The binomial distribution
Chapter 11 The binomial distribution
Exercise 11A—The binomial distribution
1 a No, since there are more than 2 possible outcomes.
b Yes, since there are 10 independent trials and two
possible outcomes, three or non–three.
c No, since there are more than 2 possible outcomes.
d Yes, since there are 15 independent trials and two
possible outcomes, tail or non–tail (head).
e No, since the probability of success is not fixed.
f Yes, since they are independent, and two possible
outcomes, black or red.
g No, since the probability of success is not fixed and
‘success’ is not defined.
2 a 7C2(0.4)2(0.6)5 = 21 × 0.16 × 0.077 76
= 0.2613
b 9C3(0.1)3(0.9)6 = 84 × 0.001 × 0.531 441
= 0.0446
c 10C5(0.5)5(0.5)5 = 252 × 0.031 25 × 0.031 25
= 0.2461
d 8C5(0.2)5(0.8)3 = 56 × 0.000 32 × 0.512
= 0.0092
7
2
1
2
1
4
e 9C7 --- --- = 36 × ------------ × --3
3
2187 9
f
3 a
b
c
16
= -----------2187
= 0.0073
10
C0(0.85)10(0.15)0 = 1 × 0.196 87 × 1
= 0.1969
n=5
p = 0.3
Pr(X = x) = 5Cx(0.3)x(0.7)5 − x
Pr(X = 0) = 5C0(0.3)0(0.7)5
= 1 × 1 × 0.168 07
= 0.168 07
Pr(X = 1) = 5C1(0.3)1(0.7)4
= 5 × 0.3 × 0.2401
= 0.360 15
Pr(X = 2) = 5C2(0.3)2(0.7)3
= 10 × 0.09 × 0.343
= 0.3087
Pr(X = 3) = 5C3(0.3)3(0.7)2
= 10 × 0.027 × 0.49
= 0.1323
Pr(X = 4) = 5C4(0.3)4(0.7)1
= 5 × 0.0081 × 0.7
= 0.028 35
Pr(X = 5) = 5C5(0.3)5(0.7)0
= 1 × 0.002 43 × 1
= 0.002 43
x
0
1
2
3
4
5
Pr(X = x) 0.168 07 0.360 15 0.3087 0.1323 0.028 35 0.002 43
4 Pr(X = x) = nCx pxqn−x
n = 50
x=5
p = 0.07
q = 0.93
Pr(X = 5) = 50C5(0.07)5(0.93)45
= 2 11 8 760 × 0.000 001 680 7 × 0.038 170 925
= 0.135 93
5 Pr(X = x) = nCx pxqn−x
n=5
x=4
p = 0.4
q = 0.6
Pr(X = 4) = 5C4(0.4)4(0.6)1
= 5 × 0.0256 × 0.6
= 0.0768
6 Pr(X = x) = nCx pxqn−x
a n=4
x=1
1
p = --5
4
q = --5
1 1 4 3
Pr(X = 1) = 4C1 --- ---
5
5
1
64
= 4 × --- × --------5 125
256
= --------625
b n=4
x=2
1
p = --5
4
q = --5
1 2 4 2
Pr(X = 2) = 4C2 --- ---
5 5
1
16
= 6 × ------ × -----25 25
96
= --------625
c n=4
x = 1, 2, 3 or 4
1
p = --5
4
q = --5
Pr(X = 1, 2, 3 or 4)
= 1 − Pr(X = 0)
1 0 4 4
= 1 − 4C0 --- ---
5 5
256
= 1 − 1 × 1 × --------625
256
= 1 − --------625
369
= --------625
1
1 1 1
7 a Pr(HHTT) = --- × --- × --- × --2
2 2 2
1
= -----16
= 0.0625
1 1 1 1
b Pr(HHHH) = --- × --- × --- × --2 2 2 2
1
= -----16
= 0.0625
c Pr(X = x) = nCx pxqn−x
n=4
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MQ-12_Sol_MM_Ch 11.fm Page 287 Thursday, August 31, 2006 7:07 PM
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MQ-12_Sol_MM_Ch 11.fm Page 288 Thursday, August 31, 2006 7:07 PM
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The binomial distribution
x=2
1
p = --2
1
q = --2
1 2 1 2
Pr(X = 2) = 4C2 --- ---
2 2
1 1
= 6 × --- × --4 4
6
= -----16
3
= --8
= 0.375
8 Pr(T) = 0.6
a Pr(TTTTH) = (0.6)4 × (0.4)
= 0.1296 × 0.4
= 0.0518
b n=5
x=4
p = 0.6
q = 0.4
Pr(X = 4) = 5C4(0.6)4(0.4)1
= 5 × 0.1296 × 0.4
= 0.2592
9 Pr(X = x) = nCx pxqn−x
n=8
p = 0.55
q = 0.45
a x=4
Pr(X = 4) = 8C4(0.55)4(0.45)4
= 70 × 0.0915 × 0.0410
= 0.262 66
≈ 0.2627
b x=8
Pr(X = 8) = 8C8(0.55)8(0.45)0
= 1 × 0.008 37 × 1
= 0.008 37
≈ 0.0084
c x=5
Pr(X = 5) = 8C5(0.55)5(0.45)3
= 56 × 0.050 33
× 0.091 25
= 0.256 82
≈ 0.2568
d Three opposing is the same as
five supporting so, same as
c = 0.2568.
10 Pr(X = x) = nCx pxqn−x
n = 52
x = 26
p = 0.4
q = 0.6
Pr(X = 26) = 52C26(0.4)26(0.6)26
= 0.038 098
≈ 0.0381
11 Pr(X = x) = nCx pxqn−x
n = 20
x = 10
5
p = --8
3
q = --8
5 10 3 10
Pr(X = 10) = 20C10 --- ---
8 8
= 0.092 41
≈ 0.0924
12 a Pr(X = x) = nCx pxqn−x
n = 10
x=6
p = 0.39
q = 0.61
Pr(X = 6) = 10C6(0.39)6(0.61)4
= 0.1023
b Pr(X = x) = nCx pxqn−x
n = 10
x=4
p = 0.3
q = 0.7
Pr(X = 4) = 10C4(0.3)4(0.7)6
= 0.2001
13 Pr(X = x) = nCx pxqn−x
n = 10
x=1
p = 0.03
q = 0.97
Pr(X = 1) = 10C1(0.03)1(0.97)9
= 0.2281
14 Pr(X = x) = nCx pxqx
n = 200
x = 12
p = 0.08
q = 0.92
Pr(X = 12) = 200C12(0.08)12(0.92)188
= 0.0653
15 Pr(X = x) = nCx pxqn−x
n = 20
600
p = ----------------- = 0.02
30 000
q = 0.98
a x=2
Pr(X = 2) = 20C2(0.02)2(0.98)18
= 0.0528
b x=0
Pr(X = 0) = 20C0(0.02)0(0.98)20
= 0.6676
16 a Pr(X = x) = nCx pxqn−x
n=8
x=4
1
p = --2
1
q = --2
1 4 1 4
Pr(X = 4) = 8C4 --- ---
2
2
= 0.2734
b Pr(X = x) = nCx pxqn−x
n=8
x=2
p = 0.3
q = 0.7
Pr(X = 2) = 8C2(0.3)2(0.7)6
= 0.2965
c n=8
x=0
p = 0.2
q = 0.8
Pr(X = 0) = 8C0(0.2)0(0.8)8
= 0.1678
17 p = 0.78
q = 0.22
n = 10
Pr(X < 3) = Pr(X = 0) + Pr(X = 1) +
Pr(X = 2)
= 10C0(0.78)0(0.22)10 +
10
C1(0.78)(0.22)9 +
10
C2(0.78)2(0.22)8
The answer is C.
18 Pr(X = x) = nCx pxqn−x
n=7
x=4
p = 0.6
q = 0.4
Pr(X = x) = 7C4(0.6)4(0.4)3
= 35 × (0.6)4(0.4)3
The answer is E.
19 Pr(X = x) = nCx pxqn−x
n = 20
x = 11
1
p = --4
3
q = --4
11
20
21
22
23
9
3
1
Pr(X = 11) = 20C11 --- ---
4
4
The answer is B.
p = 0.02
q = 0.98
n = 40
Pr(X ≥ 1) = 1 − Pr(X = 0)
= 1 − 40C0(0.02)0(0.98)40
= 1 − (0.98)40
The answer is D.
Pr(X = x) = nCx pxqn−x
n = 10
x=3
p = 0.3
q = 0.7
Pr(X = 3) = 10C3(0.3)3(0.7)7
= 0.2668
The answer is C.
p = 0.6
q = 0.4
n=5
a i Pr(X = 0) = 5C0(0.6)0(0.4)5
= (0.4)5
= 0.010 24
= 0.0102
ii Pr(X ≥ 1) = 1 − Pr(X = 0)
= 1 − 0.010 24
= 0.989 76
⯝ 0.9898
b Pr(X ≥ 1) = 1 − Pr(X = 0)
0.7 = 1 − nC0(0.6)0(0.4)n
0.7 = 1 − 0.4n
0.4n = 0.3
log 0.3
n = -----------------log 0.4
n = 1.313 963 748
⯝2
n=?
p = 0.05
q = 0.95
Pr(X ≥ 1) = 1 − Pr(X = 0)
0.6 = 1 − nC0(0.05)0(0.95)n
0.6 = 1 − 0.95n
0.95n = 1 − 0.6
0.95n = 0.4
ln 0.95n = ln 0.4
n ln 0.95 = ln 0.4
The binomial distribution
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27 a i
ln 0.4
n = ----------------ln 0.95
= 17.86
= 18 tickets
24 Pr(X = x) = nCx pxqn−x
n = unknown
x = 1, 2, 3, …, n (X ≥ 1)
p = 0.1
q = 0.9
Pr(X ≥ 1) = 0.9
Pr(X ≥ 1) = 1 − Pr(X = 0)
0.9 = 1 − Pr(X = 0)
Pr(X = 0) = 1 − 0.9
= 0.1
Pr(X = x)
0.3
0.2
0.1
0.0
ii
b i
The graph is positively skewed.
Pr(X = x)
0.3
Pr(X = 0) = nC0(0.1)0(0.9)n−0
0.2
0.1 = 0.9n
0.1
ln(0.1) = ln(0.9)n
= n ln(0.9)
0.0
ln ( 0.1 )
n = ----------------ln ( 0.9 )
ii
c i
– 2.3026
= --------------------– 0.1054
0 1 2 3 4 5 6 7 8 910 x
The graph is symmetrical or normally distributed.
Pr(X = x)
= 21.85
She needs to take 22 turns.
25 a Pr(X = x) = nCx pxqn−x
n = unknown.
x = 1, 2, 3, …, n
0.3
0.2
0.1
0.0
1
p = -----------------------8 145 060
0 1 2 3 4 5 6 7 8 910 x
ii
28 a i
ii
8 145 059
q = -----------------------8 145 060
Pr(X = 1, 2, 3, …, n) = 1 − Pr(X = 0)
0.5 = 1 − Pr(X = 0)
Pr(X = 0) = 1 − 0.5
= 0.5
0
1
Pr(X = 0) = nC0 ------------------------
8 145 060
145 059-
8---------------------- 8 145 060
b
29 a
n–0
8 145 059 n
0.5 = ------------------------
8 145 060
8 145 059
ln(0.5) = ln ------------------------
8 145 060
0 1 2 3 4 5 6 7 8 910 x
n
8 145 059
= n ln ------------------------
8 145 060
8 145 059
n = ln 0.5 ÷ ln ------------------------
8 145 060
= 5 645 725.212
The number of games to be played is 5 645 726
5 645 726
b Number of tickets = -----------------------16
= 352 858
Therefore 352 858 tickets need to be bought.
c Cost = 352 858 × 4.10
= $1446717.80
Therefore the tickets would cost $1 446 717.80
26 a x = 2, since it has the greatest probability.
b The graph is positively skewed.
b
30 a
b
31 a
b
c
d
The graph is negatively skewed.
The graph is positively skewed since p < 0.5.
The graph is symmetrical or normally distributed
since p = 0.5.
iii The graph is negatively skewed since p > 0.5.
‘p’ effects the skewness of the graph.
The graph is symmetrical or normally distributed since
p = 0.5.
The graph is still symmetical but the vertical columns
are not as high if ‘n’ is doubled.
The graph is positively skewed, since p < 0.5.
The graph is negatively skewed, since p > 0.5.
Symmetrical or normally distributed.
Positively skewed.
Symmetrical or normally distributed.
Negatively skewed.
Exercise 11B—Problems involving the
binomial distribution for multiple
probabilities
C3(0.4)3(0.6) + 4C4(0.4)4(0.6)0
= 4 × 0.064 × 0.6 + 1 × 0.0256 × 1
= 0.1536 + 0.0256
= 0.1792
b 5C3(0.6)3(0.4)2 + 5C4(0.6)4(0.4) + 5C5(0.6)5(0.4)0
= 10 × 0.216 × 0.16 + 5 × 0.1296 × 0.4 + 1 × 0.077 76 × 1
= 0.3456 + 0.2592 + 0.077 76
= 0.6826
2 a Pr(X ≥ 4) = Pr(X = 4) + Pr(X = 5)
= 0.048 77 + 0.005 25
≈ 0.0540
b Pr(X > 0) = Pr(X = 1) + … + Pr(X = 5)
= 1 − Pr(X = 0)
= 1 − 0.116 03
≈ 0.8840
1 a
4
answers
MQ-12_Sol_MM_Ch 11.fm Page 289 Thursday, August 31, 2006 7:07 PM
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MQ-12_Sol_MM_Ch 11.fm Page 290 Thursday, August 31, 2006 7:07 PM
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The binomial distribution
c Pr(X ≤ 4) = Pr(X = 0) + Pr(X = 1) + … + Pr(X = 4)
= 1 − Pr(X = 5)
= 1 − 0.005 25
≈ 0.9948
d Pr(X < 2) = Pr(X = 0) + Pr(X = 1)
= 0.116 03 + 0.312 39
≈ 0.4284
3 a Pr(X ≥ 2) = Pr(X = 2) + Pr(X = 3) + … + Pr(X = 5)
= 1 − Pr(X = 0) − Pr(X = 1)
= 1 − 5C0(0.3)0(0.7)5 − 5C1(0.3)1(0.7)4
= 1 − 0.168 07 − 0.360 15
≈ 0.4718
b Pr(X < 4) = Pr(X = 0) + Pr(X = 1) + … + Pr(X = 3)
= 1 − Pr(X = 4) − Pr(X = 5)
= 1 − 5C4(0.3)4(0.7) − 5C5(0.3)5(0.7)0
= 1 − 0.028 35 − 0.002 43
≈ 0.9692
4 a Pr(X ≥ 4) = Pr(X = 4) + Pr(X = 5)
= 5C4(0.6)4(0.4)1 + 5C5(0.6)5(0.4)0
= 0.2592 + 0.077 76
≈ 0.3370
b Pr(X ≥ 4) = Pr(X = 4) + Pr(X = 5) + Pr(X = 6)
= 6C4(0.5)4(0.5)2 + 6C5(0.5)5(0.5) + 6C6(0.5)6
= 0.234 38 + 0.093 75 + 0.015 63
≈ 0.3438
c Pr(X ≥ 4) = Pr(X = 4) + … + Pr(X = 7)
= 7C4(0.2)4(0.8)3 + 7C5(0.2)5(0.8)2
+ 7C6(0.2)6(0.8)1 + 7C7(0.2)7(0.8)0
= 0.028 672 + 0.0043 + 0.000 358 4
+ 0.000 012 8
≈ 0.0333
5 a Pr(X ≥ 3) = Pr(X = 3) + Pr(X = 4) + Pr(X = 5) + Pr(X = 6)
Pr(X = x) = nCx pxqn−x
= 28 × 0.015 625 × 0.25 + 8 × 0.007 812 5 × 0.5
+ 0.003 906 25 × 1
= 0.109 375 + 0.031 25 + 0.0039
= 0.1445
d Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)
Pr(X = x) = nCx pxqn−x
Pr(X ≤ 2) = 8C0(0.5)0(0.5)8 + 8C1(0.5)1(0.5)7 +
8
C2(0.5)2(0.5)6
= 1 × 1 × 0.0039 + 8 × 0.5 × 0.0078 + 28 × 0.25
× 0.015 625
= 0.0039 + 0.031 25 + 0.109 375
≈ 0.1445
2--1
7 n = 6, p = , q = --3
3
n
x n−x
Pr(X = x) = Cx p q
2 6 1 0
a Pr(X = 6) = 6C6 --- ---
3 3
≈ 0.0878
b Pr(X ≥ 2) = Pr(X = 2) + Pr(X = 3) + Pr(X = 4) + Pr(X = 5)
+ Pr(X = 6)
= 1 − Pr(X = 0) − Pr(X = 1)
2 0 1 6 6
2 1 1 5
= 1 − 6C0 --- --- – C1 --- ---
3
3
3
3
·
= 1 − 0.001 37 − 6 × 0. 6 × 0.004 115
= 0.982 17
≈ 0.9822
2 0 1 6 6
2 1 1 5
c Pr(X = 0) + Pr(X = 1) = 6C0 --- --- + C1 --- ---
3 3
3 3
8 a
6
1 3 5 3 6
1 4 5 2
Pr(X ≥ 3) = C3 --- --- + C4 --- ---
6 6
6 6
6
1 5
+ C5 ---
6
b
6 a
b
c
1
6
5--- + 6C 1---
6 6
6
5---
6
0
1
125
1
25
= 20 × --------- × --------- + 15 × ------------ × ------ + 6
216 216
1296 36
1
5
1
× ------------ × --- + 1 × ---------------- × 1
7776 6
46 656
= 0.005 358 + 0.0080 + 0.0006 + 0.000 02
≈ 0.0623
Pr(X ≥ 4) = Pr(X = 4) + Pr(X = 5) + Pr(X = 6)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 4) = 6C4(0.5)4(0.5)2 + 6C5(0.5)5(0.5)1
+ 6C6(0.5)6(0.5)0
= 15 × 0.0625 × 0.25 + 6 × 0.031 25 × 0.5 + 1
× 0.015 625 × 1
= 0.234 375 + 0.093 75 + 0.015 625
= 0.3438
Pr(X = x) = nCx pxqn−x
Pr(X = 3) = 8C3(0.5)3(0.5)5
= 56 × 0.125 × 0.031 25
≈ 0.2188
Pr(X ≥ 7) = Pr(X = 7) + Pr(X = 8)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 7) = 8C7(0.5)7(0.5)1 + 8C8(0.5)8(0.5)0
= 8 × 0.007 812 5 × 0.5 + 1 × 0.003 906 25 × 1
= 0.031 25 + 0.0039
≈ 0.0352
Pr(X ≥ 6) = Pr(X = 6) + Pr(X = 7) + Pr(X = 8)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 6) = 8C6(0.5)6(0.5)2 + 8C7(0.5)7(0.5)1
+ 8C8(0.5)8(0.5)0
b
9 a
b
10 a
b
= 0.001 37 + 0.016 46
≈ 0.0178
Pr(X ≥ 5) = Pr(X = 5) + Pr(X = 6) + Pr(X = 7) + Pr(X = 8)
+ Pr(X = 9) + Pr(X = 10)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 5) = 10C5(0.4)5(0.6)5 + 10C6(0.4)6(0.6)4
+ 10C7(0.4)7(0.6)3 + 10C8(0.4)8(0.6)2 +
10
C9(0.4)9(0.6) + 10C10(0.4)10(0.6)0
= 0.200 66 + 0.111 48 + 0.042 47 + 0.010 62
+ 0.001 57 + 0.000 10
≈ 0.3669
Pr(X ≥ 9) = Pr(X = 9) + Pr(X = 10)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 9) = 10C9(0.6)9(0.4)1 + 10C10(0.6)10(0.4)0
= 0.040 31 + 0.006 05
≈ 0.0464
Pr(X ≥ 3) = Pr(X = 3) + Pr(X = 4) + Pr(X = 5) + Pr(X = 6)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 3) = 6C3(0.8)3(0.2)3 + 6C4(0.8)4(0.2)2
+ 6C5(0.8)5(0.2)1 + 6C6(0.8)6(0.2)0
= 0.081 92 + 0.245 76 + 0.393 216 + 0.262 144
≈ 0.9830
Pr(X < 3) = 1 − Pr(X ≥ 3)
= 1 − 0.9830
= 0.0170
Pr(X ≥ 3) = Pr(X = 3) + Pr(X = 4) + Pr(X = 5)
Pr(X = x) = nCx pxqn−x
Pr(X ≥ 3)
1 3 1 2 5
1 4 1 1 5
1 5 1 0
= 5C3 --- --- + C4 --- --- + C4 --- ---
2 2
2 2
2 2
= 0.3125 + 0.156 25 + 0.031 25
= 0.5000
Pr(X ≥ 2) = 1 − Pr(X = 0) − Pr(X = 1)
5 1 1 4
5 0 1 5 5
= 1 − 5C0 --- --- – C1 --- ---
6 6
6 6
= 1 − 0.000 128 6 − 0.003 215 02
≈ 0.9967
The binomial distribution
c Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)
0
5
1
4
2
3
5
5
5
1
1
1
= 5C0 --- --- + C1 --- --- + C2 --- ---
6
6
6
6
6
6
= 0.401 877 6 + 0.401 877 6 + 0.160 751 0
≈ 0.9645
11 a Pr(X ≥ 7) = Pr(X = 7) + Pr(X = 8)
= 8C7(0.7)7(0.3)1 + 8C8(0.7)8(0.3)0
= 0.197 650 + 0.057 648
≈ 0.2553
Pr ( X > 7 ∩ X > 5 )
b Pr(X > 7|X > 5) = -------------------------------------------Pr ( X > 5 )
Pr ( X > 7 )
= ----------------------Pr ( X > 5 )
Pr(X > 7) = Pr(X = 8)
= 8C8(0.7)8(0.3)0
= 0.057 648
Pr(X > 5) = Pr(X = 6) + Pr(X = 7) + Pr(X = 8)
= 8C6(0.7)6(0.3)2 + 0.2553
= 0.296 48 + 0.2553 (from part (a))
= 0.551 774
648Pr
(
X
>
7
)
----------------------- = 0.057
----------------------Pr ( X > 5 ) 0.551 774
= 0.1045
12 a Pr(X ≥ 8) = Pr(X = 8) + Pr(X = 9) + Pr(X = 10)
+ Pr(X = 11) + Pr(X = 12)
= 12C8(0.49)8(0.51)4 + 12C9(0.49)9(0.51)3
+ 12C10(0.49)10(0.51)2 + 12C11(0.49)11(0.51)1
+ 12C12(0.49)12(0.51)0
= 0.111 29 + 0.047 52 + 0.013 698 + 0.002 393
+ 0.000 19
≈ 0.1751
Pr ( X ≥ 10 ∩ X ≥ 8 )
b Pr(X ≥ 10|X ≥ 8) = ----------------------------------------------Pr ( X ≥ 8 )
Pr
(
X
≥ 10 )
= -------------------------Pr ( X ≥ 8 )
Pr(X ≥ 10) = Pr(X = 10) + Pr(X = 11) + Pr(X = 12)
= 0.013 70 + 0.002 39 + 0.000 19
≈ 0.016 28
Pr(X ≥ 8) ≈ 0.1751
28
Pr
( X ≥ 10 ) 0.016
-------------------------= --------------------0.1751
Pr ( X ≥ 8 )
≈ 0.0930
13 Pr(X < 2) = Pr(X = 0) + Pr(X = 1)
= 5C0(0.7)0(0.3)5 + 5C1(0.7)1(0.3)4
5
5
= (0.3)5 + 5C1(0.7)(0.3)4
The answer is A.
