chapter 10 — the normal distribution - weebly...the normal distribution mb12 qld-10 153 exercise...
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T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 153
Exercise 10A — The standard normal distribution 1 Use table on page 392 or graphics calculator
a P( 1) 0.8413Z ≤ =
Normal cdf (−∞, 1, 0, 1) b P( 2.3) 0.9893Z < =
Normal cdf (−∞, 2.3, 0, 1) c P( 1.52) 0.9357Z ≤ =
Normal cdf (−∞, 1.52, 0.1) d P( 0.74) 0.7703Z ≤ =
Normal cdf (−∞, 0.74, 0, 1) e P( 1.234) 0.8914Z < =
Normal cdf (−∞, 1.234, 0, 1)
f P( 2.681) 0.9963Z < =
Normal cdf (−∞, 2.681, 0, 1)
2 Use table on page 358 or graphics calculator
a
P( 2) 1 P( 2)
1 0.97720.0228
Z Z> = − <= −=
Normal cdf (2, ∞, 0, 1)
b
P( 1.5) 1 P( 1.5)
1 0.93320.0668
Z Z≥ = − ≤= −=
Normal cdf (1.5, ∞, 0, 1)
c
P( 1.22) 1 P( 1.22)
1 0.88880.112
Z Z≥ = − ≤= −=
Normal cdf (1.22, ∞, 0, 1)
d
P( 0.16) 1 P( 0.16)
1 0.56360.4364
Z Z> = − <= −=
Normal cdf (0.16, ∞, 0, 1)
Chapter 10 — The normal distribution
M B 1 2 Q l d - 1 0 154 T h e n o r m a l d i s t r i b u t i o n
e
P( 1.111) 1 P( 1.111)
1 0.86670.1333
Z Z> = − <= −=
Normal cdf (1.111, ∞, 0, 1)
f
P( 2.632 1 P( 2.632)
1 0.99570.0043
Z Z≥ = − ≤= −=
Normal cdf (2.632, ∞, 0, 1)
3 a
P( 2) P( 2)1 P( 2)1 0.9772)0.0228
Z ZZ
≤ − = ≥= − ≤= −=
Normal cdf (−∞, −2, 0, 1)
b
P( 1.3) P( 1.3)1 P( 1.3)1 0.90320.0968
Z ZZ
< − = >= − <= −=
Normal cdf (−∞, −1.3, 0, 1)
c
P( 1.75) P( 1.75)1 P( 1.75)1 0.95990.0401
Z ZZ
< − = >= − <= −=
Normal cdf (−∞, −1.75, 0, 1)
d
P( 2.23) P( 2.23)1 P( 2.23)1 0.9871)0.0129
Z ZZ
< − = >= − <= −=
Normal cdf (−∞, −2.23, 0, 1)
e
P( 2.317) P( 2.317)1 P( 2.317)1 0.98980.0102
Z ZZ
≤ − = ≥= − ≤= −=
Normal cdf (−∞, −2.317, 0, 1)
f
P( 0.669) P( 0.669)1 P( 0.669)1 0.74830.2517
Z ZZ
≤ − = ≥= − ≤= −=
Normal cdf (−∞, −0.669, 0, 1)
4 a
P( 3) P( 3)0.9987
Z Z≥ − = ≤=
Normal cdf (−3, ∞, 0, 1)
b
P( 2.1) P( 2.1)0.9821
Z Z≥ − = ≤=
Normal cdf (−2.1, ∞, 0, 1)
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 155
c
P( 1.77) P( 1.77)0.9616
Z Z> − = <=
Normal cdf (−1.77, ∞, 0, 1)
d
P( 2.71) P( 2.71)0.9966
Z Z> − = <=
Normal cdf (−2.71, ∞, 0, 1)
e
P( 1.139) P( 1.139)
0.8708 0.00190.8727
