

, q = 3—5

= 2—5

(a) P(X = 1) = 5C1 p1q4

= 5C1 1 2—5 21 3—

5 24

= 0.2592

(b) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 5C0 p

0q5 + 5C1 pq4 + 5C2 p2q3

= 5C0 1 2—5 2

0

1 3—5 2

5 + 5C1 1 2—

5 21 3—5 2

4

+ 5C2 1 2—5 2

2

1 3—5 2

3

= 0.6826

(c) P(X > 1) = 1 – P(X = 0) = 1 – 5C0 p

0q5

= 1 – 5C0 1 2—5 2

0

1 3—5 2

5

= 0.9222

4. 2(y + 2y) = 1 3y = 1—

2 y = 1—

6

5. p = 1—5

, q = 4—5

, n = 50

(a) Mean = np = 50 × 1—

5 = 10

(b) Variance = npq

= 10 × 4—5

= 8 Standard deviation = AB8

CHAPTER

19 Probability Distributions

22



6. Let X represents the number of goals scored.p = 0.85, q = 0.15(a) P(X = 10) = 10C10 p

10q0

= 10C10 (0.85)10(0.15)0

= 0.1969(b) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 10C0 p

0q10 + 10C1 p1q9 + 10C2 p

2q8 + 10C3 p3q7

= 10C0(0.85)0(0.15)10 + 10C1(0.85)1(0.15)9 + 10C2(0.85)2(0.15)8 + 10C3(0.85)3(0.15)7

= 0.0001346

7. np = 120 ............................... 1npq = 40 ................................. 2

Substitute 1 into 2,120q = 40 q = 1—

3 p = 1 – q = 2—

3Substitute p = 2—

3 into 1,

n1 2—3 2 = 120

n = 120 × 3—2

= 180

8. Y = {y : 29°C < y < 39°C}

9. (a) P(Z < 1) = 1 – 0.1587 = 0.8413

f(z)

z0

0.8413

1

(b) P(–1.2 , Z , – 0.52) = P(Z , –0.52) – P(Z , –1.2) = 0.3015 – 0.1151 = 0.1864

f(z)

z0

0.1864

–0.52–1.2

(c) P(Z , 1.043) = P(–1.043 , Z , 1.043) = 1 – 2P(Z . 1.043) = 1 – 2 × 0.1485 = 0.703

f(z)

z0

0.703

–1.043 1.043

(d) P(Z . 0.416) = P(Z , –0.416 or Z . 0.416) = P(Z , –0.416) + P(Z . 0.416) = 0.3387 + 0.3387 = 0.6774

10. µ = 50, s = 5

(a) z = x – µ–––––s

= 65 – 50–––––––5

= 3

(b) z = 45 – 50–––––––5

= –1

(c) z = 52.4 – 50–––––––––5

= 0.48

11. µ = 1.5, s = 0.3Let X represents the diameters of ball bearing(a) P(X . 1.8)

(b) P(X , 1.2)

(c) P(1.2 , X , 1.8)

12. (a) P(X . 1.8) = P1Z . 1.8 – 1.5––––––––0.3 2

= P(Z . 1) = 0.1587

(b) P(X , 1.2) = P1Z , 1.2 – 1.5––––––––0.3 2

= P(Z , –1) = 0.1587

(c) P(1.2 , X , 1.8)

= P1 1.2 – 1.5––––––––0.3 , Z , 1.8 – 1.5––––––––

0.3 2 = P(–1 , Z , 1) = 1 – 2P(Z . 1) = 1 – 2(0.1587) = 0.6826

33



13. µ = m gram, s = 0.2 gramLet X be the weight of the prawn. P(X , 2) = 0.1

P1Z , 2 – m––––––0.2 2 = 0.1

zm = –1.282

2 – m––––––0.2 = –1.282

m = 2 + 1.282 × 0.2 = 2.256

f(z)

z0

0.1

zm

14. µ = 12 cm, s = 0.08 cmLet X be the length of straw. P(X . t) = 0.8

P1Z . t – 12––––––0.08 2 = 0.8

zt = – 0.842

t – 12––––––0.08 = – 0.842

t = 12 – 0.842 × 0.08 = 11.93

f(z)

z0

0.8

zt

15. (a) µ = 25 cm, s = m cm Let X represents the diameter of the wheel.