1
14 --- of 80 = 20
4
Pr(X = x) = nCxpxqn − x
x = 20, n = 80
Pr(X = 20) = 80C20p20q60
= 80C20p20(1 − p)60
The answer is D.
15 n = 10, p = k
X ∼ Bi (10, p)
Pr(X ≥ 1) = 1 − Pr(X = 0)
= 1 − 10C0p0(1 − p)10
= 1 − (1 − p)10
p = k, ⇒ 1 − (1 − k)10
The answer is A.
16 Pr(X ≥ 3) = 1 − Pr(X = 0) − Pr(X = 1) − Pr(X = 2)
= 1 − 10C0(0.03)0(0.97)10 − 10C1(0.03)1(0.97)9
− 10C2(0.03)2(0.97)8
17
18
19
20
21
MM12-11
291
= 1 − 0.7374 − 0.2281 − 0.0317
= 0.0028
The answer is D.
Pr(X ≥ 8) = Pr(X = 8) + Pr(X = 9) + Pr(X = 10)
= 10C8(0.6)8(0.4)2 + 10C9(0.6)9(0.4)1
+ 10C10(0.6)10(0.4)0
= 0.1209 + 0.0403 + 0.0060
= 0.1672
The answer is C.
Pr ( X ≥ 11 ∩ X ≥ 10 )
Pr(X ≥ 11|X ≥ 10) = -------------------------------------------------Pr ( X ≥ 10 )
Pr ( X ≥ 11 )
= -------------------------Pr ( X ≥ 10 )
Pr(X ≥ 11) = Pr(X = 11) + Pr(X = 12)
= 12C11(0.95)11(0.05)1 + 12C12(0.95)12(0.05)0
= 0.341 28 + 0.540 36
= 0.881 64
Pr(X ≥ 10) = Pr(X = 10) + Pr(X = 11) + Pr(X = 12)
= 12C10(0.95)10(0.05)2 + 0.881 64
= 0.098 791 6 + 0.881 64
= 0.980 43
64Pr
(
X
≥
11
)
-------------------------- = 0.881
-------------------Pr ( X ≥ 10 ) 0.980 43
= 0.8992
The answer is D.
Pr(X ≥ 10) = Pr(X = 10) + Pr(X = 11) + Pr(X = 12)
+ Pr(X = 13) + Pr(X = 14) + Pr(X = 15)
= 15C10(0.5)10(0.5)5 + 15C11(0.5)11(0.5)4
+ 15C12(0.5)12(0.5)3 + 15C13(0.5)13(0.5)2
+ 15C14(0.5)14(0.5)1 + 15C15(0.5)15(0.5)0
= 0.091 64 + 0.041 66 + 0.013 89
+ 0.003 20 + 0.000 46 + 0.000 03
= 0.1509
a Pr(X ≤ 3) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3)
= 30C0(0.05)0(0.95)30 + 30C1(0.05)1(0.95)29
+ 30C2(0.05)2(0.95)28 + 30C3(0.05)3(0.95)27
= 0.214 64 + 0.338 90 + 0.258 64 + 0.127 05
≈ 0.9392
b Pr(X ≥ 3) = 1 − Pr(X < 3)
= 1 − 0.214 64 − 0.338 90 − 0.258 64
≈ 0.1878
a Pr(X ≤ 1) = Pr(X = 0) + Pr(X = 1)
= 50C0(0.01)0(0.99)50 + 50C1(0.01)1(0.99)49
= 0.605 006 1 + 0.305 558 6
≈ 0.9106
b Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)
= 50C0(0.02)0(0.98)50 + 50C1(0.02)(0.98)49
+ 50C2(0.02)2(0.98)48
= 0.364 169 7 + 0.371 601 7 + 0.185 800 9
≈ 0.9216
c Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)
= 50C0(0.04)0(0.96)50 + 50C1(0.04)1(0.96)49
+ 50C2(0.04)2(0.96)48
= 0.129 885 8 + 0.270 595 4 + 0.276 232 8
≈ 0.6767
d Pr(X ≤ 3) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3)
= 50C0(0.07)0(0.93)50 + 50C1(0.07)1(0.93)49
+ 50C2(0.07)2(0.93)48 + 50C3(0.07)3(0.93)47
= 0.026 555 1 + 0.099 938 + 0.184 295 1
+ 0.221 946 7
= 0.5327
1 0 2 10
22 a Pr(X = 0) = 10C0 --- ---
3
3
= 0.0173
answers
MQ-12_Sol_MM_Ch 11.fm Page 291 Thursday, August 31, 2006 7:07 PM
answers
MQ-12_Sol_MM_Ch 11.fm Page 292 Thursday, August 31, 2006 7:07 PM
MM12-11
292
The binomial distribution
= 0.165 146 + 0.225 194 + 0.225 199 + 0.155 907
+ 0.066 82 + 0.013 363
= 0.8516
b p = 0.8516
q = 0.1484
Pr(X = 8) = 10C8(0.8516)8(0.1484)2
= 0.2741
26 The prime numbers obtainable are 2, 3, 5.
3
1
Pr(prime) = --- = --6
2
a Pr(X ≥ 4)
= Pr(X = 4) + Pr(X = 5) + Pr(X = 6)
1 4 1 2 6
1 5 1 1 6
1 6 1 0
= 6C4 --- --- + C5 --- --- + C6 --- ---
2 2
2 2
2 2
= 0.234 375 + 0.093 75 + 0.015 625
= 0.3438
b Pr(X = 5) = 8C5 (0.3438)5 (0.6562)3
= 0.0760
27 a Pr(X < 2) = Pr(X = 0) + Pr(X = 1)
= 6C0(0.05)0(0.95)6 + 6C1(0.05)1(0.95)5
= 0.735 09 + 0.232 13
= 0.9672
b Pr(X ≥ 22)
= Pr(X = 22) + Pr(X = 23) + Pr(X = 24)
= 24C22(0.9672)22(0.0328)2 + 24C23(0.9672)23(0.0328)1
+ 24C24(0.9672)24(0.0328)0
= 0.142 57 + 0.365 56 + 0.449 15
= 0.9573
1 4 2 6
b Pr(X = 4) = 10C4 --- ---
3 3
= 0.2276
c Pr(X ≥ 2) = 1 − Pr(X < 2)
= 1 − Pr(X = 0) − Pr(X = 1)
1 1 2 9
= 1 − 0.0173 − 10C1 --- ---
3 3
= 1 − 0.0173 − 0.086 708
≈ 0.8960
23
6
5
1st die 4
3
2
1
5
4
3
2
1
0
4
3
2
1
0
1
3
2
1
0
1
2
2
1
0
1
2
3
1
0
1
2
3
4
0
1
2
3
4
5
1 2 3 4 5 6
2nd die
6
Pr(a difference of zero) = -----36
1
= --6
10
Pr(a difference of 1) = -----36
5
= -----18
2
Pr(a difference of 5) = -----36
1
= -----18
a Pr(X ≥ 1) = 1 − Pr(X = 0)
Exercise 11C—Markov chains
1 a
1 0 5 5
= 1 − C0 --- ---
6
6
= 1 − 0.401 876
= 0.5981
b Pr(X ≥ 2) = Pr(X = 2) + Pr(X = 3) + Pr(X = 4) + Pr(X = 5)
= 1 − Pr(X = 0) − Pr(X = 1)
5
1
5
5
13
13
5
= 1 − 5C0 ------ ------ – C1 ------ ------
18
18
18
18
= 1 − 0.196 496 − 0.377 877
= 0.4256
c Pr(X ≥ 1) = 1 − Pr(X = 0)
0
Pr(X = 9) =
Thursday
0.35
10
C9 ( 0.6778 ) 9 ( 0.3222 ) 1
= 10 × 0.030 19 × 0.3222
= 0.0973
25 n = 15, p = 0.75, q = 0.25
a Pr(X ≥ 10)
= Pr(X = 10) + Pr(X = 11) + …. + Pr(X = 15)
= 15C10(0.75)10(0.25)5 + 15C11(0.75)11(0.25)4
+ 15C12(0.75)12(0.25)3 + 15C13(0.75)13(0.25)2
+ 15C14(0.75)14(0.25)1 + 15C15(0.75)15
Good
0.35
Poor
0.55
Good
0.45
Poor
Poor
b
4
Friday
Saturday
0.65
0.65
0.35
0.65
Good
0.35
Poor
0.55
Good
0.45
Poor 0.55 0.35
Good
Thursday
5
1
17
= 1 − 5C0 ------ ------
18
18
= 1 − 0.751 418 8
= 0.2486
24 n = 10 p = 0.8
a Pr(X ≥ 8) = Pr(8) + Pr(9) + Pr(10)
= 10C8(0.8)8(0.2)2 + 10C9(0.8)9(0.2)1
+ 10C10(0.8)10(0.2)0
= 0.301 99 + 0.268 44 + 0.107 37
= 0.6778
b (0.6778 + 0.3222)10
0.65
Good
0.65
5
0
Saturday
Friday
Poor
0.55
0.45
0.35
0.65
0.45
Sunday
Good
Poor
Good
Poor
Good
Poor
Good
Poor
<Pr(good on Sunday)
= > 0.653 + 0.65 × 0.35 × 0.55 + 0.35 × 0.55 × 0.65
+ 0.35 × 0.45 × 0.55
= 0.6115
2 a <Pr(good on Sundaypoor on Thursday) >
= 0.55 × 0.65 × 0.65 + 0.55 × 0.35 × 0.55 + 0.45 × 0.55
× 0.65 + 0.45 × 0.45 × 0.55
= 0.6105
b <Pr(poor on Sundaypoor on Thursday) >
= 1 − 0.6105
= 0.3895
3 a
0.63
0.28
0.72
0.63
R
0.27
R'
0.48
R
0.52
R' 0.48
0.52
R
0.48
R'
0.27
0.52 0.63
0.27
R
R'
R
R'
R
R'
R
R'
The binomial distribution
0.38
0.38
A
0.62
A'
0.28
A
0.72
A' 0.28
0.72
A
0.47
0.53
0.62
0.28
A'
0.72 0.38
0.62
A
A'
A
A'
A
A'
A
A'
4
0.25
Success
0.75
Film 1 0.25
S
0.75
F
Film 2
S
F
0.38
S
0.62
F
0.25
0.75
0.38
0.62
0.25
0.75
0.38
0.62
Film 3
S
F
S
b = 0.6t + 0.7b
t = 10 500 − b
= 10 500 − 2t
500
3t- = 10
------------------3
3
t = 3500
t + b = 10 500
b = 10 500 − 3500
= 7000
7 a 10%P → Sam 20% S → P
90%P → P
80% S → S
Pi + 1 = 0.9Pi + 0.2Si Si + 1 = 0.1Pi + 0.8Si
S
F
S
June
July
F
Kingfisher
Best Match
0.70
Kingfisher
0.30
Best Match
Best Match
The probability the club chooses Best Match next season
is 0.3
b
0.3
B
0.7
K
0.2
B
0.8
K
B
0.3
Best
Match
0.7
K
0.3 B
0.7 K
0.2 B
0.8 K
0.3 B
0.7 K
0.2 B
0.8 K
0.3 × 0.3 × 0.7 = 0.063
0.3 × 0.7 × 0.8 = 0.168
0.7 × 0.2 × 0.7 = 0.098
0.7 × 0.8 × 0.8 = 0.448
0.063 + 0.168 + 0.098 + 0.448 = 0.777.
Probability the club will be using Kingfisher in 3 years
time is 0.777.
6 a 30% bus → train 60% train → bus
70% bus → bus
40% train → train
ti + 1 = 0.40t1 + 0.30b1 bi + 1 = 0.60t1 + 0.70b1
January
Feburary
March
April
May
Train
Bus
4900
t = 0.40 × 4900 + 0.30
× 5600
= 3640
t = 0.4 × 3640 + 0.3 ×
6860
= 3514
t = 0.40 × 3514 + 0.3 ×
6986
= 3501.4
t = 0.4 × 3501.4 + 0.3 ×
6998.6
= 3500.14
⯝ 3500
5600
b = 0.60 × 4900 + 0.70
× 5600
= 6860
b = 0.6 × 3640 + 0.7 ×
6860
= 6986
b = 0.6 × 3514 + 0.7 ×
6986
= 6998.6
b = 0.6 × 3501.4 + 0.7
× 6998.6
= 6999.86
⯝ 7000
b b + t = 10 500
t = 0.4t + 0.3b ⇒
Aug
Sep
Oct
Kingfisher
0.20
0.6t = 0.3b
b = 2t
293
F
a Pr(second film fails, previous success)
= 0.25 × 0.75 + 0.75 × 0.62
= 0.6525
b Pr(third film fails, previous success)
= 0.25 × 0.25 × 0.75 + 0.25 × 0.75 × 0.62 + 0.75 × 0.38
× 0.75 + 0.75 × 0.62 × 0.62
= 0.665 175
5 a
0.80
MM12-11
2800 Pete
P = 0.9 × 2800 + 0.2
× 3100
= 3140
P = 0.9 × 3140 + 0.2
× 2760
= 3378
P = 0.9 × 3378 + 0.2
× 2522
= 3544.6
P = 0.9 × 3544.6 + 0.2
× 2355.4
= 3661.22
⯝3661
3100 Sam
S = 0.1 × 2800 + 0.8
× 3100
= 2760
S = 0.1 × 3140 + 0.8
× 2760
= 2522
S = 0.1 × 3378 + 0.8
× 2522
= 2355.4
S = 0.1 × 3544.6 + 0.8
× 2355.4
= 2238.78
⯝2239
b P + S = 5900
P = 0.9P + 0.2S
S = 0.1P + 0.8S
0.1P = 0.2S
P = 2S
S = 5900 − P
= 5900 − 2S
3S = 5900
5900
S = -----------3
2
= 1966 --3
⯝1967
P + S = 5900
P = 5900 − S
1
= 3933 --3
⯝ 3933
8
G
A
G 0.9 0.75
A 0.1 0.25
a and b
0.9 0.75
0.1 0.25
50
G 200
A 100
200
100
= 264.71
35.29
Gym 265, Aerobics 35
9
0.7
Make top
8
0.3
0.4
Don’t make
top 8
The answer is C.
0.6
Make top
8
Don’t make
top 8
Make top
8
Don’t make
top 8
answers
MQ-12_Sol_MM_Ch 11.fm Page 293 Thursday, August 31, 2006 7:07 PM
answers
MQ-12_Sol_MM_Ch 11.fm Page 294 Thursday, August 31, 2006 7:07 PM
MM12-11
294
The binomial distribution
Exercise 11D—Expected value, variance
and standard deviation of the binomial
distribution
1 a E(X) = np
= 10 × 0.6
=6
b E(X) = np
= 8 × 0.2
= 1.6
c E(X) = np
= 100 × 0.5
= 50
d E(X) = np
3
= 50 × --4
= 37.5
2 a Var(X) = npq
n = 20, p = 0.6, q = 0.4
Var(X) = 20 × 0.6 × 0.4
= 4.8
b Var(X) = npq
= 15 × 0.9 × 0.1
= 1.35
c Var(X) = npq
= 25 × 0.4 × 0.6
=6
d Var(X) = npq
1 3
= 20 × --- × --4 4
3--=3
4
3 a Var(X) = npq
= 10 × 0.2 × 0.8
= 1.6
SD(X) =
npq
= 1.6
= 1.26
b Var(X) = npq
= 30 × 0.5 × 0.5
= 7.5
SD(X) =
npq
= 7.5
= 2.74
c Var(X) = npq
= 50 × 0.7 × 0.3
= 10.5
SD(X) =
npq
= 10.5
= 3.24
d Var(X) = npq
2 3
= 72 × --- × --5 5
= 17.28
SD(X) =
npq
= 17.28
= 4.16
4 a E(X) = np
1
n = 10, p = --2
1
E(X) = 10 × --2
=5
b Var(X) = npq
1 1
= 10 × --- × --2 2
= 2.5
c SD(X) =
npq
= 2.5
= 1.58
5 a E(X) = np
3
n = 20, p = -----13
3
E(X) = 20 × -----13
= 4.62
b Var(X) = npq
3
10
= 20 × ------ × -----13 13
= 3.55
c SD(X) =
npq
= 3.55
= 1.88
6 a E(X) = np
n = 20, p = 0.6
E(X) = 20 × 0.6
= 12
b Var(X) = npq
= 20 × 0.6 × 0.4
= 4.8
c SD(X) =
npq
= 4.8
= 2.19
7 a E(X) = np
1
n = 10, p = --6
1
E(X) = 10 × --6
= 1.67
b Pr(X > 1.67) = Pr(X = 2) + Pr(X = 3) + ···· + Pr(X = 10)
= 1 − [Pr(X = 0) + Pr(X = 1)]
Pr(X = x) = nCx p x q n− x
1 0 5 10 10
1 1 5 9
C0 --- --- + C1 --- ---
6 6
6 6
= 1 − 0.161 51 − 0.323 01
= 0.5153
Pr(X >1.67) = 1 −
10
8 a n = 15
1
p = --- (primes are 2, 3, 5)
2
1
E(X) = 15 × --2
= 7.5
b Pr(X > 7.5) = Pr(X ≥ 8)
= Pr(X = 8) + …… + Pr(X = 15)
= 1 − Pr(X ≤ 7)
= 1 − 0.5000 (From the graphics calculator)
= 0.5
1
9 a E(X) = 27 × --5
= 5.4
b Pr(X > 5.4) = Pr(X ≥ 6)
= 1 − Pr(X ≤ 5)
= 1 − 0.5387 (From the graphics calculator)
= 0.4613
10 a E(X) = 80 × 0.35
= 28
b Pr(X > 28) = Pr(X ≥ 29)
= 1 − Pr(X ≤ 28)
= 1 − 0.5512 (From the graphics calculator)
= 0.4488
11 a E(X) = np
n = 120
p = 0.8
The binomial distribution
b
12 a
b
13 a
E(X) = 120 × 0.8
= 96
96 rabbits are expected to die.
120 − 96
= 24
24 rabbits are expected to live.
E(X) = 10 so np = 10 (i)
Var(X) = 5 so npq = 5 (ii)
Substitute (i) into (ii).
npq = 5
10q = 5
5
q = -----10
= 0.5
q=1−p
0.5 = 1 − p
p = 0.5
np = 10 where p = 0.5
n × 0.5 = 10
10
n = ------0.5
n = 20
E(X) = 12 so np = 12
Var(X) = 3 so npq = 3
npq = 3
12q = 3
3
q = -----12
1
q = --4
q=1−p
1--=1−p
4
3
p = --4
3
b np = 12 where p = --4
1 10
c Pr(X = 10) = 27C10 ---
3
15
16
17
18
3
n × --- = 12
4
3
n = 12 ÷ --4
n = 16
14 a E(X) = 9 so np = 9
Var(X) = 6 so npq = 6
npq = 6
9q = 6
6
q = --9
2
q = --3
q=1−p
2--=1−p
3
1
p = --3
b np = 9
1
n × --- = 9
3
1
n = 9 ÷ --3
n = 27
19
20
2--- 17
3
= 8 436 285 × 1.693 508 8 × 10−5
× 1.014 959 23 × 10−3
= 0.1450
a E(X) = 3 so np = 3
Var(X) = 2.4 so npq = 2.4
npq = 2.4
3q = 2.4
2.4
q = ------3
q = 0.8
q=1−p
0.8 = 1 − p
p = 0.2
b np = 3
n × 0.2 = 3
3
n = ------0.2
n = 15
c Pr(X = 10) = 15C10(0.2)10(0.8)5
= 0.0001
d Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1)
+ Pr(X = 2)
= 15C0(0.2)0(0.8)15
+ 15C1(0.2)1(0.8)14
+ 15C2(0.2)2(0.8)13
= 0.035 18 + 0.1319 + 0.2309
= 0.3980
1
E(X) = 20 × --2
= 10
The answer is C.
7
E(X) = 40 × -----10
= 28
The answer is E.
Var(X) = npq
3
= 28 × -----10
= 8.4
The answer is C.
E(X) = 200 × 0.8
= 160
Var(X) = npq
= 160 × 0.2
= 32
The answer is E.
E(X) = 10 so np = 10
Var(X) = 6 so npq = 6
npq = 6
10q = 6
q = 0.6
2
p = 0.4 = ---
5
np = 10
n × 0.4 = 10
n = 25
The answer is D.
21 Pr(X = 3) = 5C3 p3q2
= 10p3(1 − p)2
The answer is D.
22 a E(X) = 16
np = 16 where n = 20
20p = 16
MM12-11
295
16
p = -----20
p = 0.8 = 4---
5
b Var(X) = npq
= 20 × 0.8 × 0.2
= 3.2
c SD(X) = 3.2
= 1.79
23 a Pr(X = 10) where n = 20, p = 0.2
= 20C10(0.2)10(0.8)10
= 0.0020
b Pr(X ≥ 3) = Pr(X = 3) + ····
+ Pr(X = 20)
= 1 − [Pr(X = 0) + Pr(X = 1)
+ Pr(X = 2)]
= 1 − [20C0(0.2)0(0.8)20
+ 20C1(0.2)1(0.8)19
+ 20C2(0.2)2(0.8)18]
= 1 − [0.011 529 2
+ 0.057 646 08 + 0.136 909 43]
= 1 − 0.2061
= 0.7939
24 E(X) = 25 × 0.04
=1
Pr(X > 1) = Pr(X = 2) + ····
+ Pr(X = 25)
= 1 − [Pr(X = 0) + Pr(X = 1)]
= 1 − [25C0(0.04)0(0.96)25
+ 25C1(0.04)1(0.96)24]
= 1 − [0.360 396 7 + 0.375 413 2]
= 1 − 0.7358
= 0.2642
25 a E(X) = 3
np = 3
10p = 3
p = 0.3
Pr(X = 2) = 10C2(0.3)2(0.7)8
= 0.2335
b Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1)
+ Pr(X = 2)
= 10C0(0.3)0(0.7)10
+ 10C1(0.3)1(0.7)9 + 0.2335
= 0.0282 + 0.121 06 + 0.2335
= 0.3828
26 E(X) = np
= 500 × 0.8
= 400
27 a n = 15, p = 0.5
Pr(X ≥ 8) = Pr(X = 8) + Pr(X = 9)
+ ···· + Pr(X = 15)
= 15C8(0.5)8(0.5)7
+ 15C9(0.5)9(0.5)6 + ····
+ 15C15(0.5)15(0.5)0
= 0.196 38 + 0.152 74 + 0.091 64
+ 0.041 66 + 0.013 89
+ 0.003 20 + 0.000 46
+ 0.000 03
= 0.5000
answers
MQ-12_Sol_MM_Ch 11.fm Page 295 Thursday, August 31, 2006 7:07 PM
answers
MQ-12_Sol_MM_Ch 11.fm Page 296 Thursday, August 31, 2006 7:07 PM
MM12-11
296
The binomial distribution
b n = 15, p = 0.55
Pr(X ≥ 8) = 15C8(0.55)8(0.45)7 + 15C9(0.55)9(0.45)6
+ ···· + 15C15(0.55)15(0.45)0
= 0.201 34 + 0.191 40 + 0.140 36 + 0.077 98
+ 0.031 77 + 0.008 96 + 0.001 56 + 0.000 13
= 0.6535
c n = 15, p = 0.45
Pr(X ≥ 8) = 15C8(0.45)8(0.55)7 + 15C9(0.45)9(0.55)6
+ ···· + 15C15(0.45)15(0.55)0
= 0.164 74 + 0.104 83 + 0.051 46 + 0.019 14 +
0.005 22 + 0.000 99 + 0.000 12 + 0.000 006
= 0.3465
28 a E(X) = np
1
n = 100 000, p = --2
1
E(X) = 100 000 × --2
= 50 000
b Var(X) = npq
1
1
n = 100 000, p = --- , q = --2
2
1--- 1--Var(X) = 100000 × ×
2 2
= 25 000
c SD(X) =
29 a
b
30
a
b
c
d
e
31 a
b
Chapter review
Multiple choice
3
1 n = 20, p = --7
3 8 4 12
Pr(X = 8) = 20C8 --- ---
7 7
= 0.1738
The answer is C.