Z Z≥ − = ≤= +=
Normal cdf (−1.139, ∞, 0, 1)
f
P( 0.642) P( 0.642)
0.7389 0.00060.7395
Z Z> − = <= +=
Normal cdf (−0.642, ∞, 0, 1)
5 a
P(1 2) P( 2) P( 1)
0.9772 0.84130.1359
Z Z Z< < = < − <= −=
Normal cdf (1, 2, 0, 1)
b
P(1.6 2.5) P( 2.5) P( 1.6)0.9938 0.94520.0486
Z Z Z≤ ≤ = ≤ − ≤= −=
Normal cdf (1.6, 2.5, 0, 1)
c
P(0.42 1.513) P( 1.513) P( 0.42)0.9349 0.66280.2721
Z Z Z< < = < − <= −=
Normal cdf (0.42, 1.513, 0, 1)
6 a
[ ]
[ ]
P( 1.6 1.4) P( 1.4) P( 1.6)P( 1.4) P( 1.6)
P( 1.4) 1 P( 1.6)
0.9192 1 0.9452
0.9192 0.05480.8644
Z Z ZZ Z
Z Z
− ≤ ≤ = ≤ − ≤ −= ≤ − ≥
= ≤ − − ≤
= − −
= −=
Normal cdf (−1.6, 1.4, 0, 1)
b
[ ]
[ ]
P( 2.21 0.34) P( 0.34) P( 2.21)P( 0.34) P( 2.21)
P( 0.34) 1 P( 2.21)
0.6331 1 0.9864
0.6331 0.01360.6195
Z Z ZZ Z
Z Z
− < < = < − < −= < − >
= < − − <
= − −
= −=
Normal cdf (−2.21, 0.34, 0, 1)
M B 1 2 Q l d - 1 0 156 T h e n o r m a l d i s t r i b u t i o n
c
[ ][ ]
P( 0.645 0.645) P( 0.645) P( 0.645)P( 0.645) P( 0.645)P( 0.645) 1 P( 0.645)
0.7405 1 0.74050.7405 0.25950.4810
Z Z ZZ ZZ Z
− ≤ ≤ = ≤ − ≤ −= ≤ − ≥= ≤ − − ≤
= − −= −=
Normal cdf (−0.645, 0.645, 0, 1)
d
[ ] [ ][ ] [ ]
P( 0.72 0.41) P( 0.41) P( 0.72)P( 0.41) P( 0.72)1 P( 0.41) 1 P( 0.72)
1 0.6591 1 0.76420.3409 0.23580.1051
Z Z ZZ Z
Z Z
− ≤ ≤ − = ≤ − − ≤ −= ≥ − ≥= − ≤ − − ≤
= − − −= −=
Normal cdf (−0.72, −0.41, 0, 1) 7 a 40 if (25,25)X X N= ∼
225, 25, 5
40 255
155
3
xZ
µ σ σµ
σ
= = =−=
−=
=
=
b 2
12 if (17, 9)
17, 9, 3
12 173
531.667
X X N
xZ
µ σ σµ
σ
=
= = =−=
−=
−=
= −
∼
c 2
15 if (12, 6.25)
12, 6.25, 2.5
15 122.5
32.51.2
X X N
xZ
µ σ σµ
σ
=
= = =−=
−=
=
=
∼
8 a 9, 310
10 93
130.3
XxZ
µ σ
µσ
= ==
−=
−=
=
=
b 7.5
7.5 93
0.5
XzZ µ
σ
=−=
−=
= −
c 12.4
12.4 93
1.13
XxZ µ
σ
=−=
−=
=
9 a 40, 7( 42)P X
µ σ= =>
42 407
270.2857
xZ µσ−=
−=
=
=
P( 42) P( 0.2857)
1 P( 0.2857)1 (0.6103 0.0023)1 0.61260.3874
X ZZ
> = >= − <= − += −=
Normal cdf (42, 1E99, 40, 7) b ( 30)P X ≥
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 157
30 407107
1.429
Z −=
−= −
= −
P( 30) P( 1.429)
P( 1.429)0.9235
X ZZ
≥ = ≥ −= ≤=
Normal edf (30, ∞, 40, 7) c P( 45)X <
45 40
70.714
Z −=
=
P( 45) P( 0.714)0.7623
X Z< = <=
Normal cdf ( , 45, 40, 7)−∞ d P( 27)X ≤