P(X . 26) = 1––––100

P1Z . 26 – 25–––––––m 2 = 0.01

zm = 2.326

26 – 25–––––––m = 2.326

m = 1–––––––2.326

= 0.4299

f(z)

z0

0.01

zm

1. (a) Probability of picking a white marble, p = 1—5

\ Number of white marbles are expected to be picked

= n × p

= 20 × 1—5

= 4

(b) Let X represent the number of time the white marble is picked for the remaining 10 trials.

(i) P(X = 4) = 10C41 1—5 2

4

1 4—5 2

6

= 0.08808

(ii) P(X 12) = 1 – P(X = 0) – P(X = 1)

= 1 – 10C01 1—5 2

0

1 4—5 2

10 – 10C11 1—

5 21

1 4—5 2

9

= 0.6242

2. (a) P(Z . k) = 0.5 – 0.697–––––2

= 0.1515 k = 1.03

(b) µ = 80, s 2 = 9 s = 3

zk = x – µ–––––s 1.03 = x – 80––––––3 x = 80 + 3 × 1.03 = 83.09

3. (a) The mean of the distribution of success = np

= 8 × 11 – 1—4 2

= 8 × 3—4

= 6

(b) P(X > 1) = 1 – P(X = 0)

= 1 – 8C0 1 3—4 201 1—

4 28

= 1 – 1 1—4 28

= 1 – 1––––––

65 536

= 65 535––––––65 536

44



4. µ = 6.3, s = ABBB1.21 = 1.1

(a) z = x – µ–––––s = 5.9 – 6.3––––––––1.1 = – 0.3636

(b) P(5.9 < X < 6.3)

= P1 5.9 – 6.3––––––––1.1 < Z < 6.3 – 6.3––––––––1.1 2 = P(–0.3636 < Z < 0) = 0.5 – P(Z > 0.3636) = 0.5 – 0.3581 = 0.1419

f(z)

z0

0.1419

–0.3636

5. (a) 4k2 + 2t + 12

= 1

2t = 1 – 12

– 4k2

= 12

– 4k2

t = 14

– 2k2

(b) When k = 14

,

t = 14

– 21 14 22

= 18

4k2 = 41 14 22

= 14

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

= 4k2 + t + 12

= 14

+ 18

+ 12

= 78

6. p = 0.3, q = 0.7Let X be the number of girls selected.(a) P(X > 3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)] = 1 – [8C0 p

0q8 + 8C1 p1q7 + 8C2 p

2q6]

= 1 – [8C0(0.3)0(0.7)8 + 8C1(0.3)1(0.7)7 + 8C2(0.3)2(0.7)6]

= 0.4482

(b) npq = 212

n(0.3)(0.7) = 212

n = 212––––––––(0.3)(0.7)

= 2100 The population of the pupils is 2100.

7. (a) n = 8, mean = 3.2 (i) Mean = 3.2 np = 3.2 8p = 3.2 p = 0.4

(ii) q = 1 – 0.4 = 0.6 Let X be the number of times hitting the

target. P(X > 1) = 1 – P(X = 0) = 1 – 8C0 p

0q8

= 1 – 8C0(0.4)0(0.6)8

= 0.9832

(b) µ = 55.5 kg, s = 12 kg Let X represents the mass of worker. (i) P(X , 45)

= P1Z , 45 – 55.5–––––––––12 2

= P(Z , – 0.875) = 0.1908

f(z)

z0

0.1908

–0.875

(ii) P(X . k) = 0.15

P1Z . k – 55.5––––––––12 2 = 0.15

zk = 1.036

k – 55.5––––––––12 = 1.036

k = 12(1.036) + 55.5 = 67.93

f(z)

z0

0.15

zk

55



8. µ = 12 g, s = 4 g(a) P(Grade B) = P(15 , X < 20)

= P1 15 – 12–––––––4 , Z < 20 – 12–––––––

4 2 = P(0.75 , Z < 2) = P(Z . 0.75) – P(Z . 2) = 0.2266 – 0.0228 = 0.2038

f(z)

z0 0.75

0.2038

2

(b) P(Grade A) = P(X . 20)

= P1Z . 20 – 12–––––––4 2

= P(Z . 2) = 0.0228

Number of eggs of grade A = 0.0228 × 20 000 = 456

(c) P(X , m) = 0.75

P1Z , m – 12–––––––4 2 = 0.75

m – 12–––––––4 = zm

= 0.674 m = 4 × 0.674 + 12 = 14.696

f(z)

z0

0.75

zm

1. (a) P(X = 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1 – 0.1 – 0.2 – 0.3 = 0.4