3
2 n = 12, p = --8
3
4
5
6
7
3 3 5 9
Pr(X = 3) = 12C3 --- ---
8 8
= 0.1688
The answer is B.
B because it has more than two outcomes.
The answer is B.
n = 7, p = 0.65
Pr(X = 3) = 7C3(0.65)3(0.35)4
= 35 × 0.653 × 0.354
The answer is D.
n = 10, p = 0.5
Pr(X = 4) = 10C4(0.5)4(0.5)6
= 0.2051
The answer is B.
n = 10, p = 0.5
Pr(X ≥ 8) = Pr(X = 8) + Pr(X = 9) + Pr(X = 10)
= 10C8(0.5)8(0.5)2 + 10C9(0.5)9(0.5)1
+ 10C10(0.5)10(0.5)0
= 0.043 945 + 0.009 765 6 + 0.000 976 56
= 0.0547
The answer is C.
1
n = 100, p = -----50
Pr(X ≥ 3) = Pr(X = 3) + ···· + Pr(X = 100)
= 1 − Pr(X < 3)
= 1 − [Pr(X = 0) + Pr(X = 1) + Pr(X = 2)]
= 1−
Var ( X )
= 144
= 12
µ = E(X) = 160
σ = SD(X) = 12
µ + 2σ = 160 + 24
= 184
µ − 2σ = 160 − 24
= 136
c There is a probability of 0.95 that between 136 to 184
students will have a reading level that is inadequate to
cope with high school.
32 a E(X) = 1800 × 0.70
= 1260
b Var(X) = 1260 × 0.3
= 378
Var ( X )
= 378
= 19.44
µ = E(X) = 1260, σ = SD(X) = 19.44
µ + 2σ = 1260 + 38.88
= 1298.88
µ − 2σ = 1260 − 38.88
= 1221.12
c There is a probability of 0.95 that between 1222 and
1298 patients will be cured.
npq
= 25 000
= 158.11
E(X) = np
n = 30, p = 0.2
E(X) = 30 × 0.2
=6
Yoghurt B is a popular product of the company and more
popular than expected.
n = 30, p = 0.02
Pr(X = 0) = 30C0(0.02)0(0.98)30
= 0.5455
Pr(X = 1) = 30C1(0.02)1(0.98)29
= 0.3340
E(X) = 30 × 0.02
= 0.6
Pr(X ≤ 1) = Pr(X = 0) + Pr(X = 1)
= 0.5455 + 0.3340
= 0.8795
n = 10, p = 0.8795
Pr(X = 10) = 10C10(0.8795)10(0.1205)0
= 0.2769
E(X) = 1600 × 0.10
= 160
Var(X) = npq
= 160 × 0.9
= 144
SD(X) =
SD(X) =
8
9
100 100
99 100
98
1- 0 49
1- 1 49
1- 2 49
------ + C1 ---------- + C2 ----------
C0 ----50 50
50 50
50 50
100
= 1 − [0.132 619 6 + 0.270 652 + 0.273 414]
= 1 − 0.6767
= 0.3233
The answer is A.
Pr(X ≥ 1) = 1 − Pr(X = 0)
3 0 4 20
= 1 − 20C0 --- ---
7 7
4 20
= 1 − ---
7
The answer is A.
n = 15
p = 0.67
q = 0.33
Pr(X ≥ 7) = 15C7(0.67)7(0.33)8 + 15C8(0.67)8(0.33)7 + ….
+ 15C15(0.67)15
The answer is D.
The binomial distribution
1 12
c Pr(X = 12) = 12C12 ------
20
1
10 --- of 60 = 12
5
C12(k)12(1 − k)48
The answer is E.
11 Two state Markov chain only has two
possible options. There are four
options given with alternative C.
The answer is C.
12
60
Thursday
Wednesday
0.83 Good
Tuesday
Good
0.35
0.17 Poor
Poor
0.35 Good
0.65
Poor
0.65 Poor
Poor on Thursday given that it was
poor on Tuesday
= 0.65 × 0.65 + 0.35 × 0.17
= 0.4225 + 0.0595
= 0.4820
The answer is B.
1 1
13 n = 60, p = --- × --2 2
1
= --4
1
E(X) = 60 × --4
= 15
The answer is B.
1
14 n = 50, p = --- = q
2
1
1
npq = 50 × --- × --2
2
25
= -----2
The answer is D.
1
15 E(X) = 50 × --2
= 25
The answer is E.
16 A: npq = 20 × 0.6 × 0.4
= 4.8 No
B: npq = 15 × 0.9 × 0.1
= 1.35 Yes
The answer is B.
1 5 5 0
+ 5C 5 --- ---
6 6
= 0.032 15 + 0.003 215
+ 0.000 129
= 0.0355
1
b n = 5, p = --2
Pr(X ≥ 2) = 1 − [Pr(X = 0)
+ Pr(X = 1)]
Short answer
1 n = 7, p = 0.5
a Pr(X = 3) = 7C3(0.5)3(0.5)4
= 0.2734
b Pr(X = 7) = 7C7(0.5)7
= 0.0078
1
2 n = 12, p = -----20
1 2
a Pr(X = 2) = C2 ------
20
12
=1−
19
------
20
10
19
------
20
6
= 0.0988
1 6
b Pr(X = 6) = 12C6 ------
20
= 2.4414 × 10−16
3 n = 300, p = 0.01
a Pr(X = 0) = 300C0(0.01)0(0.99)300
= 0.0490
b Pr(X = 1) = 300C1(0.01)1(0.99)299
= 0.1486
4 a n = 40, p = 0.6
Pr(X = 10) = 40C10(0.6)10(0.4)30
= 5.9093 × 10−6
b n = 40, p = 0.2
Pr(X = 10) = 40C10(0.2)10(0.8)30
= 0.1075
5 a i n = 6, p = 0.65, q = 0.35
Pr(X = 0) = 6C0(0.65)0(0.35)6
= 0.0018
ii Pr(X ≥ 1) = 1 − Pr(X = 0)
= 1 − 0.0018
= 0.9982
b Pr(X ≥ 1) = 1 − Pr(X = 0)
0.75 = 1 − nC0(0.65)0(0.35)n
= 1 − 0.35n
n
0.35 = 1 − 0.75
0.35n = 0.25
log 0.25
n = -------------------log 0.35
= 1.3205
= 2 shots
6 p = 0.08, q = 0.92
Pr(X ≥ 1) = 1 − Pr(X = 0)
= 1 − nC0(0.08)0(0.92)n
0.7 = 1 − 0.92n
0.92n = 1 − 0.7
0.92n = 0.3
log 0.3
n = -------------------log 0.92
= 14.44
= 15 tickets.
1
7 a n = 5, p = --6
Pr(X > 2) = Pr(X = 3) + Pr(X = 4)
+ Pr(X = 5)
1 3 5 2
1 4 5 1
= 5C3 --- --- + 5 C 4 --- ---
6 6
6 6
= 0.000 01 (or 1.0613 × 10−5)
1 0 1 5 5
1 1 1 4
C0 --- --- + C1 --- ---
2 2
2 2
5
= 1 − [0.031 25 + 0.156 25]
= 1 − 0.1875
= 0.8125
c n = 5, p = 0.5
Pr(X ≤ 4) = 1 − Pr(X = 5)
= 1 − 5C5(0.5)5(0.5)0
= 1 − 0.031 25
= 0.9688
MM12-11
297
8 n = 20, p = 0.4
a Pr(X < 5) = Pr(X = 0) + Pr(X = 1)
+ Pr(X = 2) + Pr(X = 3) + Pr(X = 4)
= 20C0(0.4)0(0.6)20
+ 20C1(0.4)1(0.6)19
+ 20C2(0.4)2(0.6)18
+ 20C3(0.4)3(0.6)17
+ 20C4(0.4)4(0.6)16
= 0.000 036 6 + 0.000 487
+ 0.003 087 + 0.012 35
+ 0.034 99
= 0.0510
b Pr(X ≥ 15) = Pr(X = 15)
+ Pr(X = 16) + Pr(X = 17)
+ Pr(X = 18) + Pr(X = 19)
+ Pr(X = 20)
= 20C15(0.4)15(0.6)5
+ 20C16(0.4)16(0.6)4
+ 20C17(0.4)17(0.6)3
+ 20C18(0.4)18(0.6)2
+ 20C19(0.4)19(0.6)1
+ 20C20(0.4)20(0.6)0
= 0.001 29 + 0.000 27 + 0.000 04
+ 0.000 004 + 0.000 000 329 9
+ 0.000 000 010 995
= 0.0016
9 n = 10, p = 0.8
a Pr(X ≥ 5) = 1 − Pr(X < 5)
= 1 − [Pr(X = 0) + Pr(X = 1)
+ Pr(X = 2) + Pr(X = 3)
+ Pr(X = 4)]
= 1 − [10C0(0.8)0(0.2)10
+ 10C1(0.8)1(0.2)9
+ 10C2(0.8)2(0.2)8
+ 10C3(0.8)3(0.2)7
+ 10C4(0.8)4(0.2)6]
= 1 − [0.000 000 1 + 0.000 004
+ 0.000 07 + 0.000 79
+ 0.005 505]
= 1 − 0.636 94
= 0.9936
b Pr(X ≥ 8) = Pr(X = 8) + Pr(X = 9)
+ Pr(X = 10)
= 10C8(0.8)8(0.2)2
+ 10C9(0.8)9(0.2)1
+ 10C10(0.8)10(0.2)0
= 0.301 989 9 + 0.268 435
+ 0.107 374
= 0.6778
10 n = 50, p = 0.2
a Pr(X ≥ 4) = 1 − Pr(X < 4)
= 1 − [Pr(X = 0) + Pr(X = 1)
+ Pr(X = 2) + Pr(X = 3)]
= 1 − [50C0(0.2)0(0.8)50
+ 50C1(0.2)1(0.8)49
+ 50C2(0.2)2(0.8)48
+ 50C3(0.2)3(0.8)47]
= 1 − [0.000 01 + 0.000 178
+ 0.001 09 + 0.000 44]
= 1 − 0.005 66
= 0.9943
answers
MQ-12_Sol_MM_Ch 11.fm Page 297 Thursday, August 31, 2006 7:07 PM
answers
MQ-12_Sol_MM_Ch 11.fm Page 298 Thursday, August 31, 2006 7:07 PM
298
MM12-11
The binomial distribution
b Pr(X > 15) = 1 − Pr(X ≤ 15)
= 1 − [50C0(0.2)0(0.8)50 + 50C1(0.2)1(0.8)49
+ 50C2(0.2)2(0.8)48 + ···· + 50C15(0.2)15(0.8)35
= 1 − [0.000 01 + 0.000 178 4 + 0.001 09 + 0.004 37
+ 0.012 84 + 0.029 53 + 0.055 37 + 0.087 01
+ 0.116 92 + 0.1364 + 0.139 82 + 0.1271 + 0.103 28
+ 0.075 47 + 0.049 86 + 0.0299]
= 1 − 0.9692
= 0.0308
11 a n = 20, p = 0.68, q = 0.32
Pr(X ≥ 12) = 20C12(0.68)12(0.32)8 + …..
+ 20C20(0.68)20(0.32)0
= 1 − Pr(X ≤ 11)
= 1 − 0.1568 (From the graphics calculator)
= 0.8432
b p = 0.8432, q = 0.1568, n = 10
Pr(X = 9) = 10C9(0.8432)9(0.1568)
= 0.3378
12 a
Regular
Regular
Skinny
Skinny
Regular
Skinny
Skinny
b
Thur
Fri
Sat
0.8
0.8
R
0.2
S
R
0.65
Skinny
0.65
0.35
R
S
0.35
Sun
R
0.2
0.65
S 0.65 × 0.8 × 0.2 = 0.104
R
0.35
0.8
S 0.65 × 0.2 × 0.35 = 0.0455
R
0.2
0.65
S 0.35 × 0.65 × 0.2 = 0.0455
R
0.35
S 0.35 × 0.35 × 0.35 = 0.042 875
S
Total probability is 0.237 875
13 a and b
G
A
G 0.7 0.4
A 0.3 0.6
300
100
G = gym, A = aerobics
0.7 0.4
0.3 0.6
50
× 300 = 228.57
100
171.43
229 will use the gym and 171 will attend an aerobics
class.
14 a i E(X) = 100 × 0.5
= 50
ii Var(X) = 100 × 0.5 × 0.5
= 25
iii SD(X) = 25
=5
b i E(X) = 50 × 0.7
= 35
ii Var(X) = 50 × 0.7 × 0.3
= 10.5
iii SD(X) = 10.5
= 3.24
c i E(X) = 80 × 0.2
= 16
ii Var(X) = 80 × 0.2 × 0.8
= 12.8
iii SD(X) = 12.8
= 3.58
1
15 a i E(X) = 10 × --3
·
= 3.3
1
ii E(X) = 10 × --4
= 2.5
5
iii E(X) = 10 × -----12
= 4.17
b Var(X) = npq
1 2
= 10 × --- × --3 3
= 2.22
c SD(X) = 2.22
= 1.49
16 a E(X) = 10 so np = 10
Var(X) = 8 so npq = 8
npq = 8
10q = 8
q = 0.8
q=1−p
= 1 − 0.8
p = 0.2
b np = 10
n × 0.2 = 10
n = 10 ÷ 0.2
= 50
1
4
17 a n = 40, p = --- , q = --5
5
1--E(X) = 40 ×
5
=8
b Pr(X > 8) = 1 − Pr(X ≤ 8)
= 1 − 0.5931
= 0.4069
18 a n = 2400, p = 60%
= 0.60
E(X) = 2400 × 0.6
= 1440
b Var(X) = npq
= 1440 × 0.4
= 576
SD(X) =
Var ( X )
= 576
= 24
µ = E(X) = 1440, σ = SD(X) = 24
µ + 2σ = 1440 + 48
= 1488
µ − 2σ = 1440 − 48
= 1392
c There is a probability of 0.95 that between 1392 and
1488 patients will be cured.
Analysis
1 a n = 150, p = 0.9
E(X) = 150 × 0.9
= 135
b Probability of a free pizza = q
=1−p
= 1 − 0.9
= 0.1
c Late deliveries = 150 − 135
= 15
Loss = 15 × 4
= 60
Saverio expects to lose $60 this night.
The binomial distribution
2 a n = 10, p = 0.1
E(x) = np
= 10 × 0.1
=1
b Pr(X = 5) = 10C5(0.1)5(0.9)5
= 0.0015
c Pr(X ≥ 2) = 1 − Pr(X < 2)
= 1 − [Pr(X = 0) + Pr(X = 1)]
= 1 − [10C0(0.1)0(0.9)10
+ 10C1(0.1)1(0.9)9]
= 1 − [0.348 68 + 0.387 42]
= 1 − 0.7361
= 0.2639
d n = 1000, p = 0.01
E(X) = 1000 × 0.01
= 10
3 a n = 20, p = 0.05
E(X) = 20 × 0.05
=1
b Pr(X > 1) = 1 − Pr(X ≤ 1)
= 1 − [Pr(X = 0) + Pr(X = 1)]
= 1 − [20C0(0.05)0(0.95)20 + 20C1(0.05)1(0.95)19
= 1 − [0.358 49 + 0.377 35]
= 1 − 0.735 84
= 0.26416
≈ 0.2642
c Pr(X > 5) = 1 − Pr(X ≤ 5)
= 1 − [Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3)
+ Pr(X = 4) + Pr(X = 5)]
= 1 − [0.7358 + 20C2(0.05)2(0.95)18 + 20C3(0.05)3(0.95)17
+ 20C4(0.05)4(0.95)16 + 20C5(0.05)5(0.95)15]
= 1 − [0.735 84 + 0.188 68 + 0.059 58 + 0.013 33
+ 0.002 24]
= 1 − 0.9997
= 0.0003
MM12-11
299
340
4 --------- = 85% Tellya → Tellya
400
15% Tellya → Yodacall
90
--------- = 90% Yodacall → Yodacall
100
10% Yodacall → Tellya
a ti + 1 = 0.85ti + 0.1yi
yi + 1 = 0.15ti + 0.9yi
400
b Tellya = --------500
= 80% of the market
100
Yodacall = --------500
= 20% of the market
c ti + 1 = 0.85 × 400 + 0.1 × 100
= 350
yi + 1 = 0.15 × 400 + 0.9 × 100
= 150
d
0.85 0.10
0.15 0.90
4
400 = 263.28
100
236.72
52.6% Tellya 47.4% Yodacall
e
0.85 0.10
0.15 0.90
7
400 = 226.696 777 3
273.303 222 7
100
45.4% Tellya 54.6% Yodacall
f
0.85 0.10
0.15 0.90
12
400 = 206.335 270 4
293.664 729 6
100
41.2% Tellya 58.8% Yodacall
50
g
0.85 0.10
400 = 200
300
0.15 0.90
100
40% Tellya 60% Yodacall
Both will be viable however Tellya’s share has dropped
by half to 40% while Yodacall’s have increased, tripled
in fact.
answers
MQ-12_Sol_MM_Ch 11.fm Page 299 Thursday, August 31, 2006 7:07 PM
answers
MQ-12_Sol_MM_Ch 12.fm Page 300 Wednesday, August 30, 2006 7:30 PM
MM12-12
300
Continuous distributions
Chapter 12 Continuous distributions
Exercise 12A—Probability
density functions
h
1
= 2 × --2
1 a
y
= 1
f(x) may be a pdf.
y
( 1− , 2)
2
(0, 2)
2
0
e
y
Area =
0
x
1
−
2
1
0
p
−−
4
∫
1--2
Area = 2 ×
= 2x
0
= 1
f(x) may be a pdf.
1 , 0)
(−
2
x
∫
π
--4
cos 2x
0
= 4--- – 0
3
Area is not equal to 1, so f(x) is
not a pdf.
dx
π
--4
2 a
0
∫ a ( 2x – 1 )dx = 1
2
1
( 4a – 2a ) – ( a – a ) = 1
2a = 1
1
a = --2
(−3p, 2)
2
b
2
∫
1
−2p
y
(−1, 2)
− 3p −p
2
1
Area =
x
(1, 0)
x2
= x – ----2
x
−p 0
2
3π
− -----2
2 sin x
–2 π
1
c
∫
π
0
= 2
Area is not equal to 1, so f(x) is
not a pdf.
d
g
y
1
∫ ae
3
0
ae x – 2
Area =
0
1
x
e1
1
e
e
∫
-------- dx
0 2
x3
= 2 × ----2
1
0
= 1
1a = ---------e–1
= log e x
Area = 2 ×
3
a ( e1 – 1 ) = 1
x
e
∫ --x- dx
1
1 3x 2
dx = 1
ae 1 – ae 0 = 1
(1, −32 )
3
−
2
= 1
0
2
1)
(e, −
e
1
x–2
2
(1, 1)
1−
e
y
π
a--- – − a--- = 1
2
2
a = 1
d
−1
a--- sin x dx = 1
2
– a--- cosx
2
– 2π
= 2
Area is not equal to 1, so f(x) is
not a pdf.
–1
= 1
−1
1a = ----18
dx
– 3-----π2
1
9a – – 9 a = 1
18a = 1
= 0 – –2
1
= 1--- – −1 --2
2
(−1, −23 )
a ( 10 – 3x ) dx = 1
a ( 10x – x 3 )
= – 2 cosx
( 1 – x ) dx
–1
∫
2
–1
c
1
= 1
1
y
∫
2
ax 2 – ax
f
f(x) is not positive or 0 over the
whole interval, so it is not a pdf.
Area =
x – 1 dx
1
= 1
f(x) may be a pdf.
(0, −2)
0
2
1
1
= 2 × --- – 0
2
y
−1
∫2
2 × 2( x – 1)
= ------------------------------3
x
p
−
4
1
= 2 × --- sin2x
2
b
0
x
(1, 0) 2
3--- 2
2
1
--2
2 dx
0
Area =
(2, 2)
= 1–0
= 1
f(x) may be a pdf.