27 40
71.857
Z −=
= −
P( 27) P( 1.857)P( 1.857)1 P( 1.857)1 0.96830.0317
X ZZ
Z
≤ = ≤ −= ≥= − ≤= −=
e P(25 45)X≤ ≤
25 407
2.14345 40
70.714
Z
Z
−=
= −−=
=
P(25 45) P( 2.143 0.714)P( 0.714) P( 2.143)P( 0.714) P( 2.143)P( 0.714) [1 P( 2.143)]0.7623 [1 0.9839]0.7623 0.01610.7462
X ZZ ZZ ZZ Z
≤ ≤ = − ≤ ≤= ≤ − ≤ −= ≤ − ≥= ≤ − − ≤= − −= −=
Normal cdf (25, 45, 40, 7) 10 a (20, 25)X N∼
P( 27)X >
27 205
1.4
xZ µσ−=
−=
=
M B 1 2 Q l d - 1 0 158 T h e n o r m a l d i s t r i b u t i o n
P( 27) P( 1.4)1 P( 1.4)1 0.91920.0808
X ZZ
> = >= − <= −=
Normal cdf (27, ∞, 20, 5) b P( 18)X ≥
18 20
50.4
Z −=
= −
P( 18) P( 0.4)
P( 0.4)0.6554
X ZZ
≥ = ≥ −= ≤=
Normal cdf (18, ∞, 20, 5) c P( 8)X ≤
8 20
52.4
Z −=
= −
P( 8) P( 2.4)P( 2.4)1 P( 2.4)1 0.99180.0082
X ZZ
Z
≤ = ≤ −= ≥= − ≤= −=
Normal cdf ( , 8, 20, 5)−∞ d P(7 12)X≤ ≤
7 20
52.6
Z −=
= −
12 20
51.6
Z −=
= −
[ ] [ ][ ] [ ]
P(7 12) P( 2.6 1.6)P( 1.6) P( 2.6)P( 1.6) P( 2.6)1 P( 1.6) 1 P( 2.6)
1 0.9452 1 0.99530.0548 0.00470.0501
X ZZ ZZ Z
Z Z
≤ ≤ = − ≤ ≤ −= ≤ − − ≤ −= ≥ − ≥= − ≤ − − ≤
= − − −= −=
Normal cdf (7, 12, 20, 5) e P( 17 | 25)X X< ≤
17 205
0.625 20
51P( 17 | 25)P( 17 25)
P( 25)P( 17)P( 25)P( 0.6)
P( 1)P( 0.6)P( 1)
1 P( 0.6)P( 1)
1 0.72570.8413
0.27430.84130.3260
xZ
Z
X XX X
XXXZ
ZZZ
ZZ
µσ−=
−=
= −−=
== < ≤
< ∩ ≤=≤
<=≤< −=
≤>=≤
− <=≤
−=
=
=
normal cdf ( , 17, 20, 5)normal cdf ( , 25, 20, 5)
−∞−∞
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 159
f P( 17 | )X X µ< <
P( 17 | 20)
17 205
0.6
X XxZ µ
σ
< <−=
−=
= −
20 205
0P( 17 | 20)P( 17 20)
P( 20)P( 17)P( 20)P( 0.6)
P( 0)P( 0.6)P( 0)
1 P( 0.6)P( 0)
1 0.72570.5
0.27430.5
0.5486
Z
X XX X
XXXZ
ZZZ
ZZ
−=
== < <
< ∩ <=<
<=<< −=
<>=<
− <=<
−=
=
=
normal cdf ( , 17, 20, 5)normal cdf ( , 20, 20, 5)
−∞−∞
11 a 125, 11µ σ= = P( 140)X >
140 12511
1.364
xZ µσ−=
−=
=
P( 140) P( 1.364)1 P( 1.364)1 0.91370.0863
X ZZ
> = >= − <= −=
Normal cdf (140, ∞, 125, 11)
b P( 100)X <
100 125
112.273
Z −=
= −
P( 100) P( 2.273)P( 2.273)1 P( 2.273)1 0.98850.0115
X ZZ
Z
< = < −= >= − <= −=
Normal cdf ( , 100, 125, 11)−∞ c P(100 140)X≤ ≤
P(100 140)P( 2.273 1.364)P( 1.364) P( 2.273)P( 1.364) P( 2.273)P( 1.364) [1 P( 2.273)]0.9137 [1 0.9885]0.9137 0.01150.9022