(b) P(X > 1) = P(X = 1) + P(X = 2) + P(X = 3) = 0.2 + 0.3 + 0.4 = 0.9

2. Let P be the probability running below 15 seconds.p = 0.6, q = 0.4n = 8

P(exactly 5 persons ran below 15 seconds)= 8C5 p

5q3

= 8C5(0.6)5(0.4)3

= 0.2787

3. Let p represents the probability a family has at least one car, p = 0.9, q = 0.1, n = 10Let X represents the number of families having at least one car.P(X > 8) = P(X = 8) + P(X = 9) + P(X = 10) = 10C8 p

8q2 + 10C9 p9q1 + 10C10 p

10q0

= 10C8(0.9)8(0.1)2 + 10C9(0.9)9(0.1) + 10C10(0.9)10(0.1)0

= 0.9298

4. Let p represents the probability that the bulb is not defective.p = 0.8 and q = 0.2, n = 8

Let X be the number of bulbs that are not defective.P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 8C0 p

0q8 + 8C1 p1q7 + 8C2 p

2q6

= 8C 0(0.8) 0(0.2) 8 + 8C 1(0.8) 1(0.2) 7 + 8C2(0.8)2(0.2)6

= 0.001231

5. P(Z . k) = 0.1 k = 1.282 P(0 , Z , k) = 2P(m , Z , 0) 0.5 – 0.1 = 2P(m , Z , 0) P(m , Z , 0) = 0.2 m = – 0.525

f(z)

z0

0.2

0.3 m

6. (a) P(Z , k) = 0.3 k = – 0.524

(b) z = x – µ–––––s

– 0.524 = x – 63––––––3 x = 63 + 3(– 0.524) = 61.428

66



7. z = x – µ–––––s

0.5 = 58 – µ––––––2 µ = 58 – 2(0.5) = 57

8. µ = 60 kg, s = 5 kg(a) P(60 , X , k) = 0.35

P1 60 – 60–––––––5

, Z , k – 60––––––5 2 = 0.35

P10 , Z , k – 60––––––5 2 = 0.35

k – 60––––––5

= 1.036 k = 60 + 5(1.036) = 65.18

f(z)

z0

0.35

5–k – 60

(b) P(X . k) = P(X . 65.18)

= P1Z . 65.18 – 60––––––––––5 2

= P(Z . 1.036) = 0.15

The number of washing machines that have mass greater than k

= 1200 × 0.15 = 180

9. Median = mode = 15 cm, s = 1 cmLet X be the length of ruler.(a) P(X , 14.5)

= P1Z , 14.5 – 15––––––––1 2

= P(Z , –0.5) = 0.3085

(b) P(X . 14.5) = 1 – 0.3085 = 0.6915

Number of rulers that have length more than 14.5 cm

= 0.6915 × 50 000 = 34 575

10. (a) p = probability on target = 0.7 q = 1 – 0.7 = 0.3 n = 8

Let X represents the number of tries that hit the target.

(i) P(X . 6) = P(X = 7) + P(X = 8) = 8C7 p

7q1 + 8C8 p8q0

= 8C7(0.7)7(0.3)1 + 8C8(0.7)8(0.3)0

= 0.2553

(ii) P(X′ = 3) = 8C3 p5q3

= 8C3(0.7)5(0.3)3

= 0.2541

(b) µ = mean = median = 500 ml, s = 10 ml Let X be the volume of bottle. (i) P(488 , X , 502)

= P1 488 – 500–––––––––10

, Z , 502 – 500–––––––––10 2

= P(–1.2 , Z , 0.2) = 1 – P(Z , –1.2) – P(Z . 0.2) = 1 – 0.1151 – 0.4207 = 0.4642

f(z)

z0

0.4642

0.2–1.2

(ii) P(500 , X , t) = 0.4

P1 500 – 500–––––––––10

, Z , t – 500–––––––10 2 = 0.4

P10 , Z , t – 500–––––––10 2 = 0.4

t – 500–––––––10

= 1.282

t = 1.282 × 10 + 500 = 512.82

10–t – 500

f(z)

z0

0.4

11. (a) p = The probability of family having one child. Mean = 300, variance = 100 Mean = 300 np = 300 .................... 1

Variance = 100 npq = 100 ............... 2

Substitute 1 into 2, 300q = 100

q = 1—3

p = 1 – q

= 2—3

77



Substitute p = 2—3

into 1,

n1 2—3 2 = 300

n = 300 × 3—2

= 450 Therefore, p = 2—

3 and n = 450.