2 ax 3
∫ -------4 - dx = 1
0
4
ax
-------16
2
= 1
0
16a
--------- = 1
16
a = 1
MQ-12_Sol_MM_Ch 12.fm Page 301 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
∫
–2
f
a 2 – x dx = 1
–7
4
3--- – 2
2
–----------------------------2a ( 2 – x ) 3
∫
0
−1
a ( x + 2 )dx +
x 2- + 2x
a ---2
= 1
–7
∫
–1
–4
ax
–2
0
–1
2
0
−1
2a = ----13
2 ----( x + 2 ), – 1 ≤ x ≤ 0
13
f(x) = 2 1
- --- x + 2 , 0 ≤ x ≤ 2
----
13 2
0, elsewhere
= 1
−4
y
∫ ax ( x – 1 )dx = 1
a x – x dx = 1
∫
6
—
13
4
3
4
—
13
4 2
3
2
—
13
4
=1
3
–1 0
a 64
------ – 8 – 9 – 9--- = 1
3
2
5
a 40
------ – 9--- = 1
3 2
∫
0
a dx +
ax
∫
2
1
∫ mx dx + ∫ m( 12 – x ) dx = 1
3
2
4
1
3
--------3
x 2+ m 12x – ---2
1
=1
3
26m
---------- + m 40 – 63
------ = 1
3
2
26m
---------- = 1
---------- + 17m
2
3
2
ax 2
+ -------2
0
4
9m – m
---- + m ( 48 – 8 ) – 36 – 9--- = 1
3
2
ax dx = 1
1
x
2
mx 3 3
53
a × ------ = 1
6
6a = ----53
1
=1
0
13a- = 1
-------2
15a- = 1
-------4
4a = ----15
3 x2
a x----- – ----3 2
2
1
+ a --- x 2 + 2x
4
3a
------ + 5a = 1
2
( a – a ) – a--- – 4a = 1
4
3
∫ a --2- x + 2 dx = 1
+ a dx = 1
– ax − 1 + ax
h
1
301
3a
------ + a ( 1 + 4 – 0 ) = 1
2
−16a
------------- – – 18 a = 1
3
38a- = 1
-------3
3a = ----38
g
MM12-12
= 1
1
52m
+ 51m- = 1
-------------------------6
a + 2a – a--- = 1
2
103m = 6
5a
------ = 1
2
a = 2--5
6
m = -------103
6- 2
-------x ,1≤x≤3
103
f( x) = 6
- ( 12 – x ), 3 ≤ x ≤ 4
-------103
0, elsewhere
2---, 0 ≤ x ≤ 1
5
f( x) = 2
--5- x, 1 < x ≤ 2
0, elsewhere
y
y
54
—–
103
48
—–
103
4–
5
2–
5
6
—–
103
0
1
2
x
0
1
2 3
4
x
answers
MQ-12_Sol_MM_Ch 12.fm Page 302 Wednesday, August 30, 2006 7:30 PM
302
MM12-12
6
∫
0
– x dx +
–1
--2
0
2
– x----2
–1
--2
Continuous distributions
∫ x dx = 1
k
Pr ( 1.2 < X < 1.75 )
b
0
=
k
x2
+ ----2
= 1
1--- + k----2- = 1
8 2
Pr ( X > 1.5 -)
= --------------------------Pr ( X > 1.2 )
Pr ( X > 1.2 ) = 0.84
= 0.5625
---------------0.84
----- dx = 1
4
n
= 1
–1
= 0.6696
2 a Pr ( X < 3 )
1
n3
------ – − ------ = 1
12
12
∫ --9- ( 4 – x ) dx
= 1--- ( 16 – 8x + x ) dx
9∫
n 3- = 11
---------12
12
x3
= 1--- 16x – 4x 2 + ----9
3
n 3 = 11
= 1--- ( 48 – 36 + 9 ) – 16 – 4 + 1---
9
3
n = 11
Let the length of the rectangle = x and the width of the
rectangle = y
b
c
= 1--- x 2 + x
4
2
1.5
2
∫ --9- ( 4 – x ) dx
31
2
2
x3
= 1--- 16x – 4x 2 + ----3
9
3
2
2
= 1--- 21 – 18 --3
9
= 0.2593
d
1--- ( 2x + 1 )dx
4
Pr ( 2 < X < 3 )
=
Exercise 12B—Using a probability density
function to find probabilities of continuous
random variables
1.5 4
2
2
= 0.2963
p = 5
2
41
= 1--- 64 – 64 + 64
------ – 32 – 16 + 8---
9
3
3
1 - = ----1-------------30 – p
25
∫
∫ --9- ( 4 – x ) dx
x 3= 1--- 16x – 4x 2 + ---9
3
1
So, the width must be equal to -----------b–a
Pr ( X > 1.5 )
1
Pr ( X > 2 )
=
1 = b–a×w
1
------------, a ≤ x ≤ b
f( x) = b – a
0, elsewhere
3
= 0.9630
A = L×W
=
2
1
x = b–a
Area must equal 1
1 a
2
1
3
3
b
31
=
1n 3- = 1 – --------12
12
8 a
1.2
( X > 1.5 ∩ X > 1.2 )= Pr
---------------------------------------------------Pr ( X > 1.2 )
n x2
x 3----12
1.75
Pr ( X > 1.5 X > 1.2 )
k = ------72
–1
--- ( 2x + 1 )dx
4
= 0.543 125
c
= 7--4
∫
1.2
= 1--- [ 4.8125 – 2.64 ]
4
k 2- = 7-----2
8
7
1.75 1
= 1--- x 2 + x
4
0
14
k 2 = -----8
∫
Pr ( 2 < X < 3 X < 3 )
( 2 < X < 3 ∩ X < 3 )= Pr
---------------------------------------------------Pr ( X < 3 )
( 2 < X < 3 )= Pr
-------------------------------Pr ( X < 3 )
1
= --- [ 6 – 3.75 ]
4
= 0.2593
---------------0.9630
= 0.5625
= 0.2693
MQ-12_Sol_MM_Ch 12.fm Page 303 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
Pr ( X > 2 X < 3 )
e
4 a
b
( X > 2 ∩ X < 3 )= Pr
------------------------------------------Pr ( X < 3 )
y
(0, 1
−)
4
( 2 < X < 3 )= Pr
-------------------------------Pr ( X < 3 )
=
b A =
= 0.2693
∫
a x
∫
2
0
ax ( x – 2 ) dx = 1
2 2
0
=
– 2x dx = 1
=
2
3
a x----- – x 2
3
0
1.5
k
c i
lim
k→∞
x 3- – x 2
= –3
--- ---4 3
ii
0
( −e – 0.25k
– –1 )
∫
2
0
0.25e
– 0.25x
∫ 0.25e
– 0.25x
= 0.2212
∫ x----- dx
1
3
1 2
0
iii Pr ( 2 < X < 3 ) =
iii Pr ( X > 1 X < 2 )
∫
( 1 < X < 2 )= Pr
-------------------------------Pr ( X < 2 )
1.5
0.4
=
= 0.739 75
∫
2
1
1
=
[ −e – 0.5
–
– e – 0.25 ]
= 0.1722
( 0.4 < X < 1.5 )= Pr
----------------------------------------Pr ( X < 1.5 )
Pr ( X > 1 X < 2 ) = 0.1722
---------------0.3935
0.739 75 = --------------------------Pr ( X < 1.5 )
= 0.4378
∫
1.5
0
f ( x )dx
= 0.843 75
5 a
1
= --3
v Pr ( X < 3 X > 2 )
( 2 < X < 3 -)
= Pr
-------------------------------Pr ( X > 2 )
(1, 1)
75= 0.739
-------------------0.843 75
0
1--6
= -----1--2
( X < 3 ∩ X > 2 )= Pr
------------------------------------------Pr ( X > 2 )
y
1
1
dx
3
2
= 1
--6
( 2 < X < 3 )= Pr
-------------------------------Pr ( X > 2 )
= – e – 0.25x
Pr ( X > 0.4 ∩ X < 1.5 )= ---------------------------------------------------Pr ( X < 1.5 )
1
2 2
Pr ( 2 < X < 3 ∩ X > 2 )= ---------------------------------------------------Pr ( X > 2 )
f ( x ) dx
2
iv Pr ( X > 0.4 X < 1.5 )
3
iv Pr ( 2 < X < 3 X > 2 )
Pr ( 1 < X < 2 )
= – 3--- [ −1.125 – – 0.1387 ]
4
∫ ----x -
= – 1--- + 1--3 2
( X > 1 ∩ X < 2 )= Pr
------------------------------------------Pr ( X < 2 )
3
– --- x ( x – 2 ) dx
0.4 4
1
= – 1--x
= 0.7788
1.5
3
= 2--3
Pr ( X > 1 ) = 1 – 0.2212
iii Pr ( 0.4 < X < 1.5 )
= 0.8767
1
= – 1--- – – 1
3
1
= 0.104
Pr ( X < 1.5 ) =
2
= – 1--x
dx
∫
= – 3--- ----- – x 2
4 3
1
1
= – --x
0
x3
1
----- dx
x2
2
ii Pr ( X < 3 ) =
= ( −e – 0.25 + 1 )
=
∫
Pr ( X > 2 ) = 0.5
0
0.4
1
----- dx
2
x
= 0.5
0
= – e – 0.25x
x 3- – x 2
= – 3--- ---4 3
2
1
= − --- – – 1
2
= 0.156 25
0
∞
∫
c i Pr ( X > 2 ) =
ii Pr ( X > 1 ) = 1 – Pr ( X < 1 )
3
– --- x ( x – 2 ) dx
4
1
Since f ( x ) ≥ 0 and the total area
under the curve is 1, f(x) is a pdf.
dx
= 0.3935
8
--- – 4 – ( 1.125 – 2.25 )
3
k
= 1
Pr ( X < 2 ) =
= ( −e – 0.5 – – 1 )
1.5
1 2
1
1
lim – --- – − --1
k→∞ k
2
2
1
k
= 1 – Pr ( X < 2 )
1
0.4
dx
k
lim – e – 0.25x
k→∞
= – e – 0.25x
3
– --- x ( x – 2 ) dx
4
= – 3--4
–0.25x
0
∫ ----x - dx
1
lim – --k→∞ x
Since f ( x ) ≥ 0 and the total area
under the curve is 1, f(x) is a pdf.
3
a = − --4
2
∫ 0.25e
= 1
4a
− ------ = 1
3
∫
lim
k→∞
= 0+1
=1
a 8--- – 4 = 1
3
b i
x
0
= 0.2593
---------------0.9630
3 a
A = lim
k→∞
=
303
MM12-12
x
= 1--3
answers
MQ-12_Sol_MM_Ch 12.fm Page 304 Wednesday, August 30, 2006 7:30 PM
304
MM12-12
Continuous distributions
Pr ( X ≤ a ) = 0.36
--- x + 1--- dx = 0.36
13
2
6
6 – 22Using the quadratic formula, a = -----------------2
∫
a1
x----2- + 1--- x
6 2
6 + 22
(Disregard ------------------- )
2
a
= 0.36
1
9 a
a2
----- + a
--- 1 1
6 2 – --6- + --2- = 0.36
f (q)
2
1)
(0, −
2
8- = 0.36
a - + a--- – -------(multiply by 12)
6 2 12
2a 2 + 6a – 8 = 4.32
p, 0)
(− −
2
2a 2 + 6a – 12.32 = 0
Using the quadratic formula, a = 1.4
(Disregard a = −4.4 )
1
Pr ( X ≥ b ) = --3
7
∫
4
b
4
----- dx = 1--3
x2
– 4--x
4
∫
∫
a
sin θ
1
( sina – sin 0 ) = --2
1
sin a = --2
4--- = 4--b
3
1
a = sin − 1 ---
2
b = 3
0.4 quartile
∫ f( x ) dx = 0.4
--a = π
6
a
0
∫
a
0
2--- ( 3 – x ) dx = 0.4
5
2--x 23x – ---5
2
= 1--2
0
−1 + 4--- = 1--b
3
8 a
10 a
y
a
= 0.4
1)
(0, −
8
0
0.4 × 5
3a – ----- = ---------------2
2
a2
2
3a – a----- = 1
2
a 2 – 6a + 2 = 0
b
(Disregard – 3 – 7 )
b 0.85 quantile
∫
0
– 19Using the quadratic formula, a = 6-----------------2
6 + 19
(Disregard ------------------- as it is beyond the acceptable
2
domain.)
c 70th percentile
∫
a
0
f ( x ) dx = 0.70
From part a,
2
0.70 × 5
3a – a----- = ------------------2
2
a 2 – 6a + 3.5 = 0
0
2
– 10 --3
3-------x + 1--- dx +
256
8
x
(4,0)
∫ -----164
1
4 – x dx
0
3--- 4
0
2- ( 4 – x ) 2
+ – ----48
2
– 10 --3
0
= 2--- + 1--3 3
= 1
From part a,
a 2 – 6a + 4.25 = 0
∫
3x 2- + 1--- x
= -------512 8
f ( x ) dx = 0.85
2
0.85 × 5
3a – a----- = ------------------2
2
0
2 , 0)
(−10 −
3
Using the quadratic formula, a = – 3 + 7
a
(p
− , 0) q
2
b Given the symmetrical nature of the graph,
Pr ( – a ≤ θ ≤ a ) = 2 × Pr ( 0 ≤ θ ≤ a )
a 1
1
2×
--- cos θ d θ = --0 2
2
a
1
cos θ d θ = --0
2
= 1--3
b
0.5
a
−a 0
c
i Pr ( X < – 1 ) =
∫
–1
2
– 10 --3
3-------x + 1--- dx
256
8
3x 2- + 1x
= ------------512 8
−1
2
–10 --3
= – 0.1192 – – 0.6667
= 0.5475
4 1
ii Pr ( X > 2 ) =
------ 4 – x dx
2 16
∫
3--- 4
2
2
= − ------ ( 4 – x )
48
= 0 – – 0.1179
= 0.1179
2
MQ-12_Sol_MM_Ch 12.fm Page 305 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
iii Pr ( X > – 5 ) =
∫
0
–5
3-------x + 1--- dx +
256
8
2
1x
3x
= --------- + -----512 8
0
∫
1- 4 – x dx
----16
4
0
1
+ --3
–5
c
= 0.4785 + 1--3
∫
= 0.8118
iv Pr ( X < 3 ) = 1 – Pr ( X > 3 )
4 1
= 1 – ------ 4 – x dx
3 16
= 1--- sin
2
v Pr ( – 2 < X < 3 ) = Pr ( X < 3 ) – Pr ( X < – 2 )
–2
3 -x 1
-------= 0.9583 –
+ --- dx
2
8
–10 --- 256
= 3--4
−2
2
− 10 --3
1
π
1
π
ii Pr X < – --- = --- sin × 2 × − --- + --
6 2
2
6
= 0.9583 – ( – 0.2266 – – 0.6667 )
= 0.9583 – 0.4401
= 1--- sin – π
--- 1
3 + --22
= 0.5182
vi Pr ( – 2 < X < 1 X < 3 )
1
= --- × – ------3- + 1--2
2 2
( –2 < X < 1 ∩ X < 3 )
= Pr
-------------------------------------------------------Pr ( X < 3 )
Pr ( −2 < X < 1 )
= -----------------------------------Pr ( X < 3 )
–2
3 + 2= –-------------------4
3-------x + 1--- dx +
256
8
1x3x 2- + ----= -------512 8
π
Pr X < − --- ∩ Pr ( X < 0 )
6
π
iii Pr X < − --- X < 0 = -----------------------------------------------------------
6
Pr ( X < 0 )
∫
1- 4 – x dx
----0 16
1
3--- 1
0
2
2
+ − ------ ( 4 – x )
48
−2
---
Pr X < – π
6
= ----------------------------Pr ( X < 0 )
0
= ( 0 – 0.2266 ) + ( – 0.2165 – 0.3333 )
–-------------------3 + 24 = -------------------1
--2
= 0.2266 + 0.1168
= 0.3434
0.3434
Pr ( – 2 < X < 1 X < 3 ) = ---------------0.9583
= 0.3584
3 + 2= –-------------------2
11 a
y
12 a
(0, 1)
x
f(x) = a sin ---, 0 ≤ x ≤ π
2
π
∫ asin --2x
dx = 1
0
(− −p, 0) 0
4
b
π
--4
cos 2xdx
–π
--4
∫
= 2×
( −p, 0)
4
π
--4
cos
0
∫
2x dx
1
= 2 × --- sin2x
2
= sin2x
x
x
– a 2cos --2
x
– 2 a cos --2
π
= 1
0
π
=1
0
1 cos π
--- – cos 0 = – ----
2
2a
π
--4
10 – 1 = – ----2a
0
1– 1 = – ----2a
= sin π
--- – sin 0
2
= 1
π
--- + 1-- 6 2
1 1
= --- × --- + 1--2 2 2
3
3x 2 1x
= 0.9583 – --------- + -----512 8
∫
–π
--4
π
π
1
1
d i Pr X < ------ = --- sin 2 × ------ + --
12 2
12 2
3
= 1 – ( 0 – – 0.0417 )
= 0.9583
∫
x
= 1--- sin2x + 1--2
2
3 4
---
Pr ( – 2 < X < 1 ) =
cos 2x dx
– --π4
= 1--- sin 2x – – 1--2
2
∫
0
x
1
= --- sin2x
2
2
2
= 1 – − ------ ( 4 – x )
48
305
π
π
− --- < x < --4
4
X<x
Pr ( X < x ) =
(from part b)
MM12-12
a = 1--2
answers
MQ-12_Sol_MM_Ch 12.fm Page 306 Wednesday, August 30, 2006 7:30 PM
MM12-12
306
Continuous distributions
1 x
--- sin ---, 0 ≤ x ≤ π
b f( x) = 2 2
0, elsewhere
x
= – cos --2
π
= – cos π
--- – – c os ---
4
3
y
0
π
c i Pr X < --- =
2
1 1
= – --- + -------
2
2
(p, 1− )
2
1−
2
p
1 + 2= –-------------------2
∫
π
π
Pr X > --- = 1 – Pr X < ---
2
2
π
--2
0
= – cos π
--- – – cos 0
4
1- – −1
= – -----2
1= 1 – -----2
(from part i)
+ 2= 2--- – 2--------------2
2
π
–1 + 2
Pr X < 2π
------ X > --- = ----------------------
2
2
3
---------------------------22
π1
x
--- sin --- dx
π2
2
--3
∫
x
= – cos --2
– 2= 1 – 2--------------2
= ------22
– 2= 2--------------2
π
ii Pr X > --- =
3
1
= − --- + ------22 2
x
π
--21
--- sin --x- dx
02
2
x
= – cos --2
2 π----3
π
--2
–1 + 2
= --------------------2
π
2 + 2= –-------------------2
π
--3
π
= – cos π
--- – – c os ---
6
2
= 0 – – ------3-
2
a
13 f ( x ) = --1- ----------------- , x ∈ R and a > 0
π a2 + x2
a
y
= ------32
a=1
2π
π
iii Pr --- ≤ X ≤ ------ =
3
3
∫
2 π----3
π
--3
x
= – cos --2
2 π----3
π
--3
π
= – cos π
--- – – cos ---
6
3
= – 1--- + ------3-
2 2
3 – 1= --------------2
2π
π
Pr X < ------ ∩ X > ---
3
2
π
iv Pr X < 2π
------ X > --- = ------------------------------------------------
2
3
π
Pr X > ---
2
2π
π
Pr --- < X < ------
2
3
= -------------------------------------π
Pr X > ---
2
2π
π
Pr --- < X < ------ =
2
3
∫
a=2
x
1--sin --- dx
2
2
2 π----x
3 1
--- sin --- dx
π
2
--- 2
2
0
a=3
x
As a increases, the graph becomes ‘flatter.’
1
1
b i Pr ( X < 1 ) f ( x ) = --- × -----------π 1+x
From the graphics calculator, Pr ( X < 1 ) = 0.75
ii Pr ( – 1 X < 1 ) = 2 × Pr ( 0 < X < 1 )
Pr ( 0 < X < 1 ) = 0.25
Pr ( – 1 < X < 1 ) = 2 × 0.25
= 0.5
Pr ( – 1 < X < 1 ∩ X > – 1 )
iii Pr ( – 1 < X < 1 X > – 1 ) = -----------------------------------------------------------Pr ( X > −1 )
Pr
(
–
1
< X < 1 )= -----------------------------------Pr ( X > −1 )
0.5= --------0.75
1
--= --23--4
2
= --3
MQ-12_Sol_MM_Ch 12.fm Page 307 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
14 a
Median:
y
∫
1
– 1--- xdx = --2
5
m
– 11
2 m
(0,
– 1x
-------10
1
)
2 p
Median:
∫
m
0
2
1
3x dx = --2
= 1--2
– 11
x3
b A=
∞
∫
–∞
= 2×
∫
∞
From the graphics calculator,
area = 0.5 .
c For g(x) to be a pdf, area under
the curve must be 1
f( x) = 2 × g(x )
2
1
= 2 × ---------- e − x
2 π
2
1
f ( x ) = ------- e − x , x ∈ R
π
d i Pr ( X > 0.5 )
From the graphics calculator,
= 0.2398
ii Pr ( X < 1 ) = 0.9214
iii Pr ( X > – 0.32 ) = 0.6746
iv Pr ( – 0.5 < X < 1.09 ) = 0.6987
Exercise 12C—Measures of
central tendency and
spread
∫
=
∫
1.5
1 a Mean =
–2
x f ( x )dx
1.5
–2
m = ± 6
– 1.5
= 2.25 – 4
∫
m
–2
∫
2m + 4 = 1--2
2m = – 3.5
m = – 1.75
Mode: f(x) = 2. There is no
x-value that determines the
maximum value, so there is no
mode.
–1
b Mean =
x × – 1--- xdx
– 11
5
∫
– x3 – 1
= --------15
– 11
1- – 11
11-
= ------------------ 15
15
– 2.37
Median:
–2
∫
2
– 1--- ( 2 – x ) 3
8
−2
∫
=
∫ 3x dx
d Mean =
1
0
x × 3x 2 dx
1
3
0
3x 4
= -------4
= 3--4
1
0
3
−4
m = 2 – 3 −4
Mode: f(x) is increasing over the
interval, so the maximum value
occurs when x = 4.
Mode = 4.
m 2 + m + 1 = 1-------4
2
m = −2 + 2
Mode: Since f(x) is an increasing
function, the maximum value
occurs at the right–hand end point
of the interval, that is, x = 0 .
Mode is 0.
= 1--2
2–m =
2
m
------ + m – ( 1 – 2 ) = 1-- 4
2
m 2 + 4m + 2 = 0
Using the quadratic formula,
2
( 2 – m )3 = – 4
= 1--2
2
m
------ + m + 1--- = 0
4
2
m
– 1--- ( 2 – m ) 3 = 1---
8
2
1--x + 1 dx = 1--2
x----2- + x
4
2
--- ( 2 – x ) 2 dx = 1--8
2
–2 2
m
4
m3
2
= − --3
2 dx = 1--2
[ 2x ] –m2 = 1--2
∫
= 3.5
0
8
= 0 – − --- + 2
6
= – 1.75
Median:
2 3x 3
--------- – 12x
----------- + ------- dx
2 8
8
8
4 12x
= ( 12 – 32 + 24 ) – 3 – 4 + 3
---
2
1 2
--- x + x dx
–2 2
0
m
2
6x 2 4x 3 3x 4
= -------- – -------- + -------8
8
32
∫
2x dx
= [ x 2 ] –2
=
1
c Mean =
x --- x + 1 dx
–2 2
∫
2
2
2
0
Median:
3
4
4
the mode is – 11 .
x3 x2
= ----- + ----6 2
1--2
∫ x × --8- ( 2 – x ) dx
3x
=
∫ -----8- ( 4 – 4x + x )dx
e Mean =
m = – 6
Mode: f(x) is a decreasing
function over the interval, so the
maximum value occurs at the
left–hand end of the interval. So,
=
3
Mode: f(x) is increasing over this
interval, so the maximum value
occurs at the right–hand end point
of the interval, that is, x = 1.
Mode is 1.
2
6m
------ = ----10
10
g ( x ) dx
0
m =
m 2 = 11
----------- – 1--10
10 2
g ( x )dx
= 1--2
1
m 3 = --2
m2
– ------ + 11
------ = 1--10 10
2
x
m
0
m 2 – − 11
– ----------- = 1-- 10
10
2
0
307
MM12-12
f
∫ --5- x ( x – 1 )dx
6
6
=
∫ --5- x – --5- x dx
Mean =
26 2
1
2
3
2
1
4
3
-------- – 2x
-------= 6x
20
5
2
1
= ( 4.8 – 3.2 ) – ( 0.3 – 04 )
= 1.7
Median:
∫
∫
m6
1
--- x ( x – 1 )dx = 1--5
2
m6 2
--- x
1 5
1
– 6--- xdx = --2
5
2
2x 3- – 3x
-------------5
5
m
1
= 1--2
2
2m 3- – 3m
------------------ – 2--- – 3--- = 1-- 5
5 5 5
2
answers
MQ-12_Sol_MM_Ch 12.fm Page 308 Wednesday, August 30, 2006 7:30 PM
MM12-12
308
Continuous distributions
2m 3 – 3m 2 + 1- = 1------------------------------------5
2
3
2
2m – 3m + 1 = 5--2
3
3
2
2m – 3m – --- = 0
2
m = 1.746
Mode: The function is increasing
over the given interval, so the
maximum value occurs at x = 2 .
Mode is 2.