XZ
Z ZZ ZZ Z
≤ ≤= − ≤ ≤= ≤ − ≤ −= ≤ − ≥= ≤ − − ≤= − −= −=
Normal cdf (100, 140, 125, 11) 12 2152, 49, 7µ σ σ= = =
a P( 159)159 152
71
X
Z
≥−=
=
M B 1 2 Q l d - 1 0 160 T h e n o r m a l d i s t r i b u t i o n
P( 159) P( 1)1 P( 1)1 0.84130.1587
X ZZ
≥ = ≥= − ≤= −=
Normal cdf (159, ∞, 152, 7) b P( 150)X <
150 152
70.2857
Z −=
= −
P( 150) P( 0.2857)P( 0.2857)1 P( 0.2857)1 0.61260.3874
X ZZ
Z
< = < −= >= − <= −=
Normal cdf ( , 150, 152, 7)−∞ c P(145 159)X< <
145 152
71
Z −=
= −
[ ]
159 1527
1P(145 159) P( 1 1)
P( 1) P( 1)P( 1) P( 1)P( 1) 1 P( 1)0.8413 [1 0.8413]0.8413 0.15870.6826
Z
X ZZ ZZ ZZ Z
−=
=< < = − < <
= < − < −= < − >= < − − <= − −= −=
Normal cdf (145, 159, 152, 7) d P(140 160)X< <
140 1527
1.714160 152
71.143
P(140 160) P( 1.714 1.143)P( 1.143) P( 1.714)P( 1.143) P( 1.714)P( 1.143) [1 P( 1.714]0.8735 [1 0.9567]0.8735 0.04330.8302
Z
Z
X ZZ ZZ ZZ Z
−=
= −−=
=< < = − < <
= < − < −= < − >= < − − <= − −= −=
Normal cdf (140, 160, 152, 7) e P(145 150 | 140)
P(145 150 140P( 140)
P(145 150)P( 140)
X XX X
XX
X
< < >< < ∩ >=
>< <=
>
145 1527
1150 152
70.2857
Z
Z
−=
= −−=
= −
140 1527
1.714P(145 150) P( 1 0.2857)
P( 0.2857) P( 1)P( 0.2857) P( 1)[1 P( 0.2857)] [1 P( 1)][1 0.6126] [1 0.8413]0.3874 0.15870.2287
Z
X ZZ ZZ Z
Z Z
−=
= −< < = − < < −
= < − − < −= > − >= − < − − <= − − −= −=
P( 140) P( 1.714)P( 1.714)0.9567
P(145 150) 0.2287P( 140) 0.9567
0.2391normal cdf (145, 150, 152, 7)normal cdf (140, , 152, 7)
X ZZ
XX
> = > −= <=
< < =>
=
∞
13 P( 1.251) 1 P( 1.251)Z Z> = − <
1 0.89460.1054
= −=
Answer is A 14 (0,1)Z N∼
P( 0.25) P( 0.25)1 P( 0.25)1 0.59870.4013
Z ZZ
< − = >= − <= −=
Answer is B 15 20, 6µ σ= =
29 206
1.5
xZ µ−=σ−=
=
Answer is E 16 1.4, 0.1µ σ= =
P( 1.25)1.25 1.4
0.11.5
X
Z
>−=
= −
P( 1.25) P( 1.5)
P( 1.5)0.9332
X ZZ
> = > −= <=
Answer is E 17 270, 12, 12
P( 77)77 703.464
2.021
X
Z
µ σ σ= = =>
−=
=
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 161
P( 77) P( 2.021)1 P( 2.021)1 0.97830.0217
X ZZ
> = >= − <= −=
Answer is A 18 12 2µ σ= =
P( 9)9 12
21.5
X
Z
<−=
= −
P( 9) P( 1.5)P( 1.5)1 P( 1.5)1 0.93320.0668
X ZZ
Z
< = < −= >= − <= −=
Answer is A 19 (16, 4)X N∼
16, 2P( 11.5)Xµ σ= =