(b) Let X be the marks. (i) Let m be the minimum mark to score band

1. P(X > m) = 0.2

P1Z . m – 70–––––––5 2 = 0.2

zm = 0.842

m – 70–––––––5 = 0.842

m = 70 + 5 × 0.842 = 74.21

(ii) Let t be the maximum mark for band 3. P(X , t) = 0.1

P1Z , t – 70––––––5 2 = 0.1

t – 70––––––5 = –1.282

t = 70 + 5(–1.282) = 63.59

zm

f(z)

z0zt

0.1 0.2

12. (a) Mean = 1500 months, standard deviation = 30 months

Let X be the life span. (i) P(X , 1480)

= P1Z , 1480 – 1500–––––––––––30 2

= P(Z , – 0.667) = 0.2524

f(z)

z0–0.667

0.2524

(ii) P(X . n) = 0.08

P1Z . n – 1500––––––––30 2 = 0.08

n – 1500––––––––30 = 1.405

n = 1500 + 30(1.405) = 1542.15

f(z)

z0 z = 1.405

0.08

Therefore, n = 1542

(b) p = probability of hitting the target

= 1—2

q = 1—2

(i) Let X be the number of times hitting the target.

P(X = 1) = P(X = 2) nC1 p

1qn – 1 = nC2 p2qn – 2

n1 1—2 21 1—

2 2n – 1

= n(n – 1)–––––––2 1 1—

2 22

1 1—2 2

n – 2

n1 1—2 2

n = n(n – 1)–––––––

2 1 1—2 2

n

n = n(n – 1)–––––––2

2n = n2 – n n2 – 3n = 0 n(n – 3) = 0 n = 0, 3 Therefore, n = 3.

(ii) Mean = np = 50

r1 1—2 2 = 50

r = 100

13. (a) p = 0.8, q = 0.2 Let X be the number of scores.

(i) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 8C0 p

0q8 + 8C1 p1q7 + 8C2 p

2q6 + 8C3 p3q5

= 8C0(0.8)0(0.2)8 + 8C1(0.8)(0.2)7 + 8C2(0.8)2(0.2)6 + 8C3(0.8)3(0.2)5

= 0.01041

88



(ii) P(X > 1) = 0.92 1 – P(X = 0) = 0.92 1 – nC0 p

0qn = 0.92 1 – (1)(1)(0.2)n = 0.92 (0.2)n = 1 – 0.92 = 0.08 n log10 0.2 = log10 0.08

n = log10 0.08

–––––––––log10 0.2

= 1.569

Therefore, n = 2

(b) (i) z = x – µ–––––s

–1 = 10 – µ––––––s 10 – µ = –s µ – s = 10 ......................... 1

2 = 12 – µ––––––s 2s = 12 – µ µ + 2s = 12 ........................ 2

2 – 1, 3s = 2 s = 2—

3

Substitute s = 2—3

into 1,

µ – 2—3

= 10

µ = 10 + 2—3

= 10 2—3

(ii) P(9 , X , 11)

= P1 9 – 10 2—3––––––––

2—3

, Z , 11 – 10 2—

3–––––––––2—3

2 = P(–2.5 , Z , 0.5) = 1 – P(Z , –2.5) – P(Z . 0.5) = 1 – 0.00621 – 0.3085 = 0.6853

f(z)

z0 0.5–2.5

0.6853

Hence, the percentage is 68.53%.

14. (a) (i) Mean = 12 np = 12 ............. 1

Variance = 4 npq = 4 .......... 2

Substitute 1 into 2, 12q = 4 q = 1—

3

p = 2—3

In 1, n1 2—3 2 = 12

n = 12 × 3—2

= 18

Therefore, p = 2—3

and the number of trials, n = 18

(ii) P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8) = 8C6 p

6q2 + 8C7 p7q1 + 8C8 p

8q0

= 8C6 1 2—3 2

6

1 1—3 2

2 + 8C7 1 2—

3 27

1 1—3 2

1

+ 8C8 1 2—3 2

8

1 1—3 2

0

= 0.4682

(b) µ = 5.2, s = 0.5 Let X represents the time, in hour. (i) P(X , 5)

= P1Z , 5 – 5.2–––––––0.5 2

= P(Z , – 0.4) = 0.3446

(ii) P(X . t) = 0.6

P1Z . t – 5.2––––––0.5 2 = 0.6

zt = – 0.253

t – 5.2––––––0.5 = – 0.253

t = (– 0.253)(0.5) + 5.2 = 5.0735

f(z)

z0zt

0.6

99



1. (a) There are 2 vowels, a and e.