2 a Variance = E ( X 2 ) – µ 2
Mean =
10 2
Var ( X ) = 1 – ------
12
61
SD ( X ) =
3
2
2
2
2
–
e Var ( X ) = E ( X 2 ) – µ 2
– 4x dx
3
2
∫
=
∫ 2x
8= 18 – ----12
= 52
-----3
Variance = 52
------ – ( 4 ) 2
3
52
= ------ – 16
3
4
= --3
Standard deviation =
3
b Var ( X ) = E ( X 2 ) – µ 2
2
Mean = 1--- x ( 3 – x )dx
04
2
1
= --- 3x – x 2 dx
4 0
∫
∫
SD ( X ) =
∫ x × 4( x – x ) dx
=
∫ 4x – 4x dx
2
3
2
0
1
4
3
2
= 3x
-------- + 2x
-------- + 3x
-------28
7
14
∫ x × --4- ( 3 – x )dx
= 1--- 3x – x dx
4∫
1
3
0
x4 2
0
0
3- + 2--- + ----3= ----28 7 14
= 17
-----28
0
3
1 2
2
0
1
4
0
∫ x × 4( x – x ) dx
=
∫ 4x – 4x dx
1 2
E( X2 ) =
3
0
1
3
5
0
6
-------= x 4 – 2x
3
1
0
= 1 – 2--3
= 1--3
Var ( X ) = E ( X 2 ) – µ 2
∫ x × --7- ( x + 1 ) dx
3x
3x
6x
=
∫ -------7 - + -------7 - + -------7 - dx
E( X2) =
1
8= ----15
0
1
4
= 4--- – 4--3 5
var ( X )
1
2
4x 3 4x 5
= -------- – -------3
5
∫
3x
=
∫ -----7- ( x + 2x + 1)dx
3x
6x
3x
=
∫ -------7 - + -------7 - + -----7- dx
0
3
0
0
Var ( X ) = E ( X 2 ) – µ 2
1
3
Mean = x × --- ( x + 1 ) 2 dx
0
7
2
1
1
1 - or ------2-
= --------3 2 6
d
Var ( X ) = E ( X 2 ) – µ 2
Mean =
1= ----18
= 10
-----12
= 1--- [ 8 – 4 ]
4
= 1
f
43 8 2
Var ( X ) = ------ – ---
6 3
= 1--- 6 – 8--- – 0
4
3
= 1--- x 3 – ----4
4
2
= 43
-----6
4--3
2
= ------3
Var ( X )
= 0.4674
0.6837
3
32
= 40.5 – 36 – 8 + ------
3
=
2
SD ( X ) =
– 4x 2 dx
2x 4 4x 3
= -------- – -------4
3
variance
3x 2 x 3
= 1--- ------- – ----4 2
3
3
2
–2 π
Var ( X ) = 22.674 – ( −4.7124 ) 2
0.4674
x × 2 ( x – 2 ) dx
2
–π
∫
2 1
x × --- sin x dx
2
From the graphics calculator,
= 22.674
3 2
E( X2 ) =
–2 π
E( X2 ) =
= 8--3
2
–π
∫
1
x × --- sin x dx
2
From the graphics calculator,
= – 4.7124
Mean =
= 18 – 18 – 16
------ + 8
3
6
Var ( X )
= 0.0742
0.2725
µ2
2x 3
= -------- – 2x 2
3
∫
2
SD ( X ) =
x × 2 ( x – 2 ) dx
3
1
E ( X 2 ) = x × --- dx
2
4
2 2
E( X2 )
∫
=
∫ 2x
6
6 2
E( X2) =
291- ( or 0.0742 )
= ----------3920
Var ( X )
Var ( X ) =
Mean =
= 9--- – 1--2 2
= 4
x3
= -----12
31 17 2
Var ( X ) = ------ – ------
70 28
11= --------6
2
x2
= ----8
= 31
-----70
= 144
--------- – 100
--------144 144
44
= --------144
= 11
-----36
c
∫ --4- × xdx
3- + ----3- + 1--= ----35 14 7
3
2
0
3x 5 6x 4 x 3
= -------- + -------- + ----35
28
7
1
0
8 2
= 1--- – ------
3 15
11= -------225
SD ( X ) =
Var ( X )
11= --------15
MQ-12_Sol_MM_Ch 12.fm Page 309 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
3 a Mean of X =
∞
∫
0
x × 0.3e – 0.3x dx
c
E( X2 ) =
From the graphics calculator,
= 10
-----3
b Median of X:
∫
m
0.3e
0
– 0.3x
– e – 0.3x
0
= 1--2
– e – 0.3m – – e – 0.3 × 0 = 1--2
– e – 0.3m + 1 = 1--2
1
e – 0.3m = --2
1
– 0.3m = log e ---
2
– 0.3m = log e
5 a
π
0
y
+ 2= 1--------------2 2
(p
– , 1– )
2 2
1---------
0
2 2
Pr(X ≤ π--- | X ≥ --π- ) = ---------------2
4
1+ 2
x
(p, 0)
p
–
2
---------------2 2
Mean of X
π
π
1 - (or 0.4142)
= --------------1+ 2
∫ 0.5x cos x – --2- dx
0
6 a
= π
--2
∫
a - dx = 1
-----------------2 ( x – 1 )2
3
E( X2 ) =
2
π
d Pr X ≤ π--- =
2
a = 2
- dx
∫ (-----------------x – 1)
2x
2
From the graphics calculator,
– 2.3863
π-2
π
0.5 cos x – --- dx
2
0
∫
---
= 0.5sin x – π
2
= ( 0 – 0.5 )
= 0.5
π
a sin2 x + --- dx = 1
4
π
--4
π
− --4
= 1
a--- + a--- = 1
2 2
a = 1
= 0.4674
a--- = 1
2
π
− --4
− a--- × – 1 – − a--- × 1 = 1
2
2
0
Var ( X ) = 2.9348 – 1.5708 2
a
− --- + a = 1
2
2
∫ 0.5x cos x – --2- dx
From the graphics calculator,
2.9348
[ – a ( x – 1 ) − 1 ] 23 = 1
3
π
∫
π
--4
a
π
− --- cos2 x + ---
2
4
Var ( X ) = E ( X 2 ) – µ 2
a ( x – 1 ) − 2 dx = 1
b Mean of x =
1= 1--- + --------2 2 2
From the graphics calculator,
= 1.5708
Var ( X )
–π
--- dx
2
1
= 1--- – – 1--- × -------
2
2
2
a = 0.5
1
–
2
π
0.5cos x
π
--4
∫
---
= 0.5sin x – π
2
2a = 1
=
= 10
-----3
2
Pr(X ≥ --π- ) =
4
= 1
0
a – –a = 1
c
= 100
--------9
∫
π
asin π
--- – a sin – π
---
2 = 1
2
x × 0.3 e – 0.3x dx
π
-2
π
--4
1= --------2 2
∫ acos x – --2- dx = 1
∞ 2
–π
--- dx
2
1 1
= 0 – – --- × -------
2
2
2
---
asin x – π
2
π
--2
0.5 cos x
π
--4
π
= 0.5sin x – ---
2
π
∫
=
b
200 10 2
Var ( X ) = --------- – ------
3
9
3
Pr( --π- ≤ X ≤ --π- )
4
2
= 0.9720
From the graphics calculator,
200
E ( X 2 ) = --------9
4 a
π
π
Pr --- ≤ X ≤ ---
4
2
= ---------------------------------π
Pr X ≥ ---
4
2.9455
–2
= ---------------(x – 1)
d Var ( X ) = E ( X 2 ) – µ 2
e SD ( X ) =
π
π
Pr X ≤ --- ∩ X ≥ ---
2
4
= ---------------------------------------------π
Pr X ≥ ---
4
2
2
∫
m = 10
------ loge 2
3
c The mode of x.
The function is decreasing over
the given domain. The maximum
value occurs at the left–hand-end
of the domain, that is, when
x = 0.
The mode is 0.
0
3
SD ( X ) = 0.0782
0.2796
d Pr(µ − 2σ ≤ X ≤ µ + 2σ)
Pr(2.3863 − 2 × 0.2796
≤ X ≤ 2.3863 + 2 × 0.2796)
Pr(1.8271 ≤ X ≤ 2.9455)
Pr(2 ≤ X ≤ 2.9455) (as 2 is the
lower limit)
2.9455
2 - dx
Pr =
-----------------2
( x – 1 )2
2 −1
0.3m = log e 2
∫
2x 2
- dx
∫ (-----------------x – 1)
Var ( X ) = 5.7726 – 2.3863 2
0.0782
– 0.3m = – log e 2
E( X2 ) =
e Pr(X ≤ π--- | X ≥ --π- )
2
4
Var ( X )
From the graphics calculator,
– 5.7726
dx = 1--2
m
SD ( X ) =
309
MM12-12
b
y
π
--2
(0,1)
0
(−p
– , 0) 0
4
( –p , 0)
4
x
π
π
--4
answers
MQ-12_Sol_MM_Ch 12.fm Page 310 Wednesday, August 30, 2006 7:30 PM
310
MM12-12
c
Mean =
Continuous distributions
π
--4
∫
+π
--- dx = 1--4
2
m
sin2 x
π
− --4
∫
---
– 1--- cos2 x + π
4
2
m
–π
--4
= 1--2
1
1
π
– 1
--- cos 2m + --- – – --- cos 0 = --
2
2
2
2
--- + 1--- = 1--– 1--- cos 2m + π
2 2 2
2
=
4
12 – 2 – ( 4 4 – 2 )
=
4
12 – 4 4
5
8 Mean of X is − --- .
7
∫
0
–1
5
x × ( mx 2 + nx ) dx = − --7
∫
0
–1
3
5
mx + nx 2 dx = − --7
3
mx 4
--------- + nx
-------4
3
--- = 0
cos 2m + π
2
–2
k------------------(x + 2)4
4
b
∫
m
0
m
−2
= 1--2
= 1
2m – 3n = 6 [2]
120
2 × [1] ⇒ – 6m + 8n = – --------- [3]
7
3 × [2] ⇒ 6m – 9n = 18
[3] + [4]
m+2 =
4
8
m =
4
8–2
1--( x + 2 ) 3 dx = 0.25
–2 4
a
−2
= 0.25
( a + 2 )4 = 4
a =
4
4–2
∫
1
--- ( x + 2 ) 3 dx = 0.75
–2 4
a
1----( x + 2 )4
16
−1
2m – 3n- = 1
------------------6
( m + 2 )4 = 8
d
0
m
---- – n--- = 1
3 2
1--1
( x + 2 ) 3 dx = --2
1----( x + 2 )4
16
3
2
mx
--------- + nx
-------3
2
k = 1--4
11
----( m + 2 ) 4 – 0 = --16
2
a
mx + nx dx = 1
0 – – m
---- + n--- = 1
3 2
–2 4
∫
2
–1
16k
--------- = 1
4
1----( x + 2 )4
16
c
1f f(x) is a pdf, then
0
= 1
−2
60
4n − 3m = − ------ [1]
7
∫
k ( x + 2 ) 3 dx = 1
a
−2
5
= − --7
4n – 3m- = − 5--------------------7
12
m = 0
0
−1
5
n--- – m
---- = − --7
3 4
2m = 0
∫
0
5--0 – m
---- – n---
4 3 = − 7
--- = π
--2m + π
2
2
7 a
12
a = 4 12 – 2
e IQR = 0.75 Quantile − 0.25 Quantile
--- = 0
– 1--- cos 2m + π
2
2
Median is 0.
Mode: From the graph, the maximum value occurs when
x = 0 . Mode is 0.
d The distribution is symmetrical.
4
a+2 =
4
From the graphics calculator, = 0
Median:
( a + 2 ) 4 = 12
x sin2 x + π
--- dx
π
4
− ---
= 0.75
( a + 2 ) 4 = 0.75 × 16
– n = 6--7
6
n = − --7
6
Substitute n = − --- into [1]
7
6
60
4 × − --- – 3m = − -----7
7
24
60
− ------ – 3m = − -----7
7
36
– 3 m = − -----7
m = 12
-----7
6
m = 12
------ and m = − --- .
7
7
[4]
MQ-12_Sol_MM_Ch 12.fm Page 311 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
= ( −e – 0.9999 – 1 )
Exercise 12D—Applications to problem
solving
1 a
∫
10
Pr( 0 ≤ X ≤ 10) =
= 0.6321
0.15e – 0.15 x dx
0
3 a i
10
= – e – 0.15x
Pr ( X < 100 ) =
∫
100
0
=
–
= −e
–e0 )
0
= −e – 1.5 + 1
= 0.7769
∫
20
b Pr( 10 ≤ X ≤ 20) =
= −e – 0.1 + 1
= 0.0952
20
– e – 0.15x
=
= −e – 0.1 – – e 0
0.15e – 0.15 x dx
10
ii Pr( X ≤ 180) =
10
= ( −e
– 3.0
– – e − 1.5 )
= 0.1733
c Pr ( X > 30 ) = 1 – Pr ( X < 30 )
= 1–
∫
30
=
iii Pr ( X > 200 ) =
= e − 4.5
= 0.0111
∫
800
0
k e –k x dx
∫
1000
= – e – kx
– kx
1000
Pr ( X ≥ 230 ) = 1 –
9
k = ---------------10 000
∫
2000
0
ke
= 1 – – e –k x
–1
∫
230
0
0.7945
–k x
= 0.8353
= 1–
c
∫
n
9
k = ---------------10 000
–1 )
x
1 - – --------------------e 1000 dx = 0.10
0 1000
n
– 9x
∫
ke
= – e –k x
x – ----------1000
n
= 0.1
0
9x e ---------------10 000 dx
---------------0 10 000
From the graphics calculator, = 1111.11
1111 grams
0
∫
x
1 - e – ----------1000 dx = 0.90
----------1000
−e
∞
1111
Pr ( X ≥ n ) = 0.90
∞
= e − 1.8
= 0.1653
∫
Pr(X ≥ 230 | X ≥ 180) = 0.7945
---------------0.8353
= 0.9512
dx
2000
( – e – 1.8 –
–k x
dx
1111
0
x
1 - – --------------------e 1000 dx
1000
Pr ( X ≥ 180 ) = 1 – 0.1647
0
Pr ( X < 1111 ) =
e – 0.2
= + e – 0.23
= 1–
c
x - 200
– ----------1000
–e
= 1 – ( −e – 0.23 – – 1 )
= ( −e – 1.8 + e –0.9 )
= 0.2413
iii Pr ( X > 2000 ) = 1 – Pr ( X < 2000 )
Mean weight =
0
x
1 - – --------------------e 1000 dx
1000
( X ≥ 230 )
----------------------------b Pr(X ≥ 230 | X ≥ 180) = Pr
Pr ( X ≥ 180 )
–9
–9
---------------× 2000
× 1000
---------------= −e 10 000
– – e 10 000
b
∫
200
= 0.8187
dx
2000
∫
= 1+
= ( −e – 0.72 + 1 )
= 0.5132
Pr(1000 ≤ X ≤ 2000)
ke
x
1 - – --------------------e 1000 dx
200 1000
0
9
k = ---------------10 000
9
× 800
– ---------------
= −e 10 000
+ 1
2000
+1
∞
= 1–
800
0
=
−e – 0.18
= 1–
= – e –k x
ii
x - 180
– ----------1000
−e
30
= 1 – [ −e – 4.5 + 1 ]
Pr ( X < 800 ) =
0
0
0
i
∫
= 0.1647
= 1 – – e – 0.15x
2 a
x
1 - – --------------------e 1000 dx
1000
180
=
0.15 e – 0.15x dx
0
1
-x
1 - – --------------------e 1000 dx
1000
x - 100
– ----------1000
0
( −e – 1.5
MM12-12
9
k = ---------------10 000
–e
n – ----------1000
e
+ 1 = 0.1
n – ----------1000
= 0.9
– n - log
----------=
e 0.9
1000
– n = 1000 × log e 0.9
n = 105.36 hours
311
answers
MQ-12_Sol_MM_Ch 12.fm Page 312 Wednesday, August 30, 2006 7:30 PM
MM12-12
312
Continuous distributions
4 a Between 30 and 40 minutes
b i Pr ( X < 33 ) = P r ( 30 ≤ X < 33 )
=
∫
33
30
33
30
38
36
∫
10
0.1 dx
= 0.1x
0
38
= 3.8 – 3.6
= 0.2
∫
40
30
30
n
0
20
10
50k + 50 k = 1
100k = 1
1k = -------100
1 - x, 0 ≤ X ≤ 10
------- 100
f( x)= 1
- ( x – 20 ), 10 < X ≤ 20
– -------100
0, elsewhere
= 0.1
y
∫
10 (20, 0) x
0
2
∫
∫
0.5
–0.5
1
π
π-
1- sin − ----= ------- sin ------ – ----- 2 40
2 40
= 0.1110
b Missed the beginning at 6.58 pm means – 5 ≤ X ≤ – 2
−2 π
πx
------------- cos ------ dx
−5 20 2
20
∫
1
πx
= ------- sin -----2 20
(10, 10k)
10k
0.1
----------------------------- dx
–2 π ( 0.01 + x 2 )
From the graphics calculator = 0.9682
5
0.1
------------------------------ dx
ii Pr ( – 5 ≤ X ≤ 5 ) =
– 5 π ( 0.01 + x 2 )
From the graphics calculator, = 0.9847
b Pr ( X ≥ 407 ) Greater than the ordered length by
7 metres
Pr ( X ≥ 7 ) From the graphics calculator = 0.0045
c Shorter than 3.9 metres (less than 390 cm)
Pr(X ≤ −10)
From the graphics calculator = 0.0032
6 a 30 seconds = 0.5 minutes
Pr ( – 0.5 ≤ X ≤ 0.5 )
0.5
π πx
-----------cos ------ dx
=
– 0.5 20 2
20
Pr ( – 2 ≤ X ≤ 2 ) =
1
πx
= ------- sin -----2 20
= 1
10
k × 100- = 1
100k
- – 0 + 0 – –-------------------- ----------2
2
0.1n – 3 = 0.1
0.1n = 3.1
n = 31 minutes
i
3
20
−k ( x – 20 ) 2
+ ---------------------------2
30
5 a
5
– k ( x – 20 ) dx = 1
30
0.1 dx = 0.1
0.1x
3
40
= ( 80 – 45 )
= 35 minutes
d Pr ( X ≤ n ) = 0.1
n
10
∫
x × 0.1 dx
0.1x 2
= ------------2
∫
kx dx +
kx 2------2
36
c Mean of X =
5
1
π
1- sin 3π
------
= ------- sin --- – ----- 2 4
2 20
= 0.1790
7 a and b First determine the value for k.
= 3.3 – 3.0
= 0.3
∫
πx
cos ------ dx
20
π
∫ -----------20 2
1
π
= ------- sin ------ x
2 20
0.1 dx
= 0.1x
ii Pr ( 36 ≤ X ≤ 38 ) =
c Pr ( 3 ≤ X ≤ 5 ) =
–2
–5
1
π
1- sin − π
---
= ------- sin − ------ – -----10
4
2
2
= 0.2815
c
( X > 10 )
-------------------------Pr(X > 10 | X ≥ 7) = Pr
Pr ( X ≥ 7 )
20
1
Pr ( X > 10 ) =
− --------- ( x – 20 )dx
10 100
∫
1
= − --------- ( x – 20 ) 2
200
20
10
100
= 0 – – ---------
200
= 0.5
Pr ( X ≥ 7 ) =
∫
10
7
1 - xdx + 0.5
-------100
x2
= --------200
10
+ 0.5
7
49- + 0.5
= 100
--------- – -------200 200
= 0.755
0.5 Pr(X > 10 | X ≥ 7) = -----------0.755
= 0.6623
d It takes Nathan 3 minutes to deposit a cheque, so times
up to 12 minutes will enable him to return to his car on
time.
10 1
12
1 - ( x – 20 ) dx
--------- xdx +
Pr ( X ≤ 12 ) =
– -------0 100
10 100
∫
1
= 0.5 + – --------- ( x – 20 ) 2
200
64
= 0.5 + − --------- – – 0.5
200
= 0.68
∫
12
10
MQ-12_Sol_MM_Ch 12.fm Page 313 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
MM12-12
Exercise 12E—The normal distribution
Σxf
1 a x = -------n
26.5
× 7 + 27.5 × 14 + 28.5 × 29 + 29.5 × 12 + 30.5 × 8
= ------------------------------------------------------------------------------------------------------------------------------------70
185.5
+
385
+
826.5
+
354
+ 244
= ------------------------------------------------------------------------------70
1995
= -----------70
= 28.5
The mean time is 28.5 seconds.
b s=
Σ
( x – x ) 2-f
----------------------n–1
=
( 26.5 – 28.5 ) 2 × 7 + ( 27.5 – 28.5 ) 2 × 14 + ( 28.5 – 28.5 ) 2 × 29 + ( 29.5 – 28.5 ) 2 × 12 + ( 30.5 – 28.5 ) 2 × 8-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------70 – 1
=
28
+ 14 + 0 + 12 + 32--------------------------------------------------69
=
86
-----69
Number of swimmers
= 1.246 38
= 1.12
The standard deviation is 1.12 seconds.
c µ − σ = 28.5 − 1.12
= 27.38
µ + σ = 28.5 + 1.12
= 29.62
µ − 2σ = 28.5 − 2 × 1.12
= 28.5 − 2.24
= 26.26
µ + 2σ = 28.5 + 2 × 1.12
= 28.5 + 2.24
= 30.74
µ − 3σ = 28.5 − 3 × 1.12
= 28.5 − 3.36
= 25.14
µ + 3σ = 28.5 + 3 × 1.12
= 28.5 + 3.36
= 31.86
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
µ − 3σ
µ − 2σ
µ−σ
µ
µ+σ
µ + 2σ
26
27
28
29
30
31
Time (seconds)
µ + 3σ x
2 a Midpoints: 305, 315, 325, 335, 345, 355 & 365.
Σxf
b x = -------n
305
× 6 + 315 × 12 + 325 × 18 + 335 × 29 + 345 × 19 + 355 × 11 + 365 × 5
= ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------100
1830
+
3780
+
5850
+
9715
+
6555
+ 3905 + 1825
= -------------------------------------------------------------------------------------------------------------------------100
33 460
= ----------------100
= 334.6
The average longest drive is 334.6 metres.
313
MM12-12
314
Continuous distributions
Σ ( x – x ) 2-f
----------------------n–1
c s=
2
2
2
2
2
( 305 – 334.6 ) × 6 + ( 315 – 334.6 ) × 12 + ( 325 – 334.6 ) × 18 + ( 335 – 334.6 ) × 29 + ( 345 – 334.6 ) × 19
2
2
=
+ ( 355 – 334.6 ) × 11 + ( 365 – 334.6 ) × 5
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------100 – 1
=
5256.96 + 4609.92 + 1658.88 + 4.64 + 2055.04 + 4577.76 + 4620.8
-----------------------------------------------------------------------------------------------------------------------------------------------------------------99
=
22
784---------------99
= 230.1414
= 15.17
The standard deviation is 15.17 metres
µ − σ = 334.6 − 15.17
= 319.43
µ + σ = 334.6 + 15.17
= 349.77
µ − 2σ = 334.6 − 2 × 15.17
= 334.6 − 30.34
= 304.26
µ + 2σ = 334.6 + 2 × 15.17
= 334.6 + 30.34
= 364.94
µ − 3σ = 334.6 − 3 × 15.17
= 334.6 − 45.51
= 289.09
µ + 3σ = 334.6 + 3 × 15.17
= 334.6 + 45.51
= 380.11
µ−σ
320
µ
µ+σ
330 340 350
Drive (metres)
µ + 2σ
360 370
215
µ − 2σ
300 310
195
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
µ − 3σ
190
Number of golfers
d
3 a
Height (cm)
Negatively skewed.
210
205
200
185
180
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
175
Number of players
answers
MQ-12_Sol_MM_Ch 12.fm Page 314 Wednesday, August 30, 2006 7:30 PM
µ + 3σ
380
MQ-12_Sol_MM_Ch 12.fm Page 315 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
µ + 2σ = 0 + 2 × 1
=0+2
=2
µ − 3σ = 0 − 3 × 1
=0−3
= −3
µ + 3σ = 0 + 3 × 1
=0+3
=3
Number of jockeys
b
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
135 140 145 150 155 160 165
Height (cm)
Positively skewed.
4 a µ − σ = 10 − 2
=8
µ + σ = 10 + 2
= 12
µ − 2σ = 10 − 2 × 2
= 10 − 4
=6
µ + 2σ = 10 + 2 × 2
= 10 + 4
= 14
µ − 3σ = 10 + 3 × 2
= 10 − 6
=4
µ + 3σ = 10 + 3 × 2
= 10 + 6
= 16
4
b
MM12-12
6
–3 –2 –1 0
5 a i
ii
iii
b i
ii
1
2
3
µ = 15
µ = 15
µ = 45
3σ = 15
15
σ = -----3
=5
3σ = 5
σ = 5--3
2
= 1 --3
iii 3σ = 15
15
σ = -----3
=5
c i and ii
d i and iii
6 a
i
ii
8 10 12 14 16
µ − σ = 20 − 5
= 15
µ + σ = 20 + 5
= 25
µ − 2σ = 20 − 2 × 5
= 20 − 10
= 10
µ + 2σ = 20 + 2 × 5
= 20 + 10
= 30
µ − 3σ = 20 − 3 × 5
= 20 − 15
=5
µ + 3σ = 20 + 3 × 5
= 20 + 15
= 35
iii
0 10 20 30 40 50 60 70 80 90 100110120
b It widens the graph, increasing the range.
c It shifts the graph to the right.