>
11.5 162
2.25P( 11.5) P( 2.25)
P( 2.25)0.9878
Z
X ZZ
−=
= −> = > −
= <=
Answer is E 20 1.000, 0.006µ σ= =
P( 1.011| 1.004)P( 1.011 1.004)
P( 1.004)P(1.004 1.011)
P( 1.004)1.004 1
0.0060.6671.011 1
0.0061.833
P(1.004 1.011) P(0.667 1.833)P( 1.833) P( .667)0.9666 0.74770.2189
P( 1.004)
X XX X
XX
X
Z
Z
X ZZ Z
X
< >< ∩ >=
>< <=>
−=
=−=
=< < = < <= < − <= −=> = P( 0.667)
1 P( 0.667)1 0.74770.2523
ZZ
>= − <= −=
0.2189P( 1.011| 1.004)0.25230.8676
normal cdf (1.004, 1.011, 1, 0.006normal cdf (1.004, , 1, 0.006)
X X< > =
=
∞
21 82, 5µ σ= = a P(80 90)X< <
80 82
50.4
Z −=
= −
90 82
51.6
Z −=
=
P(80 90) P( 0.4 1.6)P( 1.6) P( 0.4)P( 1.6) P( 0.4)P( 1.6) [1 P( 0.4)]0.9452 [1 0.6554]0.9452 0.34460.6006
X ZZ ZZ ZZ Z
< < = − < <= < − < −= < − >= < − − <= − −= −=
Normal cdf (80, 90, 82, 5) b P( 90) costs $2
90 825
1.6P( 90) P( 1.6)
1 P( 1.6)1 .94520.0548
X
Z
X ZZ
>−=
=> = >
= − <= −=
Normal cdf (90, ∞, 82, 5) c $1.40 when P( 70)X <
70 825
2.4P( 70) P( 2.4)
P( 2.4)1 P( 2.4)1 0.99180.0082
Z
X ZZ
Z
−=
= −< = < −
= >= − <= −=
$1.60 when P(70 80)X< <
P(.70 80) P( 2.4 0.4)P( 0.4) P( 2.4)P( 0.4) P( 2.4)[1 P( 0.4)] [1 P( 2.4)][1 0.6554] [1 0.9918]0.3446 0.00820.3364
X ZZ ZZ Z
Z Z
< < = − < < −= < − − < −= > − >= − < − − <= − − −= −=
Normal cdf (70, 80, 82, 5)
Cost ($) 1.40 1.60 1.80 2.00
Probability 0.0082 0.3364 0.6006 0.0548
d ( ) 1.40 0.0082 1.60 0.3364E X = × + × + 1.80 0.6006 2.00 0.0548
$1.74× + ×
=
22 6, 0.03µ σ= = a Pr(Acceptable) P(5.93 6.07)X= < <
[ ]
5.93 60.032.33
6.07 60.03
2.33
Pr(Acceptable) P( 2.33 2.33)( 2.33) P( 2.33)
P( 2.33) P( 2.33)P( 2.33) 1 P( 2.33)0.9902 [1 0.9902]
Z
Z
ZZ Z Z
Z ZZ Z
−=
= −−=
=
= − < <= < − < −= < − >= < − − <= − −
M B 1 2 Q l d - 1 0 162 T h e n o r m a l d i s t r i b u t i o n
0.9902 0.00980.9804
= −=
Normal cdf (5.93, 6.07, 6, 0.03) b E( ) 0.9804 1000
980X = ×
=
23 32, 4µ σ= = a Pr(undersized) = P(X < 27)
Z = 27 324−
= −1.25 Pr(undersized) P( 1.25)Z= < −
P( 1.25)1 P( 1.25)1 0.89440.1056
ZZ
= >= − <= −=
Normal cdf ( , 27, 32, 4)−∞ b 20 Pr(undersized) 20 0.1056× = ×
2.1120 Pr(OK) 20 (1 0.1056)
17.88817
=× = × −
=
Exercise 10B — The inverse normal cumulative distribution 1 a P( ) 0.9Z c< =
