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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
Additional Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
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
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Additional Mathematics SPM Chapter 19
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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
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Additional Mathematics SPM Chapter 19
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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
Additional Mathematics SPM Chapter 19
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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
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Additional Mathematics SPM Chapter 19
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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
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Additional Mathematics SPM Chapter 19
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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
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Additional Mathematics SPM Chapter 19
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(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
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Additional Mathematics SPM Chapter 19
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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
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Additional Mathematics SPM Chapter 19
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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%