7 a
–3 –2 –1 0
1
2
3
b
5 10 15 20 25 30 35
c
5 10 15 20 25 30 35
c
µ−σ=0−1
= −1
µ+σ=0+1
=1
µ − 2σ = 0 − 2 × 1
=0−2
= −2
20 30 40 50 60 70 80
8 a i, since it has the smallest mean.
b The mean increases with time.
c The standard deviation increases with time.
315
answers
MQ-12_Sol_MM_Ch 12.fm Page 316 Wednesday, August 30, 2006 7:30 PM
MM12-12
316
Continuous distributions
9 a Difference = 75 − 65
= 10
Difference
---------------------------- = 10
-----10
σ
=1
µ ± σ ≈ 68%
Approximately 68% of the scores
lie between 55 and 75.
b Difference = 85 − 65
= 20
Difference 20
---------------------------- = -----10
σ
=2
µ ± 2σ ≈ 95%
Approximately 95% of the scores
lie between 45 and 85.
c Difference = 95 − 65
= 30
Difference
---------------------------- = 30
-----10
σ
=3
µ ± 3σ ≈ 99.7%
Approximately 99.7% of the
scores lie between 35 and 95.
d µ + σ = 75
65
75
µ ± σ ≈ 68%. Therefore 32% of
the scores lie outside of the µ ± σ
values. 16% of the scores will be
greater than 75.
e µ − 2σ = 45
95%
45
65
85
µ ± 2σ ≈ 95%. Therefore 5% of
the scores lie outside of the
µ ± 2σ values. 2.5% of the scores
will be less than 45.
f µ + 3σ = 95
99.7%
35
65
95
µ ± 3σ ≈ 99.7%. Therefore 0.3%
of the scores lie outside of the
µ ± 3σ values. 0.15% of the
scores are greater than 95.
10 a µ + σ = 35
68%
25
30
95%
20
30
40
µ ± 2σ ≈ 95%
2.5% of the values are above 40.
c µ + 3σ = 45
99.7%
15
30
45
µ ± 3σ ≈ 99.7%
0.15% of the values are above 45.
11 µ = 42, σ = 169
= 13
3 16 29 42 55 68 81
68%
55
b µ + 2σ = 40
35
µ ± σ ≈ 68%
16% of the values are above 35.
a µ + σ = 55
µ ± σ ≈ 68%
68% of the values are between 29
and 55.
b µ + 2σ = 68
µ ± 2σ ≈ 95%
95% of the values are between 16
and 68.
c µ + 3σ = 81
µ ± 3σ ≈ 99.7%
99.7% of the values are between
3 and 81.
d µ − σ = 29
µ ± σ ≈ 68%
16% of the values are below 29.
e µ + σ = 55
µ ± σ ≈ 68%
16% of the values are above 55.
f µ − 3σ = 3
µ ± 3σ ≈ 99.7%
0.15% of the values are below 3.
12 a µ ± σ ≈ 68%
µ − σ = 40 − 12
= 28
µ + σ = 40 + 12
= 52
68% of the values are between 28
and 52.
b µ ± 2σ ≈ 95%
µ − 2σ = 40 − 2 × 12
= 40 − 24
= 16
µ + 2σ = 40 + 2 × 12
= 40 + 24
= 64
∴ 95% for 16 ≤ X ≤ 64
c µ ± 3σ ≈ 99.7%
µ − 3σ = 40 − 3 × 12
= 40 − 36
=4
µ + 3σ = 40 + 3 × 12
= 40 + 36
= 76
99.7% of the values lie between 4
and 76.
13 a µ ± σ ≈ 68%
µ − σ = 27.2 − 1.4
= 25.8
µ + σ = 27.2 + 1.4
= 28.6
25.8 ≤ X ≤ 28.6
b µ ± 2σ ≈ 95%
µ − 2σ = 27.2 − 2 × 1.4
= 27.2 − 2.8
= 24.4
µ + 2σ = 27.2 + 2 × 1.4
= 27.2 + 2.8
= 30.0
24.4 ≤ X ≤ 30.0
c µ ± 3σ ≈ 99.7%
µ − 3σ = 27.2 − 3 × 1.4
= 27.2 − 4.2
= 23
µ + 3σ = 27.2 + 3 × 1.4
= 27.2 + 4.2
= 31.4
23 ≤ X ≤ 31.4
d
16%
16%
68%
x
25.8 27.2 28.6
16% of the values lie above 28.6.
e
2.5%
2.5%
95%
x
24.4
27.2
30.0
2.5% of the values lie above 30.
f
0.15%
23.0
99.7%
27.2
0.15%
31.4
0.15% of the values lie above
31.4.
14 a µ − σ = 16.6 − 0.6
= 16
µ + σ = 16.6 + 0.6
= 17.2
16 ≤ X ≤ 17.2
b µ − 2σ = 16.6 − 1.2
= 15.4
µ + 2σ = 16.6 + 1.2
= 17.8
15.4 ≤ X ≤ 17.8
MQ-12_Sol_MM_Ch 12.fm Page 317 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
c µ − 3σ = 16.6 − 1.8
= 14.8
µ + 3σ = 16.6 + 1.8
= 18.4
14.8 ≤ X ≤ 18.4
14.8 15.4 16 16.6 17.2 17.8 18.4
d 16% of the values are below 16.
e 2.5% of the values are below 15.4.
f 0.15% of the values are below 14.8.
15 a
µ − 3σ µ − 2σ µ − σ
µ
µ + σ µ + 2σ µ + 3σ
38.1 59.5 80.9 103.2 123.7 145.1 166.5
50% − 16%
= 34%
ii 50% − 2.5%
= 47.5%
iii 50% − 0.15%
= 49.85%
iv 100% − 16% − 2.5%
= 81.5%
b i
16 µ = 3 kg, σ =
0.36
= 0.6 kg
a µ − σ = 3 − 0.6
= 2.4
µ + σ = 3 + 0.6
= 3.6
2.4 kg ≤ X ≤ 3.6 kg
b µ − 2σ = 3 − 2 × 0.6
= 1.8
µ + 2σ = 3 + 2 × 0.6
= 4.2
1.8 kg ≤ X ≤ 4.2 kg
c µ − 3σ = 3 − 3 × 0.6
= 1.2
µ + 3σ = 3 + 3 × 0.6
= 4.8
1.2 kg ≤ X ≤ 4.8 kg
17 µ − 2σ = 36 [1]
µ + 2σ = 60 [2]
[1] + [2] 2µ = 96
µ = 48
Sub. µ = 48 into [2]
48 + 2σ = 60
2σ = 12
σ=6
The mean is 48 and standard deviation is 6.
The answer is B.
MM12-12
317
18 The mean does not have to be greater than the standard
deviation for a normal distribution.
The answer is D.
19 µ − 3σ = 19 [1]
µ + 3σ = 61 [2]
[1] + [2] 2µ = 80
µ = 40
Sub. µ = 40 into [2]
40 + 3σ = 61
3σ = 21
σ=7
The answer is C.
20 µ = 9, σ = 9
=3
µ − 2σ = 9 − 2 × 3
=3
µ + 2σ = 9 + 2 × 3
= 15
3 ≤ X ≤ 15
The answer is C.
21 µ = 100, σ = 9
µ − 2σ = 100 − 2 × 9
= 82
µ + 2σ = 100 + 2 × 9
= 118
82 ≤ X ≤ 118
The answer is B.
22 Time is a measured quantity.
The answer is B.
23 µ = 12, σ = 2
~– 1.414
a µ − σ = 12 − 1.414
= 10.586
µ + σ = 12 + 1.414
= 13.414
10.59 ≤ X ≤ 13.41
b µ − 2σ = 12 − 2 × 1.414
= 9.172
µ + 2σ = 12 + 2 × 1.414
= 14.828
9.17 ≤ X ≤ 14.83
c µ − 3σ = 12 − 3 × 1.414
= 7.758
µ + 3σ = 12 + 3 × 1.414
= 16.242
7.76 ≤ X ≤ 16.24
Exercise 12F—The standard normal
distribution
1 a
0
1
z
Pr(Z ≤ 1) = 0.8413
b
0
Pr(Z < 2.3) = 0.9893
2.3 z
answers
MQ-12_Sol_MM_Ch 12.fm Page 318 Wednesday, August 30, 2006 7:30 PM
MM12-12
318
Continuous distributions
c
c
0
1.52
b
z
0
Pr(Z ≤ 1.52) = 0.9357
1.22
z
c
d
z
Pr(Z ≤ 0.74) = 0.7703
0 0.16
e
z
–1.75
0
z
Pr(Z < −1.75) = Pr(Z > 1.75)
= 1 − Pr(Z < 1.75)
= 1 − 0.9599
= 0.0401
Pr(Z > 0.16) = 1 − Pr(Z < 0.16)
= 1 − 0.5636
= 0.4364
d
e
0
z
Pr(Z < −1.3) = Pr(Z > 1.3)
= 1 − Pr(Z < 1.3)
= 1 − 0.9032
= 0.0968
Pr(Z ≥ 1.22) = 1 − Pr(Z < 1.22)
= 1 − 0.8888
= 0.1112
d
0 0.74
–1.3 0
z
1.234
Pr(Z < 1.234) = 0.8914
f
0
0
1.111
z
Pr(Z > 1.111) = 1 − Pr(Z < 1.111)
= 1 − 0.8667
= 0.1333
2.681 z
f
–2.23
0
z
Pr(Z < −2.23) = Pr(Z > 2.23)
= 1 − Pr(Z < 2.23)
= 1 − 0.9871
= 0.0129
e
Pr(Z < 2.681) = 0.9963
2 a
0
0
2
z
Pr(Z > 2) = 1 − Pr(Z < 2)
= 1 − 0.9772
= 0.0228
2.632 z
–2.317
Pr(Z ≥ 2.632) = 1 − Pr(Z < 2.632)
= 1 − 0.9957
= 0.0043
0
z
Pr(Z ≤ −2.317) = Pr(Z > 2.317)
= 1 − Pr(Z < 2.317)
= 1 − 0.9898
= 0.0102
3 a
f
b
–2
0
1.5
z
Pr(Z ≥ 1.5) = 1 − Pr(Z < 1.5)
= 1 − 0.9332
= 0.0668
0
z
Pr(Z ≤ −2) = Pr(Z > 2)
= 1 − Pr(Z < 2)
= 1 − 0.9772
= 0.0228
–0.669 0
z
Pr(Z ≤ −0.669) = Pr(Z > 0.669)
= 1 − Pr(Z < 0.669)
= 1 − 0.7483
= 0.2517
MQ-12_Sol_MM_Ch 12.fm Page 319 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
4 a
5 a
0
–3
0
1
2
z
–0.645 0 0.645
Pr(1 < Z < 2) =
Pr(Z < 2) − Pr(Z < 1)
= 0.9772 − 0.8413
= 0.1359
b
0
d
z
0
Pr(Z ≥ −2.1) = Pr(Z < 2.1)
= 0.9821
1.6 2.5 z
Pr(1.6 ≤ Z ≤ 2.5) =
Pr(Z < 2.5) − Pr(Z < 1.6)
= 0.9938 − 0.9452
= 0.0486
c
–0.72 0–0.41
c
–1.77
0
z
Pr(Z > −1.77) = Pr(Z < 1.77)
= 0.9616
0 0.42 1.513 z
Pr(0.42 < Z < 1.513) =
Pr(Z < 1.513) − Pr(Z < 0.42)
= 0.9349 − 0.6628
= 0.2721
d
6 a
2.71
0
40 – 25
= -----------------5
–1.6
e
z
0
1.4
z
Pr(−1.6 < Z < 1.4) =
Pr(Z < 1.4) − Pr(Z < − 1.6)
= Pr(Z < 1.4) − Pr(Z > 1.6)
= Pr(Z < 1.4) − [1 − Pr(Z < 1.6)]
= 0.9192 − [1 − 0.9452]
= 0.9192 − 0.0548
= 0.8644
b
15
= -----5
=3
b µ = 17, σ = 9
=3
x–µ
z = -----------σ
12 – 17
= -----------------3
–5
= -----3
= −1.667
Pr(Z ≥ −1.139) = Pr(Z < 1.139)
= 0.8727
f
c µ = 12, σ = 6.25
= 2.5
–2.21
–0.6420
z
Pr(Z > −0.642) = Pr(Z < 0.642)
= 0.7395
0 0.34
z
Pr(−2.21 < Z < 0.34)
= Pr(Z < 0.34) − Pr(Z < −2.21)
= Pr(Z < 0.34) − Pr(Z > 2.21)
= Pr(Z < 0.34) − [1 − Pr(Z < 2.21)
= 0.6331 − [1 − 0.9864]
= 0.6331 − 0.0136
= 0.6195
z
Pr(−0.72 < Z < −0.41)
= Pr(Z < −0.41) − Pr(Z < −0.72)
= Pr(Z > 0.41) − Pr(Z > 0.72)
= [1 − Pr(Z < 0.41)]
− [1 − Pr(Z < 0.72)]
= [1 − 0.6591] − [1 − 0.7642]
= 0.3409 − 0.2358
= 0.1051
7 a µ = 25, σ = 25
=5
x–µ
z = -----------σ
z
Pr(Z > −2.71) = Pr(Z < 2.71)
= 0.9966
–1.139 0
z
Pr(−0.645 < Z < 0.645)
= Pr(Z < 0.645) − Pr(Z < − 0.645)
= Pr(Z < 0.645) − Pr(Z > 0.645)
= Pr(Z < 0.645)
− [1 − Pr(Z < 0.645)]
= 0.7405 − [1 − 0.7405]
= 0.7405 − 0.2595
= 0.4810
b
–2.1
319
c
z
Pr(Z ≥ −3) = Pr(Z < 3)
= 0.9987
MM12-12
x–µ
z = -----------σ
15 – 12
= -----------------2.5
3
= ------2.5
= 1.2
answers
MQ-12_Sol_MM_Ch 12.fm Page 320 Wednesday, August 30, 2006 7:30 PM
MM12-12
320
Continuous distributions
8 µ = 9, σ = 3
x–µ
a z = -----------σ
Pr(X ≤ 27) = Pr(Z ≤ −1.857)
= Pr(Z > 1.857)
= 1 − Pr(Z < 1.857)
= 1 − 0.9683
= 0.0317
– 10
= --------7
= −1.429
10 – 9
= --------------3
e
1
= --3
= 0.333
x–µ
b z = -----------σ
–1.429 0
7.5 – 9
= ---------------3
z
Pr(X ≥ 30) = Pr(Z > −1.429)
= Pr(Z < 1.429)
= 0.9235
1.5
= ------3
= −0.5
x–µ
c z = -----------σ
c
40 45
x
x–µ
z1 = -----------σ
25 – 40
= -----------------7
12.4 – 9
= ------------------3
3.4
= ------3
= 1.133
40 45
x
x–µ
z = -----------σ
9 a
40 42
25
– 15
= --------7
= −2.143
x–µ
z2 = -----------σ
45 – 40
= -----------------7
45 – 40
= -----------------7
5
= --7
= 0.714
5
= --7
= 0.714
x
x–µ
z = -----------σ
42 – 40
= -----------------7
2
= --7
= 0.286
0 0.714
z
Pr(X < 45) = Pr(Z < 0.714)
= 0.7623
d
0 0.286
z
Pr(X > 42) = Pr(Z > 0.286)
= 1 − Pr(Z < 0.286)
= 1 − 0.6126
= 0.3874
b
27
40
x
x–µ
z = -----------σ
27 – 40
= -----------------7
0.714
z
Pr(25 ≤ X ≤ 45)
= Pr(−2.143 ≤ Z ≤ 0.714)
= Pr(Z < 0.714) − Pr(Z < −2.143)
= Pr(Z < 0.714) − Pr(Z > 2.143)
= Pr(Z < 0.714)
− [1 − Pr(Z < 2.143)]
= 0.7623 − [1 − 0.9839]
= 0.7623 − 0.0161
= 0.7462
10 µ = 20, σ = 25
=5
27 – 20
= -----------------5
7
= --5
x
x–µ
z = -----------σ
30 – 40
= -----------------7
0
x–µ
a z = -----------σ
– 13
= ---------7
= −1.857
30 40
–2.143
–1.857
0
z
= 1.4
Pr(X > 27) = Pr(Z > 1.4)
= 1 − Pr(Z < 1.4)
= 1 − 0.9192
= 0.0808
MQ-12_Sol_MM_Ch 12.fm Page 321 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
x–µ
b z = -----------σ
18 – 20
= -----------------5
–2
= -----5
= −0.4
Pr(X ≥ 18) = Pr(Z > −0.4)
= Pr(Z < 0.4)
= 0.6554
x–µ
c z = -----------σ
8 – 20
= --------------5
– 12
= --------5
= −2.4
Pr(X ≤ 8) = Pr(Z < −2.4)
= Pr(Z > 2.4)
= 1 − Pr(Z < 2.4)
= 1 − 0.9918
= 0.0082
d normalcdf (7, 12, 20, 5) = 0.0501
x–µ
e z1 = -----------σ
17 – 20
= -----------------5
–3
= -----5
= −0.6
x–µ
z2 = -----------σ
25 – 20
= -----------------5
5
= --5
=1
Pr(X < 17 | X ≤ 25)
= Pr(Z < − 0.6 | Z < 1)
Pr ( Z < – 0.6 ∩ Z < 1 )
= ---------------------------------------------------Pr ( Z < 1 )
Pr ( Z < – 0.6 )
= -------------------------------Pr ( Z < 1 )
1 – Pr ( Z < 0.6 )
= -------------------------------------Pr ( Z < 1 )
1 – 0.7257
= ------------------------0.8413
0.2743
= ---------------0.8413
= 0.3260
x–µ
f z1 = -----------σ
x–µ
z2 = -----------σ
µ–µ
= -----------5
0
= --5
=0
Pr(X < 17 | x < µ)
= Pr(Z < − 0.6 | Z < 0)
Pr ( Z < – 0.6 ∩ Z < 0 )
= ---------------------------------------------------Pr ( Z < 0 )
Pr ( Z < – 0.6 )
= -------------------------------Pr ( Z < 0 )
1 – Pr ( Z < 0.6 )
= -------------------------------------Pr ( Z < 0 )
1 – 0.7257
= ------------------------0.5
0.2743
= ---------------0.5
= 0.5486
11 µ = 125, σ = 11
x–µ
a z = -----------σ
140 – 125
= -----------------------11
15
= -----11
= 1.364
Pr(X > 140) = Pr(Z > 1.364)
= 1 − Pr(Z < 1.364)
= 1 − 0.9137
= 0.0863
x–µ
b z = -----------σ
100 – 125
= -----------------------11
– 25
= --------11
= −2.273
Pr(X < 100) = Pr(Z < −2.273)
= Pr(Z > 2.273)
= 1 − Pr(Z < 2.273)
= 1 − 0.9885
= 0.0115
c Pr(100 < X < 140)
= Pr(−2.273 < Z < 1.364)
= Pr(Z < 1.364) − Pr(Z < −2.273)
= 0.9137 − 0.0115
= 0.9022
12 µ = 152, σ = 49
=7
x–µ
a z = -----------σ
17 – 20
= -----------------5
159 – 152
= -----------------------7
–3
= -----5
7
= --7
=1
= −0.6
MM12-12
321
Pr(X ≥ 159) = Pr(Z > 1)
= 1 − Pr(Z < 1)
= 1 − 0.8413
= 0.1587
x–µ
b z = -----------σ
150 – 152
= -----------------------7
–2
= -----7
= −0.286
Pr(X < 150) = Pr(Z < −0.286)
= Pr(Z > 0.286)
= 1 − Pr(Z < 0.286)
= 1 − 0.6126
= 0.3874
x–µ
c z1 = -----------σ
145 – 152
= -----------------------7
–7
= -----7
= −1
z2 = 1 (from a)
Pr(145 < X < 159)
= Pr(−1 < Z < 1)
= Pr(Z < 1) − Pr(Z < −1)
= Pr(Z < 1) − Pr(Z > 1)
= Pr(Z < 1) − [1 − Pr(Z < 1)]
= 0.8413 − [1 − 0.8413]
= 0.8413 − 0.1587
= 0.6826
x–µ
d z1 = -----------σ
140 – 152
= -----------------------7
– 12
= --------7
= −1.714
x–µ
z2 = -----------σ
160 – 152
= -----------------------7
8
= --7
= 1.143
Pr(140 < X < 160)
= Pr(−1.714 < Z < 1.143)
= Pr(Z < 1.143)
− Pr(Z < −1.714)
= Pr(Z < 1.143) − Pr(Z > 1.714)
= Pr(Z < 1.143)
− [1 − Pr(Z < 1.714)]
= 0.8735 − [1 − 0.9567]
= 0.8735 − 0.0433
= 0.8302
e z1 = −1 (from part c)
z2 = −0.286 (from part b)
z3 = −1.714 (from part d)
answers
MQ-12_Sol_MM_Ch 12.fm Page 322 Wednesday, August 30, 2006 7:30 PM
MM12-12
322
Continuous distributions
Pr(145 < X < 150 | X > 140)
= Pr(−1 < Z < −0.286 | Z > −1.714)
Pr ( – 1 < Z < – 0.286 ∩ Z > – 1.714 )
= --------------------------------------------------------------------------------------Pr ( Z > – 1.714 )
Pr ( – 1 < Z < – 0.286 )
= ---------------------------------------------------Pr ( Z > – 1.714 )
Pr ( Z < – 0.286 ) – Pr ( Z < – 1 )
= ------------------------------------------------------------------------Pr ( Z < 1.714 )
=
[ 1 – Pr ( Z < 0.286 ) ] – [ 1 – Pr ( Z < 1 ) ]
--------------------------------------------------------------------------------------------Pr ( Z < 1.714 )
[ 1 – 0.6126 ] – [ 1 – 0.8413 ]
= -----------------------------------------------------------------0.9567
0.3874 – 0.1587
= --------------------------------------0.9567
0.2287
= ---------------0.9567
= 0.2391
13 Pr(Z > 1.251) = 1 − Pr(Z < 1.251)
= 1 − 0.8946
= 0.1054
The answer is A.
14 Pr(Z < − 0.25) = Pr(Z > 0.25)
= 1 − Pr(Z < 0.25)
= 1 − 0.5987
= 0.4013
The answer is B.
x–µ
15 z = -----------σ
29 – 20
= -----------------6
9
= --6
= 1.5
The answer is E.
16 µ = 1.4, σ = 0.1
x–µ
z = -----------σ
1.25 – 1.4
= -----------------------0.1
– 0.15
= ------------0.1
= −1.5
Pr(X > 1.25) = Pr(Z > −1.5)
= Pr(Z < 1.5)
= 0.9332
The answer is E.
17 µ = 70, σ = 12
= 3.4641
– µx----------z=
σ
77 – 70
= -----------------3.4641
7
= ---------------3.4641
= 2.021
Pr(X > 77) = Pr(Z > 2.021)
= 1 − Pr(Z < 2.021)
= 1 − 0.9783
= 0.0217
The answer is A.
18 µ = 12, σ = 2
x–µ
z = -----------σ
9 – 12
= --------------2
–3
= -----2
= −1.5
Pr(X < 9) = Pr(Z < −1.5)
= 1 − Pr(Z < 1.5)
= 1 − 0.9332
= 0.0668
The answer is A.
19 µ = 16, σ = 4
=2
– µx----------z=
σ
11.5 – 16
= ---------------------2
– 4.5
= ---------2
= −2.25
Pr(X > 11.5) = Pr(Z > −2.25)
= Pr(Z < 2.25)
= 0.9878
The answer is E.