P( ) 0.91.282
Z cc
< ==
Inv Norm (0.9, 0, 1) b P( ) 0.6Z c< =
P( ) 0.60.253
Z cc
< ==
Inv Norm (0.6, 0, 1) c P( ) 0.75Z c≤ =
P( ) 0.750.674
Z cc
≤ ==
Inv Norm (0.75, 0, 1)
d P( ) 0.52Z c≤ =
P( ) 0.52
0.050Z c
c≤ =
=
Inv Norm (0.52, 0, 1) 2 a
P( ) 0.3Z c≤ =
1
1
1
P( ) 0.3P( ) 0.7
0.524for P( ) 0.3
0.524
Z cZ c
cZ c
c
≥ =≤ =
=≤ =
= −
Inv Norm (0.3, 0, 1) b
P( ) 0.2Z c< =
1
1
1
P( ) 0.2P( ) 0.8
0.842for P( ) 0.2
0.842
Z cZ c
cZ c
c
> =< =
=< =
= −
Inv Norm (0.2, 0, 1) c
P( ) 0.35Z c≤ =
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 163
1
1
1
P( ) 0.35P( ) 0.65
0.385for P( ) 0.35
0.385
Z cZ c
cZ c
c
≥ =≤ =
=≤ =
= −
Inv Norm (0.35, 0, 1) d
P( ) 0.42Z c< =
1
1
1
P( ) 0.42P( ) 0.58
0.202for P( ) 0.42
0.202
Z cZ c
cZ c
c
> =< =
=< =
= −
Inv Norm (0.42, 0, 1) 3 a
P( ) 0.8Z c≥ =
1
1
P( ) 0.80.842
for P( ) 0.80.842
Z cc
Z cc
≤ ==
≥ == −
b
P( ) 0.65Z c≥ =
1
1
P( ) 0.650.385
for P( ) 0.650.385
Z cc
Z cc
≤ ==
≥ == −
c
P( ) 0.54Z c> =
1
1
P( ) 0.540.100
for P( ) 0.540.100
Z cc
Z cc
< ==
> == −
d
P( ) 0.72Z c> =
1
1
P( ) 0.720.583
for P( ) 0.720.583
Z cc
Z cc
< ==
> == −
4 a
P( ) 0.3P( ) 0.7
0.524
Z cZ c
c
> =< =
=
M B 1 2 Q l d - 1 0 164 T h e n o r m a l d i s t r i b u t i o n
b
P( ) 0.45P( ) 0.55
0.126
Z cZ c
c
≥ =≤ =
=
c
P( ) 0.22P( ) 0.78
0.772
Z cZ c
c
> =< =
=
5 P( ) 0.6826c Z c− < < =
a
Unshaded area 1 0.68262
−=
2
2
1
0.1587P( ) 0.6826 0.1587
0.84131.0
1.0P( 1 1) 0.6826
1
Z c
cc
Zc
=< = +
=== −
− < < ==
b P( ) 0.5c Z c− ≤ ≤ =
Unshaded area 1 0.52
−=
2
2
1
0.25P( ) 0.5 0.25
0.750.674
0.674
Z c
cc
=≤ = +
=== −
P( 0.674 0.674) 0.50.674
Zc
− ≤ ≤ ==
c P( ) 0.2c Z c− < < =
Unshaded area 1 0.22
−=
2
2
1
0.4P( ) 0.2 0.4
0.60.253
0.253P( 0.253 0.253) 0.2
0.253
Z c
cc
Zc
=< = +
=== −
− < < ==
d P( ) 0.38c Z c− ≤ ≤ =
Unshaded area 1 0.382
−=
2
2
1
0.31P( ) 0.38 0.31
0.690.496
0.496
Z c
cc
=≤ = +
=== −
P( 0.496 0.496) 0.380.496
Zc
− ≤ ≤ ==
6 a
0.25 quantile
25% shaded therefore 75% unshaded
1
1
P( ) 0.750.675
P( ) 0.250.675
Z cc
Z cc
< ==
< == −
b
40th percentile
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 165
40% shaded therefore 60% unshaded
1
1
P( ) 0.60.253
P( ) 0.40.253
Z cc
Z cc
< ==