P(vowel) = 2—8

= 1—4

(b) (i) 10C81 1—4 2

8

1 3—4 2

2 = 0.0003862

(ii) P(X > 2) = 1 – [P(X = 0) + P(X = 1)]

= 1 – 310C01 1—4 2

0

1 3—4 2

10 + 10C11 1—4 2

1

1 3—4 2

9

4 = 0.756

(iii) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 10C01 1—4 2

0

1 3—4 2

10 + 10C11 1—

4 21

1 3—4 2

9

+ 10C21 1—4 2

2

1 3—4 2

8

= 0.5256

2. (a) µ = 55, s = 10 Let X be the mark.

P(X > 35) = P1Z . 35 – 55–––––––10 2

= P(Z . –2) = 1 – 0.0228 = 0.9772

0

0.9772

f(z)

z–2

(b) 0.9772 × 500 = 488

(c) P(Z . t) = 0.13 t = 1.127

0

13%

f(z)

zt

x – 55––––––10 = 1.127

x = 11.27 + 55 = 66.27

Therefore, the minimum mark is 66.27.

3. Let p be the percentage who fail and q be the percentage who pass.p = 40%, q = 60% = 0.4 = 0.6

Let X be the number of candidates who fail.

P(X > 8) = P(X = 8) + P(X = 9) + P(X = 10) = 10C8(0.4)8(0.6)2 + 10C9(0.4)9(0.6)1 + 10C10(0.4)10(0.6)0

= 0.01229

4. Let p be the percentage who fail to get a credit and q be the percentage who pass with a creditp = 10% q = 90% = 0.1 = 0.9

Let X be the number of students who fail to get a credit.

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 8C0(0.1)0(0.9)8 + 8C1(0.1)1(0.9)7 + 8C2(0.1)2(0.9)6

= 0.9619

5. Let p be the percentage of pencils which are defective and q be the percentage of pencils which are not defective.p = 10% q = 90% = 0.1 = 0.9

Let X be the number of defective pencils.

(a) P(X = 0) = 20C0(0.1)0(0.9)20

= 0.1216

(b) P(X = 2) = 20C2(0.1)2(0.9)18

= 0.2852

6. p = probability of arrival on time = 0.9q = probability of arrival not on time = 0.1

(a) Let X be the number of flights on time. P(X > 5) = P(X = 5) + P(X = 6) = 6C5(0.9)5(0.1)1 + 6C6(0.9)6(0.1)0

= 0.8857

(b) Let Y be the number of ‘Good service’ award. P(Y > 9) = P(Y = 9) + P(Y = 10) + P(Y = 11)

+ P(Y = 12) = 12C9(0.8857)9(0.1143)3 + 12C10(0.8857)10(0.1143)2

+ 12C11(0.8857)11(0.1143) + 12C12(0.8857)12(0.1143)0

= 0.9603

1010



7. Let X be the radius of the wheel. P(X . 12.52) = 5%

P1Z . 12.52 – µ––––––––s 2 = 0.05

12.52 – µ–––––––––s = 1.645

12.52 – µ = 1.645s µ = 12.52 – 1.645s ...................1

0

5%

f(z)

z1.645

P(X , 11.02) = 1%

P1Z , 11.02 – µ––––––––s 2 = 0.01

11.02 – µ––––––––s = –2.326

11.02 – µ = –2.326s ..............................2

0

f(z)

z–2.326

1%

Substitute 1 into 2, 11.02 – (12.52 – 1.645s) = –2.326s 11.02 – 12.52 = –2.326s – 1.645s –1.5 = –3.971s

s = 1.5–––––3.971

= 0.3777

Substitute s = 0.3777 into 1,µ = 12.52 – 1.645(0.3777) = 11.9Therefore, the mean is 11.9 and the standard deviation is 0.3777.

8. (a) µ = 40 kg, s = 5 kg Let X represents the weight of the student.

P(X . 50) = P1Z . 50 – 40–––––––5 2

= P(Z . 2) = 0.0228

0

f(z)

z2

0.0228

Therefore, the percentage is 2.28%.

(b) µ = t kg, s = 5 kg P(X . 60) = 5%

P1Z . 60 – t––––––5 2 = 0.05

60 – t––––––5 = 1.645

t = 60 – 5 × 1.645 = 51.775

0

f(z)

z1.645

5%

19[Anal Add Math CD]

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