20 µ = 1.000, σ = 0.006
x–µ
z1 = -----------σ
1.011 – 1.000
= --------------------------------0.006
0.011
= ------------0.006
= 1.833
x–µ
z2 = -----------σ
1.004 – 1.000
= --------------------------------0.006
0.004
= ------------0.006
= 0.667
Pr(X < 1.011 | X > 1.004) = Pr(Z < 1.833 | Z > 0.667)
Pr ( Z < 1.833 ∩ Z > 0.667 )
= ----------------------------------------------------------------Pr ( Z > 0.667 )
Pr ( 0.667 < Z < 1.833 )
= ------------------------------------------------------Pr ( Z > 0.667 )
Pr ( Z < 1.833 ) – Pr ( Z < 0.667 )
= ---------------------------------------------------------------------------1 – Pr ( Z < 0.667 )
0.9666 – 0.7477
= --------------------------------------1 – 0.7477
0.2189
= ---------------0.2523
= 0.8676
x–µ
21 a z1 = -----------σ
80 – 82
= -----------------5
–2
= ------5
= −0.4
MQ-12_Sol_MM_Ch 12.fm Page 323 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
x–µ
z2 = -----------σ
MM12-12
c
90 – 82
= -----------------5
75%
8
= --5
= 1.6
Pr(80 < X < 90) = Pr(−0.4 < Z < 1.6)
= Pr(Z < 1.6) − Pr(Z < −0.4)
= Pr(Z < 1.6) − Pr(Z > 0.4)
= Pr(Z < 1.6) − [1 − Pr(Z < 0.4)]
= 0.9452 − [1 − 0.6554]
= 0.9452 − 0.3446
= 0.6006
b Pr(X > 90) = Pr(Z > 1.6)
= 1 − Pr(Z < 1.6)
= 1 − 0.9452
= 0.0548
c Pr(X < 70) = normalcdf (−∞, 70, 82, 5)
= 0.0082
Pr(70 < X < 80) = normalcdf (70, 80, 82, 5)
= 0.3364
See parts (a) and (b) for other Probabilities.
0
z
c
Pr(Z ≤ c) = 0.75
c = 0.6745
d
52%
z
0c
Pr(Z ≤ c) = 0.52
c = 0.0502
2 a
30%
Cost ($)
Probability
1.40
0.0082
1.60
0.3364
1.80
0.6006
2.00
0.0548
Exercise 12G—The inverse cumulative
normal distribution
z
c 0
d Average Price = 1.4 × 0.0082 + 1.6 × 0.3364
+ 1.8 × 0.6006 + 2 × 0.0548
= $1.74
22 a µ = 6, σ = 0.03
Pr(5.93 ≤ X ≤ 6.07) = normalcdf (5.93, 6.07, 6, 0.03)
= 0.9804
b E(X) = 1000 × 0.9804
= 980
23 a Pr(X < 27) = normalcdf (−1010, 27, 32, 4)
= 0.1056
b In 20 fish, he would expect 20 × 0.1056, that is, 2.11 to
be undersize.
He would take home 17 fish.
Pr(Z ≤ c) = 0.3
30%
0
z
c1
Pr(Z < c1) = 0.70
c1 = 0.5244
⇒ c = −0.5244
b
1 a
90%
20%
c
0
c z
z
0
Pr(Z < c) = 0.2
Pr(Z < c) = 0.9
c = 1.2816
b
20%
60%
0
0 c
Pr(Z < c) = 0.6
c = 0.2533
z
Pr(Z < c1) = 0.8
c1 = 0.8416
⇒ c = −0.8416
c1
z
323
answers
MQ-12_Sol_MM_Ch 12.fm Page 324 Wednesday, August 30, 2006 7:30 PM
MM12-12
324
Continuous distributions
c
b
4 a
65%
35%
30%
z
c 0
z
c0
Pr(Z ≤ c) = 0.35
0
z
c
Pr(Z < c) = 0.7
c = 0.5244
Pr(Z ≥ c) = 0.65
b
65%
35%
45%
0
z
c1
0 c
1
Pr(Z < c1) = 0.65
c1 = 0.3853
z
⇒ c = −0.3853
Pr(Z < c) = 0.55
c = 0.1257
⇒ c = −0.3853
d
z
0 c
Pr(Z < c1) = 0.65
c1 = 0.3853
c
c
54%
22%
42%
0
z
c0
z
c 0
Pr(Z < c) = 0.42
c
z
c
z
Pr(Z < c) = 0.78
c = 0.7722
Pr(Z > c) = 0.54
5 a
54%
42%
0.6826
z
0 c
1
0 c
1
Pr(Z < c1) = 0.58
c1 = 0.2019
⇒ c = −0.2019
z
–c
Pr(Z < c1) = 0.54
c1 = 0.1004
⇒ c = −0.1004
3 a
0
1 – 0.6826
Each unshaded area = ------------------------2
0.3174
= ---------------2
= 0.1587
Pr(Z < c) = 1 − 0.1587
= 0.8413
c=1
d
72%
80%
b
c
z
c 0
z
0
Pr(Z > c) = 0.72
Pr(Z ≥ c) = 0.8
0.5
–c
72%
80%
0
c
z
1 – 0.5
Each unshaded area = ---------------2
0
c1
Pr(Z < c1) = 0.80
c1 = 0.8416
⇒ c = −0.8416
z
0
c1
Pr(Z < c1) = 0.72
c1 = 0.5828
⇒ c = −0.5828
z
Pr(Z < c) = 1 − 0.25
= 0.75
c = 0.6745
0.5
= ------2
= 0.25
MQ-12_Sol_MM_Ch 12.fm Page 325 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
c
MM12-12
325
b
0.4
0.4
0.4
0.2
z
–c 0 c
0 c
1
Pr(Z < z2) = 0.6
z2 = 0.2533
⇒ z1 = −0.2533
x1 – µ
z1 = ------------σ
c
Pr(Z < c) = 1 − 0.4
= 0.6
c = 0.2533
0.72
x 1 – 10
−0.2533 = ---------------2
−0.5066 = x1 − 10
x1 = 9.4934
d
0
c
z
c
Pr(Z < c) = 0.72
c = 0.5828
0.38
–c 0
z
z10 z2
Pr(Z < c1) = 0.6
c1 = 0.2533
⇒ c = −0.2533
1 – 0.2
Each unshaded area = ---------------2
0.8
= ------2
= 0.4
z
0.63
d
z
c
0.995
0.62
= ---------2
= 0.31
Pr(Z < c) = 1 − 0.31
= 0.69
c = 0.4959
c z
0
0.63
Pr(Z < c) = 0.995
c = 2.5758
7 a
z
0 z
2
6 a
Pr(Z < z2) = 0.63
z2 = 0.3319
⇒ z1 = −0.3319
0.72
0.25
10 x1
x1 – µ
z1 = ------------σ
x
z
0
z
z1 0
1 – 0.38
Each unshaded area = ------------------2
x 1 – 10
−0.3319 = ---------------2
−0.6638 = x1 − 10
x1 = 9.3362
0.72
d
0.25
0
0
c1
z
z1
Pr(Z < z1) = 0.72
z1 = 0.5828
Pr(Z < c1) = 0.75
c1 = 0.6745
x–µ
z = -----------σ
⇒ c = −0.6745
x1 – µ
z1 = ------------σ
b
0.4
c0
z
x 1 – 10
0.5828 = ---------------2
0.5828 × 2 = x1 − 10
1.1656 = x1 − 10
x1 = 11.1656
z
0.2
0
z1
Pr(Z < z1) = 0.8
z1 = 0.8416
x1 – µ
z1 = ------------σ
x 1 – 10
0.8416 = ---------------2
1.6832 = x1 − 10
x1 = 11.6832
z
answers
MQ-12_Sol_MM_Ch 12.fm Page 326 Wednesday, August 30, 2006 7:30 PM
MM12-12
326
8 a µ = 34, σ =
Continuous distributions
16 = 4
0.55
0.1854
0
z1
z
0 z
2
0.31
z1
k–µ
z1 = -----------σ
c–µ
z1 = -----------σ
k – 22
−0.895 = -------------5
c – 34
−0.1257 = -------------4
−0.5028 = c − 34
c = 33.4972
c – 34
0.4959 = -------------4
1.9836 = c − 34
c = 35.9836
−4.475 = k − 22
k = 17.525
10 Pr(Z ≤ c) = 0.8
InvNorm (0.8, 0, 1) = 0.8416
The answer is E.
11
9 µ = 22, σ = 25
=5
a
b
z
⇒ z1 = −0.895
⇒ z1 = −0.1257
c–µ
z1 = -----------σ
z2
z2 = 0.895 (using CND Table)
z2 = 0.1257
Pr(Z < z1) = 0.69
z1 = 0.4959
0
Pr(Z < z2) = 0.8146
Pr(Z < z2) = 0.55
z
0.1854
0.75
0.7
0.7
0
0.15
z
z1
Pr(Z < z1) = 0.75
z1 = 0.6745
22 − k
22 + k
22
x
c
c – 34
0.6745 = -------------4
2.698 = c − 34
c = 36.698
0.7
b
c
0.24
0
0
z2
z
InvNorm (0.7, 0, 1) = 0.5244
∴ c = −0.5244
The answer is A.
z
z
Find k by finding InvNorm
(0.5 + 0.12, 22, 5) which gives
the X value whose area to the left
is 0.62.
InvNorm (0.62, 22, 5) = 23.5274
22 + k = 23.5274
k = 1.5274
c Pr(X < k | X < 23) = 0.32
Pr(Z < z2) = 0.79
z2 = 0.8064
⇒ z1 = −0.8064
c–µ
z1 = -----------σ
c – 34
−0.8064 = -------------4
−3.2256 = c − 34
c = 30.7744
Pr ( X < k ∩ X < 23 )----------------------------------------------= 0.32
Pr ( X < 23 )
Pr ( X < k )--------------------------= 0.32
Pr ( X < 23 )
Pr(X < k) = 0.32 Pr(X < 23)
d
0.55
z1 0
z1
Pr(Z < z1) = 0.7
0.21
–z10 z1
z1
z
Find value of k by finding
InvNorm (0.85, 22, 5) which
gives 27.182.
Therefore 22 + k = 27.182.
Hence, k = 5.182.
c–µ
z1 = -----------σ
0.21
0
23 – 22
= 0.32 Pr Z < ------------------
5
z
= 0.32 Pr(Z < 0.2)
= 0.32 × 0.5793
= 0.185 376 ≈ 0.1854
12
0.4
−1.2
0
k
z
Pr(−1.2 < Z < k) = 0.4
Pr(Z < k) − Pr(Z < −1.2) = 0.4
Pr(Z < k) − [1 − Pr(Z < 1.2)] = 0.4
Pr(Z < k) − 1 + Pr(Z < 1.2) = 0.4
Pr(Z < k) − 1 + 0.8849 = 0.4
Pr(Z < k) − 0.1151 = 0.4
Pr(Z < k) = 0.5151
k = 0.038
The answer is C.
MQ-12_Sol_MM_Ch 12.fm Page 327 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
13 Pr(Z < z1) = 0.35
Pr(Z < z2) = 0.82
z2 = 0.9154
⇒ z1 = −0.9154
0
z
z2
Pr(Z < z2) = 0.65
z2 = 0.3853
⇒ z1 = −0.3853
The answer is B.
14 Pr(Z < k | Z < 0.5) = 0.6
x1 – µ
z1 = ------------σ
k – 12
−0.9154 = -------------2
−1.8308 = k − 12
k = 10.1692
The answer is B.
17 µ = 160, σ = 8
a Pr(X < x1) = 0.95
x 1 – 45
−0.8416 = ---------------0.5
−0.4208 = x1 − 45
x1 = 44.5792
Therefore the minimum length of
nails is 44.58 mm.
b Pr(X > x1) = 0.2
or Pr(X < x1) = 0.8
0.95
Pr ( Z < k ∩ Z < 0.5 )-----------------------------------------------= 0.6
Pr ( Z < 0.5 )
0.8
Pr ( Z < k ) ---------------------------= 0.6
Pr ( Z < 0.5 )
Pr(Z < k) = 0.6 Pr(Z < 0.5)
= 0.6 × 0.6915
= 0.4149
z1 z
0
0
Pr(Z < z1) = 0.95
⇒ z1 = 1.6449
0.4149
z
k 0 z1
Pr(Z < z1) = 0.5851
z1 = 0.215
⇒ k = −0.215
The answer is A.
15 Pr(Z < z1) = 0.6
z1 = 0.2533
x 1 – 45
0.8416 = ---------------0.5
0.4208 = x1 − 45
x1 = 45.4208
Therefore the maximum nail
length is 45.42 mm.
19 µ = 50, σ = 2
Pr(X < x1) = 0.25
0.2
z1
0
z1 0
z
0
x 1 – 160
−0.8416 = ------------------8
−6.7328 = x1 − 160
x1 = 153.2672
Luisa is approximately 153.27 cm
tall.
18 µ = 45, σ = 0.5
a Pr(X < x1) = 0.2
x 1 – 50
−0.6745 = ---------------2
−1.349 = x1 − 50
x1 = 48.65
Therefore the qualifying time is
48.65 seconds, or less.
20 µ = 20
Pr(X ≥ 19) = 0.7
0.7
0.2
z2
z
x1 – µ
z1 = ------------σ
0.82
0
z2
Pr(Z < z1) = 0.25
z1 = −0.6745
x1 – µ
z1 = ------------σ
4 =2
0.25
z
z2
Pr(Z < z2) = 0.8
z2 = 0.8416
⇒ z1 = −0.8416
0.82
z1
x1 – µ
z1 = ------------σ
0.25
k – 20
0.2533 = -------------4
1.0132 = k − 20
k = 21.0132
The answer is D.
16 µ = 12, σ =
x 1 – 160
1.6449 = ------------------8
13.1592 = x1 − 160
x1 = 173.159
This is approximately 173.16 cm
tall.
b Pr(X > x1) = 0.8
or Pr(X < x1) = 0.2
0.2
k–µ
z1 = -----------σ
z
z1
Pr(Z < z1) = 0.8
z1 = 0.8416
x1 – µ
z1 = ------------σ
0.4149
327
Pr(Z < z2) = 0.8
z2 = 0.8416
⇒ z1 = −0.8416
k–µ
z1 = -----------σ
0.35
MM12-12
z
0.2
z1
0
z2
z
z1 0
z
answers
MQ-12_Sol_MM_Ch 12.fm Page 328 Wednesday, August 30, 2006 7:30 PM
MM12-12
328
Continuous distributions
0.7
0
Pr(Z < z1) = 0.7
z1 = 0.5244
x1 – µ
z1 = ------------σ
240 – µ
0.5244 = -----------------30
15.732 = 240 − µ
µ = 240 − 15.732
= 224.268
The mean puzzle completion
time is 224.27 seconds or
3 min 44.27 seconds.
– 15
σ = ------------------– 1.2816
= 11.704
The standard deviation is
11.704 grams.
z
z2
Pr(Z < z2) = 0.7
z2 = 0.5244
⇒ z1 = −0.5244
23 σ = 16
=4
Pr(X ≥ 17) = 0.99
0.99
x1 – µ
z1 = ------------σ
Chapter review
19 – 20
−0.5244 = -----------------σ
−0.5244σ = −1
–1
σ = ------------------– 0.5244
= 1.907
The standard deviation is 1.907.
21 µ = 50
Pr(X < 48) = 0.4
0.4
0.4
z
z10 z2
Pr(Z < z2) = 0.6
z2 = 0.2533
⇒ z1 = −0.2533
x1 – µ
z1 = ------------σ
48 – 50
−0.2533 = -----------------σ
−0.2533σ = −2
–2
σ = ------------------– 0.2533
= 7.896
The standard deviation is 7.896.
22 µ = 500
Pr(X < 485) = 0.1
0.1
z1
0.1
0
z2
Pr(Z < z2) = 0.9
z2 = 1.2816
⇒ z1 = −1.2816
z
z1
Multiple choice
1 II and III cannot be pdfs because
they are not positive over their given
domain.
0.99
∫
I
1--3
0
3 dx
1--3
= 3x
z2 z
0
Pr(Z < z2) = 0.99
z2 = 2.3263
⇒ z1 = −2.3263
x1 – µ
z1 = ------------σ
17 – µ
−2.3263 = --------------4
−9.3052 = 17 − µ
µ = 17 + 9.3052
= 26.3052
The mean is 26.305.
24 σ = 3
Pr(X ≥ 27) = 0.35
or Pr(X < 27) = 0.65
0
= 1
I is the only possible pdf.
The answer is A.
2 2sin 4x a ≤ x ≤ b
π
Period = --2
Sine function is positive
0, π
------ , …
--- , π
---, 3π
4 2 4
∫
3 π----4
2sin
--π2
4x dx = 1
The answer is D.
2 a
----- dx = 1
3
1 x2
∫
0.65
2
a
− --x
z
0 z
1
Pr(Z < z1) = 0.65
z1 = 0.3853
x1 – µ
z1 = ------------σ
27 – µ
0.3853 = --------------3
1.1559 = 27 − µ
µ = 27 − 1.1559
= 25.8441
The mean is 25.844.
25 Pr(X < 240) = 0.7
= 1
1
− a--- – −a = 1
2
a--- = 1
2
a = 2
The answer is C.
4 Pr ( X > 1.2 ) =
∫
2--( 4 – x )dx
5
2
1.2
= 1–
∫
The answer is C.
5 Pr ( X < a ) = 0.625
a 1
--- dx = 0.625
0 4
∫
a
x1 – µ
z1 = ------------σ
485 – 500
−1.2816 = -----------------------σ
−1.2816σ = −15
z
0
0.7
= 0.625
0.25x
0
0.25a – 0 = 0.625
0
z1
z
a = 2.5
The answer is C.
1.2
1
2--( 4 – x )dx
5
MQ-12_Sol_MM_Ch 12.fm Page 329 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
∫ --3- x
4
=
∫ --3- x
6 Mean of X =
4
4
−2
1
4
1
× x dx
− 1 dx
4
= --- log e x
3
4
1
= 4--- log e 4 – 4--- log e 1
3
3
= 4--- ( log e 4 )
3
1.8484
The answer is A.
m 4 −2
--- x dx = 1--7
1 3
2
∫
4
− --- x − 1
3
m
= 1--2
1
44
− -----– − --- = 1-- 3m
3
2
4 4
1
− ------- + --- = --3m 3
2
4 - = 5-------3m
6
24 = 15m
m = 24
-----15
8
= --5
The answer is B.
8 The graph is decreasing over the
given interval, so the maximum
value occurs at the left-hand end of
the domain. The mode is 1.
The answer is D.
9 Var ( X ) = E ( X 2 ) – µ 2
∫
=
∫
=
4
1
4
1
2
2
--- loge 4
--- x−2 dx – 4
x ×4
3
3
2
4
4--dx – --- log e 4
3
3
2
4 4 4
= --- x – --- log e 4
3
3 1
The answer is E.
10 Pr ( X > 15 ) = 1 – Pr ( X < 15 )
= 1–
∫
15
0
0.3e
= 1 − −e − 0.3x
= 1–
( −e − 4.5
− 0.3x
dx
15
0
– −e 0 )
= ( 1 + e − 4.5 – 1 )
= 0.0111
The answer is D.
Pr ( X > 15 )
11 Pr ( X > 15 X > 10 ) = -------------------------Pr ( X > 10 )
Pr ( X > 10 ) = 1 – Pr ( X < 10 )
= 1–
∫
10
0
12 µ + 2σ = 74 [1]
µ − 2σ = 26 [2]
[1] + [2]: 2µ = 100
µ = 50
So 50 + 2σ = 74
2σ = 24
σ = 12
The answer is E.
13 µ = 27, σ = 4
µ + 2σ = 27 + 2 × 4
= 35
µ − 2σ = 27 − 2 × 4
= 19
⇒ 95% lie between 19 ≤ X ≤ 35
The answer is D.
14 Pr(Z > 1.111) = 1 − Pr(Z < 1.111)
= 1 − 0.8667
= 0.1333
The answer is A.
x–µ
15 z = -----------σ
22
– 16
= -----------------5
6--=
5
= 1.2
The answer is D.
16 µ = 1.4, σ = 0.25
1.6 – 1.4
Pr(X ≥ 1.6) = Pr Z ≥ ---------------------
0.25
= Pr(Z ≥ 0.8)
= 1 − Pr(Z < 0.8)
= 1 − 0.7881
= 0.2119
The answer is C.
17 µ = 8, σ = 2
9.25 – 8
Pr(X < 9.25) = Pr Z < -------------------
2
= Pr(Z < 0.625)
= 0.7340
The answer is C.
18 µ = 13, σ = 4
Pr(Z < z1) = 0.71
⇒ z1 = 0.5534
k–µ
z1 = -----------σ
k------------– 130.5534 =
4
2.2136 = k − 13
k = 15.2136
The answer is C.
2 a When x = 0, f(0) = 0
When x = 4, f(4) = 2
y
(4, 2)
2
0
x
4
f(x) ≥ 0
1--2
∫
4
x dx
0
3--- 4
2
2x
= -------3
0
1
= 5 --3
Area beneath the curve is not
equal to 1, so f(x) is not a pdf.
b
∫
4
0
1--2
ax dx = 1
3--- 4
2
2ax
-----------3
= 1
0
16a
--------- = 1
3
3a = ----16
c
f(x) = x
a≤x≤4
Find a for f(x) to be a pdf.
∫
4
a
1--2
x dx = 1
3--- 4
2
2x
-------3
= 1
a
3--2
16
2a - = 1
------ – -------3
3
3--2
13
2a ------ = -------3
3
3
---
Short answer
1 a When x = 0,
f(0) = 4
1
When x = 1---, f --- = 0
2
2
2
13
------ = a
2
13
a = ------
2
3 a Pr ( X > 1.4 ) =
y
(0, 4)
∫
2--3
e
1.4
1--dx
x
e
f ( x )dx
1 , 0)
(−
2
= 0.0498
0.0111
Pr ( X > 15 X > 10 ) = ---------------0.0498
= 0.2231
The answer is A.
329
MM12-12
0
b
= log e x
1.4
x
f(x) ≥ 0
Area beneath f ( x ) = 1 , so f(x)
may be a pdf.
= log e e – log e 1.4
= 1 – log e 1.4
= 0.6635
answers
MQ-12_Sol_MM_Ch 12.fm Page 330 Wednesday, August 30, 2006 7:30 PM
MM12-12
330
Continuous distributions
∫
2.5
b Pr ( X < 2.5 ) =
1
1--dx
x
Pr ( 2 < X < 3 ) =
2.5
= log e x
0.1455
Pr ( X > 2 X < 3 ) = ---------------0.6321
Pr ( X > 1.4 ∩ X < 2.5 )
c Pr ( X > 1.4 X < 2.5 ) = ----------------------------------------------------Pr ( X < 2.5 )
∫
1--dx
x
2.5
1.4
= 0.2302
5
∫
1.4
= log e 2.5 – log e 1.4
= 0.5798
0.5798
Pr ( X > 1.4 X < 2.5 ) = ---------------0.9163
–4
0
= 0.5
a 0.8 quartile = 0.5 + 0.3
∫
a
0
lim
k→∞
as k → ∞
b i
=
– –e0
=
k
− --−e 3
+1
−e
From the graphics calculator, a = 1.47
b The 45th percentile
∫
Pr ( X > 2 ) = 1 – Pr ( X < 2 )
∫
= 1−
1 2
− --- x
−e 3
0
–1
= 1–
= 1−
=
∫
0
1
1--- − --3- x
e dx
3
1
− --- x
−e 3
3
0
e– 1
= 0.3679
Pr ( X > 2 ∩ X < 3 )
iii Pr ( X > 2 X < 3 ) = -------------------------------------------Pr ( X < 3 )
( 2 < X < 3 )= Pr
-------------------------------Pr ( X < 3 )
Pr ( X < 3 ) = 1 – 0.3679
= 0.6321
= 0.45
–4
a 2 + 8a + 1.6 = 0
From the graphics calculator, a = −0.21
Pr ( X > 3 ) = 1 – Pr ( X < 3 )
3
a
a 2- + 1--- a + 0.05 = 0
----32 4
= 0.5134
ii
11
----x + --- dx = 0.45
16
4
a 2- + 1--- a + 1--- = 0.45
----32 4
2
2
= 1+
–4
a2 1 1
---- 32- + --4- a – --2- – 1 = 0.45
− --30
= 1 – −e – −e
– 2--e 3
a
x 2- + 1--- x
----32 4
1
1--- − --3- x
e dx
3
= 1–
0
= 0.3
0
−a 2 + 8a – 9.6 = 0
→0 = 1
2
a
2
−a
--------- + 1--- a = 0.3
32 4
0
k
− --3
1
− ------ x + 1--- dx = 0.3
16
4
x 2- + 1--- x
– ----32 4
1
k
− --−e 3
–4
= − 1--- – 1
2
1--- − --3- x
e dx must equal 1 for f(x) to be a pdf.
3
1 k
− --- x
−e 3
0
x 2- + 1--- x
----32 4
= 0.6328
∫
1----x + 1--- dx
16
4
0
2.5
= log e x
4 a
1 3
− --- x
−e 3
= 0.1455
= 0.9163
( 1.4 < X < 2.5 )= Pr
----------------------------------------Pr ( x < 2.5 )
1
1--- – --3- x
e
dx
3
2
= log e 2.5 – log e 1
∞
2
=
1
Pr ( 1.4 < X < 2.5 ) =
∫
3
6 a
∫
4
1
a ( x – 1 ) 3 dx = 1
4
a--------------------( x – 1 ) 44
= 1
1
81a
--------- – 0 = 1
4
4a = ----81
b Mean of X =
∫
4
1
4x----( x – 1 ) 3 dx
81
4x 5 3x 4 4x 3 2x 2
= --------- – -------- + -------- – -------405 81
81 81
= 3.3975
4
1
MQ-12_Sol_MM_Ch 12.fm Page 331 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
c
Median:
∫
m
1
4----( x – 1 ) 3 dx = 0.5
81
1----( x – 1 )4
81
m
= 0.5
1
1
------ ( m – 1 ) 4 – 0 = 0.5
81
b 4 minutes and 36 seconds
= 4.6 minutes
5 minutes and 15 seconds
= 5.25 minutes
5.25 1
--- dx
=
4.6 2
∫
1
= --- x
2
( m – 1 ) 4 = 81
-----2
1---
81 4
m = ------ + 1
2
c
=
∫
2
1
2x
1- + 2 dx
------ ---
5 x2
4.6
= 0.325
∫
a
4
1--dx = 0.9
2
1--x
2
3.5227
d The graph is increasing over the
domain, so the maximum value
occurs at the right-hand side of
the interval.
The mode is 4.
7 a Mean of X =
5.25
a
= 0.9
4
1--a – 2 = 0.9
2
a = 5.8
5 minutes and 48 seconds
9a
∫ -----5x- + -----5- dx
2
2
4x
1
2
2
-------= --- log e x + 2x
5
5
5
2
6
7
8
13 a
= 1.8773 – 0.4
1
= 1.4773
b Var ( X ) = E ( X 2 ) – µ 2
E( X2 ) =
∫
=
∫
2
1
2
1
9 17 25 33 41 49
c
b
2x 2- 1
----------- + 2 dx
5 x2
2--- 4x 2
+ -------- dx
5
5
2
4x 3= --- x + ------5
15
= normalcdf(92, 1010, 68, 12)
= 0.0228
Therefore approximately 2.3%
are above 92.
c Pr(X > 104)
9 10 11
2
2
8
2
= --- log e 2 + --- – --- log e 1 + ---
5
5
5
5
2
–9 –6 –3 0
3
6
9
10 a µ = 30, σ = 8
c
1
4 32
2 4
= --- + ------ – --- + ------
5 15
5 15
d
= 2.2667
Var ( X ) = 2.2667 – 1.4773 2
= 0.084
6 14 22 30 38 46 54
e
b µ = 8, σ = 2
c SD ( X ) = Var ( X )
= 0.29
Pr ( µ – 2 σ ≤ X ≤ µ + 2 σ )
= Pr ( 0.8968 ≤ X ≤ 2.0578 )
Both end points are outside the
given domain of the pdf. so the
probability is 1.
8 a If the rice is not ready at
X = 5.5 minutes then it needs
more time of cook.
6 1
--- dx
=
5.5 2
∫
1
= --- x
2
6
5.5
= 3 – 5.5
------2
= 0.25
f
2
4
6
8 10 12 14
c µ = 100, σ = 10
14 a
b
c
d
e
f
70 80 90 100 110 120 130
11 µ = 60, σ = 9
a Pr(51 < X < 69)
= normalcdf(51, 69, 60, 9)
= 0.6827
Therefore approximately 68% of
sleepers are between 51 kg and
69 kg
331
b Pr(42 < X < 78)
= normalcdf(42, 78, 60, 9)
= 0.9544
Therefore approximately 95% are
between 42 kg and 78 kg.
c Pr(33 < X < 87)
= normalcdf(33, 87, 60, 9)
= 0.9973
Therefore approximately 99.7%
are between 33 kg and 87 kg.
12 µ = 68, σ = 12
a Pr(X > 80)
= normalcdf(80, 1010, 68, 12)
= 0.1587.
Therefore approximately 16% are
above 80.
b Pr(X > 92)
b
1
MM12-12
g
h
= normalcdf(104, 1010, 68, 12)
= 0.0013
Therefore approximately 0.13%
are above 104.
µ − σ = 19.6 − 3.1
= 16.5
µ + σ = 19.6 + 3.1
= 22.7
16.5 ≤ X ≤ 22.7
µ − 2σ = 19.6 − 2 × 3.1
= 13.4
µ + 2σ = 19.6 + 2 × 3.1
= 25.8
13.4 ≤ X ≤ 25.8
µ − 3σ = 19.6 − 3 × 3.1
= 10.3
µ + 3σ = 19.6 + 3 × 3.1
= 28.9
10.3 ≤ X ≤ 28.9
16% are above µ + σ
= 22.7
2.5% are above µ + 2σ
= 25.8
0.15% are above µ + 3σ
= 28.9
Pr(Z ≤ 2.6) = 0.9953
Pr(Z < 1.1) = 0.8643
Pr(Z ≤ 1.34) = 0.9099
Pr(Z ≤ 0.98) = 0.8365
Pr(Z > 1.5) = 1 − Pr(Z < 1.5)
= 1 − 0.9332
= 0.0668
Pr(Z ≥ 2.3) = 1 − Pr(Z < 2.3)
= 1 − 0.9893
= 0.0107
Pr(Z ≥ 1.18) = 1 − Pr(Z < 1.18)
= 1 − 0.8810
= 0.1190
Pr(Z > 0.73) = 1 − Pr(Z < 0.73)
= 1 − 0.7673
= 0.2327
answers
MQ-12_Sol_MM_Ch 12.fm Page 332 Wednesday, August 30, 2006 7:30 PM
MM12-12
332
Continuous distributions
i Pr(Z ≤ −1) = Pr(Z > 1)
= 1 − Pr(Z < 1)
= 1 − 0.8413
= 0.1587
j Pr(Z < −1.8) = Pr(Z > 1.8)
= 1 − Pr(Z < 1.8)
= 1 − 0.9641
= 0.0359
k Pr(Z < −2.18) = Pr(Z > 2.18)
= 1 − Pr(Z < 2.18)
= 1 − 0.9854
= 0.0146
l Pr(Z < −2.39) = Pr(Z > 2.39)
= 1 − Pr(Z < 2.39)
= 1 − 0.9916
= 0.0084
m Pr(Z ≥ −2) = Pr(Z < 2)
= 0.9772
n Pr(Z ≥ −2.5) = Pr(Z < 2.5)
= 0.9938
o Pr(Z ≥ −1.61) = Pr(Z < 1.61)
= 0.9463
p Pr(Z > −2.03) = Pr(Z < 2.03)
= 0.9788
15 a Pr(Z < 1.2 | Z < 1.5)
Pr ( Z < 1.2 ∩ Z < 1.5 )
= ----------------------------------------------------Pr ( Z < 1.5 )
Pr ( Z < 1.2 )
= ----------------------------Pr ( Z < 1.5 )
0.8849
= ---------------0.9332
= 0.9482
b Pr(Z < − 0.53 | Z < 0.265)
Pr ( Z < – 0.53 ∩ Z < 0.265 )
= ------------------------------------------------------------------Pr ( Z < 0.265 )
Pr ( Z < – 0.53 )
= -----------------------------------Pr ( Z < 0.265 )
1 – Pr ( Z < 0.53 )
= ----------------------------------------Pr ( Z < 0.265 )
1 – 0.7019
= ------------------------0.6045
0.2981
= ---------------0.6045
= 0.4931
c Pr(Z ≥ 2.19 | Z ≥ 1.2)
Pr ( Z ≥ 2.19 ∩ Z ≥ 1.2 )
= -------------------------------------------------------Pr ( Z ≥ 1.2 )
Pr ( Z ≥ 2.19 )
= -------------------------------Pr ( Z ≥ 1.2 )
1 – Pr ( Z < 2.19 )
= ----------------------------------------1 – Pr ( z < 1.2 )
1 – 0.9857
= ------------------------1 – 0.8849
0.0143
= ---------------0.1151
= 0.1242
d Pr(Z > 1.9 | Z < 2.368)
Pr ( Z > 1.9 ∩ Z < 2.368 )
= ----------------------------------------------------------Pr ( Z < 2.368 )
Pr ( 1.9 < Z < 2.368 )
= ------------------------------------------------Pr ( Z < 2.368 )
Pr ( Z < 2.368 ) – Pr ( Z < 1.9 )
= ---------------------------------------------------------------------Pr ( Z < 2.368 )
c Pr(Z ≤ c) = 0.1
0.9911 – 0.9713
= --------------------------------------0.9911
0.0198
= ---------------0.9911
= 0.0200
16 a µ = 22, σ = 6
26 – 22
z = -----------------6
4
= --6
= 0.6667
b µ = 16, σ = 4
18 – 16
z = -----------------4
2
= --4
= 0.5
c µ = 50.9, σ = 10
56 – 50.9
z = ---------------------10
5.1
= ------10
= 0.51
17 µ = 22, σ = 6
a Pr(X > 23)
= normalcdf(23, 1010, 22, 6)
= 0.4338
b Pr(X ≥ 11)
= normalcdf(11, 1010, 22, 6)
= 0.9666
c Pr(X < 31)
= normalcdf(−1010, 31, 22, 6)
= 0.9332
d Pr(X ≤ 19)
= normalcdf(−1010, 19, 22, 6)
= 0.3085
e Pr(20 ≤ X ≤ 26)
= normalcdf(20, 26, 22, 6)
= 0.3781
18 µ = 18, σ = 1
20.5 – 18
a Pr(X > 20.5) = Pr Z > ----------------------
1
= Pr(Z > 2.5)
= 1 − Pr(Z < 2.5)
= 1 − 0.9938
= 0.0062
19 – 18
b Pr(X < 19) = Pr Z < ------------------
1
= Pr(Z < 1)
= 0.8413
c Pr(X >19 | X > 20.5)
0.152 45
= --------------------0.993 79
= 0.1534
19 a Pr(Z < c) = 0.7
c = 0.5244
b Pr(Z ≤ c) = 0.95
c = 1.6449
0.1
0.1
0
c
z1
z
Pr(Z < Z1) = 0.9
z1 = 1.2816
⇒ c = −1.2816
d Pr(Z < c) = 0.28
0.28
0.28
0
c
z
z1
Pr(Z < z1) = 0.72
z1 = 0.5828
⇒ c = −0.5828
e Pr(Z ≥ c) = 0.71
0.71
c
0
z
0.71
0
z
z1
Pr(Z < z1) = 0.71
z1 = 0.5534
⇒ c = − 0.5534
f Pr(Z > c) = 0.89
or Pr(Z < c) = 0.11
0.89
0.11
c
0
Pr(Z < z1) = 0.89
z1 = 1.2265
⇒ c = −1.2265
z1 z
MQ-12_Sol_MM_Ch 12.fm Page 333 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
g Pr(Z > c) = 0.4
Pr(Z < z1) = 0.89
k–µ
z1 = -----------σ
0.4
k – 31
1.2265 = -------------9
z
Pr(Z < c) = 0.6
c = 0.2533
h Pr(Z ≥ c) = 0.27
Pr(Z < c) = 0.73
c = 0.6128
20 µ = 31 σ = 9
a Pr(X < k) = 0.3
0.3
11.0385 = k − 31
k = 42.0385
or k ≈ 42.04
d Pr(x ≤ k) = 0.997
0.3
0
z1
More than 10 minutes
= 1− Pr(X < 10 )
0.997
= 1–
∫
10
0.09 e − 0.09x dx
0
= 1 − −e − 0.09x
z1z
0
0.3
= 1–
Pr(Z < z1) = 0.997
0
z
z2
Pr(Z < z2) = 0.7
z2 = 0.5244
⇒ z1 = −0.5244
k–µ
z1 = -----------σ
k – 31
−0.5244 = -------------9
−4.7196 = k − 31
k = 26.2804
or k ≈ 26.28
b Pr(X > k) = 0.83
or Pr(X < k) = 0.17
k–µ
z1 = -----------σ
0.17
k – 31
2.7478 = -------------9
20
0.09 e − 0.09x dx
0
20
0
= e − 1.8
= 0.1653
iii More than half an hour
= 1− Pr(X < 30)
∫
30
0.09 e − 0.09x dx
0
= 1 − −e − 0.09x
0.1
0
z2
z
30
0
Pr(Z < z2) = 0.9
= 0.0672
b Pr ( X < 45 ( X ≥ 20 ) )
( 20 ≤ X < 45 )= Pr
-------------------------------------Pr ( X ≥ 20 )
z2 = 1.2816
Pr(Z < z2) = 0.83
z2 = 0.9542
⇒ z1 = −0.9542
k–µ
z1 = -----------σ
k – 31
−0.9542 = -------------9
−8.5878 = k − 31
k = 22.4122
or k ≈ 22.41
c Pr(X ≥ k) = 0.11
or Pr(X < k) = 0.89
⇒ z1 = −1.2816
Pr ( 20 ≤ X < 45 )
x1 – µ
z1 = ------------σ
=
25 – 28
−1.2816 = -----------------σ
–3 σ = -----------------– 1.2816
= 2.3408
The standard deviation is
2.34 cm.
z1 z
b Pr(X > 30) = normalcdf(30, 1010,
28, 2.34)
= 0.1964
c Pr(X > x1) = 0.3
InvNorm (0.3, 28, 2.34) = 26.77
The maximum length of fish
∫
45
20
0.09 e − 0.09x dx
= −e − 0.09x
−1.2816σ = −3
0.89
0
∫
= e − 2.7
z
z2
– −e 0 )
= 1 − −e − 0.09x
= 1–
z1
0
= 1–
24.7302 = k − 31
k = 55.7302
or k ≈ 55.73
21 µ = 28
a Pr(X > 25) = 0.9
or Pr(X < 25) = 0.1
0.17
z1
0
= 0.4066
More than 20 minutes
= 1− Pr(X < 20 )
ii
0.1
( −e − 0.9
10
= e − 0.9
z1 = 2.7478
z1
z
z2
Analysis
1 a i
0.3
333
which must be returned is
26.77 cm.
z1 = 1.2265
0 c
MM12-12
45
20
= 0.1479
Pr
( 20 ≤ X < 45 -) = 0.1479
----------------------------------------------------Pr ( X ≥ 20 )
0.1653
0.8946
c
∫
a
0
0.09 e − 0.09x dx = 0.95
−e − 0.09x
a
= 0.95
0
−e − 0.09a + 1 = 0.95
0.05 = e − 0.09a
log e 0.05 = −0.09 × a
answers
MQ-12_Sol_MM_Ch 12.fm Page 334 Wednesday, August 30, 2006 7:30 PM
MM12-12
334
Continuous distributions
3 a Pr(X < 180) = 0.8
log e 0.05
a = -------------------– 0.09
ii
Pr ( B ∩ A )
Pr(B|A) = ------------------------Pr ( A )
0.2
= ------0.7
a = 33.3 minutes
33.3 minutes prior to 7.00 pm to
the nearest minute is 6.26 pm.
d Pr(call is less than 10 minutes)
= 1 − 0.4066
0.8
= 0.5934
0
Pr(call is between 10 and
20 minutes) = 0.2413
Pr(call is between 20 and
30 minutes) = 0.0981
Pr(call is greater than 30 minutes)
= 0.0672
Expected cost
= 60 × 0.5934 + 140 × 0.2413
+ 200 × 0.0981 + 250 × 0.0672
= $1.06 to the nearest cent.
2 a µ = 50
Pr(47 < X < 53) = 0.9
180 – µ
0.842 = -----------------20
16.84 = 180 − µ
µ = 180 − 16.84
= 163.16
The mean is approximately
163.2 seconds.
b Pr(X > 210)
= normalcdf(210, 1010, 163.2, 20)
= 0.0096
c Pr(150 < X < 210)
= normalcdf(150, 210, 163.2, 20)
= 0.7357
d Pr(X > x1) = 0.9
or Pr(X < x1) = 0.1
InvNorm (0.1, 163.2, 20)
= 137.57
Therefore 90% of students read
the page in 137.57 seconds or
more.
z2 z
Pr(Z < z2) = 0.95
z2 = 1.645
⇒ z1 = −1.645
x1 – µ
z1 = ------------σ
Technology–free questions
Pr ( 47 < X < 53 ∩ X > 47 )
= --------------------------------------------------------------Pr ( X > 47 )
Pr ( 47 < X < 53 )
= ---------------------------------------Pr ( X > 47 )
c Events A and B are not mutually
exclusive since A ∩ B ≠ 0
3 ε = {ΗΗΗ, ΗΤΗ, ΗΗΤ, ΤΗΗ, ΗΤΤ,
ΤΗΤ, ΤΤΗ, ΤΤΤ}
2
Pr(tail first ∩ HT or TH) = --8
2--8
Pr(Head or tail | Tail first) = --1--2
1
= --2
4 a 2k + k + 4k + 2k + k = 1
10k = 1
k = 0.1
b 7k + 6k + 5k + 4k + 3k = 1
25k = 1
1
k = -----25
5 a Ε(X) = Σx Pr(X = x)
1
1
1
= 1 × --- + 2 × --- + 3 × --4
8
2
1--+4×
8
1 1 3 1
= --- + --- + --- + --4 4 2 2
8
1 a Pr(Not white) = -----10
1
= 2 --2
4
= --5
47 – 50
−1.645 = -----------------σ
–3
σ = ---------------– 1.645
= 1.823
The standard deviation is
approximately 1.82 cm.
b From part a Pr(X > 53) = 0.05
The percentage of oversize laces
is 5%.
c Pr(X < 47) = 0.05
The percentage of undersize laces
is 5%.
d Pr(47 < X < 53 | X > 47)
0.2
= ------0.3
2
= --3
x1 – µ
z1 = ------------σ
0.05
0
z
⇒ z1 = 0.842
= 0.05
(As symmetrical about mean)
if Pr(X < 47) = 0.05
z1
Pr ( B ∩ A ′ )
iii Pr(B|A′) = --------------------------Pr ( A′ )
Pr(Z < z1) = 0.8
0.1
or Pr(X < 47) = Pr(X > 53) = ------2
0.05
z1
2
= --7
7
b Pr(Blue or white) = -----10
2
c Pr(Not blue or red) = -----10
b Ε(X2) = Σx2Pr(X = x)
1
1
1
= 12 × --- + 22 × --- + 32 × --4
8
2
2 1
+ 4 × --8
1
= --5
1 4 9 16
= --- + --- + --- + -----4 8 2
8
= 7.25
2 a
A
c Var(X) = E(X2) − [E(X)]2
A′′
= 7.25 − (2.5)2
=1
B
0.2
0.2
B′′
0.5
0.1
0.6
d SD(X) =
0.7
0.3
1.0
=
b i
0.4
Pr ( A ∩ B )
Pr(A|B) = ------------------------Pr ( B )
0.9
= ---------- (From part a)
0.95
0.2
= ------0.4
= 0.9474
= 0.5
Var ( X )
1
=1
6 a
Surcharge ($)
0
1
2
3
Pr(X = x)
1--5
2--5
3- ----1----10 10
MQ-12_Sol_MM_Ch 12.fm Page 335 Wednesday, August 30, 2006 7:30 PM
Continuous distributions
54
= -----64
27
= -----32
1
2
3
1
b Mean surcharge = 0 × --- + 1 × --- + 2 × ------ + 3 × -----5
5
10
10
2
6
3
= 0 + --- + ------ + -----5 10 10
10 a
13
= -----10
= $1.30
1
2
3
1
c E(X2) = 02 × --- + 12 × --- + 22 × ------ + 32 × -----5
5
10
10
25 169
= ------ − --------10 100
81
= --------100
SD(X) =
=
Var ( X )
81-------100
9
= -----10
= 90 cents
7 a p = 0.8, q = 0.2, n = 4
Pr(X = 1) = 4C1(0.8)1(0.2)3
= 4 × 0.8 × 0.008
= 0.0256
16
= --------625
b Pr(X = 0) = C0(0.8) (0.2)
4
0
4
= (0.2)4
= 0.0016
1
= --------625
c Pr(X = 4) = 4C4(0.8)4(0.2)4
4 4
= 1 × (0.8)4 or 1 × ---
5
= 0.4096
4
4
= ---
5
8 If p < 0.5, the graph is positively skewed.
If p > 0.5, the graph is negatively skewed.
If p = 0.5, the graph is symmetrical.
a Graph a is positively skewed, so p < 0.5.
b Graph b appears symmetrical, so p ≈ 0.5.
c Graph c is negatively skewed, so p > 0.5.
9 n = 3, p = 0.75, q = 0.25
Pr(X ≥ 2) = Pr(X = 2) + Pr(X = 3)
3
= 3C2 ---
4
2
3
1--- + 3C 3---
3
4
4
9
1
27
= 3 × ------ × --- + 1 × -----16 4
64
27 27
= ------ + -----64 64
2
4
_ _
5 5
P
1
_
5 3
_
5
F
2
_
5
F
P
1
_
5
F
1
_
2
F
2
_
5
2
P
P
3
_
5
25
= -----10
Var(X) = E(X2) − [E(X)]2
4
_
5
1
_
3
5 _
5
4
_
5
P
1
_
2
2 12
9
= 0 + --- + ------ + -----5 10 10
25
13
= ------ − ------
10 10
MM12-12
b Pass the first test but fail the third
= PPF or PFF
1 4 1 1 1 2
= --- × --- × --- + --- × --- × --2 5 5 2 5 5
4
2
= ------ + -----50 50
6
= -----50
3
= -----25
1 40
11 E(X) = 40 and Var(X) = 13 --- = -----3
3
40
np = 40 and npq = -----3
40
40 × q = -----3
1--q=
3
1--2--If q = , then p = .
3
3
np = 40
2
n × --- = 40
3
3 × 40
n = --------------2
= 60
2
So, n = 60 and p = --- .
3
12 f: [2, 6] → R, f(x) = a(x − 2)
∫ a( x – 2 )dx = 1
a ( x – 2 ) dx = 1
∫
6
2
6
2
x2
a ----- – 2x
2
6
2
=1
a[(18 − 12) − (2 − 4)] = 1
a(6 − −2) = 1
1
a = --8
13 Pr(x > 5) =
6 1
∫ --2- x – 2 dx
5
x2
= ----- – 2x
4
6
5
36
25
= ------ – 12 − ------ – 10
4
4
= −3 − −3.75
= 0.75
F
P
F
P
F
335
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