< == −
c
0.72 quantile
P( ) 0.720.583
Z cc
< ==
d
0.995 quantile
P( ) 0.995
2.576Z c
c< =
=
7 10, 2µ σ= = a 1P( ) 0.72X x≤ =
1P( ) 0.72Z z≤ =
1
1
1
1
0.583
100.5832
1.166 1011.166
zxZ
x
xx
µσ
=−=
−=
= −=
b 1P( ) 0.4X x< =
1
2
P( ) 0.4P( ) 0.6
Z zZ z
< =< =
2
1
11
1
1
1
0.2530.253
100.2532
0.506 109.494
zz
xz
x
xx
µσ
== −
−=
−− =
− = −=
c 1P( ) 0.63X x> =
1
2
P( ) 0.63P( ) 0.63
Z zZ z
> =< =
2
1
11
1
1
1
0.3320.332
2100.332
20.664 10
9.336
zz
xz
x
xx
µ
== −
−=
−− =
− = −=
d 1P( ) 0.2X x≥ =
1
1
P( ) 0.2P( ) 0.8
Z zZ z
≥ =≤ =
1
11
1
0.84210
2100.842
2
zxz
x
=−=
−=
1
1
1.684 1011.684
xx
= −=
M B 1 2 Q l d - 1 0 166 T h e n o r m a l d i s t r i b u t i o n
8 34, 16µ σ= = a P( ) 0.31X c> =
P( ) 0.31P( ) 0.69
Z kZ k
> =< =
0.49634
160.496 16 347.936 34
41.936
kck
cc
c
=−=
× = −+ =
=
b P( ) 0.75X c≤ =
P( ) .75Z k≤ =
0.67434
160.674 16 34
10.784 3444.784
kck
cc
c
=−=
× = −+ =
=
c P( ) 0.21X c< =
1
2
P( ) 0.21P( ) 0.79
Z kZ k
< =< =
2
1
1
0.8070.807
3416
0.807 16 3412.912 34
21.088
kk
ck
cc
c
== −
−=
− × = −− + =
=
d P( ) 0.55X c≥ =
1
2
P( ) 0.55P( ) 0.55
Z kZ k
> =< =
2
1
1
0.1260.126
3416
0.126 16 342.016 34
31.984
kk
ck
cc
c
== −
−=
− × = −− + =
=
9 (22, 25), 22, 5X N µ σ= =∼ a P(22 22 ) 0.7k X k− ≤ ≤ + =
1 2
2
1 0.7 0.3 0.152 2
P( ) 0.7P( ) 0.7 0.15
z Z zZ z
− = =
≤ ≤ =≤ = +
2
1
2
0.851.036
1.03622 22
5
zz
kz
=== −
+ −=
1.036 55.18k
k× =
=
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 167
b P(22 22 ) 0.24k X k− < < + =
1 2
2
1 0.24 0.76 0.382 2
P( ) 0.24P( ) 0.24 0.38
z Z zZ z
− = =
< < =< = +
2
1
2
0.620.305
0.30522 22
50.305 5
1.525
zz
kz
kk
=== −
+ −=
× ==
c P( | 23) 0.32X k X< < =
P( 23) 0.32P( 23)
X k XX
< ∩ < =<
23k < 23 22
50.2
P( 23) P( 0.2)0.5793
P(( ) ( 23)) 0.32P( 23)
P( ) 0.32P( 23)
Z
X Z
X k XX
X kX
−=
=< = <
=< ∩ < =
<< =<
1
2
P( ) 0.32 0.57930.1854
P( ) 0.1854P( ) 0.8146
X k
Z zZ z
< = ×=
< =< =
2
1
0.8950.895
220.8955
4.475 2217.525
zz
k
kk
== −
−− =
− = −=
10 P( ) 0.8Z c≤ = 0.842 (CND Table)c = or inv norm (0.8, 0, 1) Answer is E
11 1P( ) 0.7Z c> =
2
2
1
P( ) 0.70.524
0.524
Z ccc
< === −
Answer is A
12 P( 1.2 ) 0.4Z k− < < = P( ) P( 1) 0.4Z k Z< − < − =
P( ) 0.4 P( 1.2)0.4 1 P( 1.2)1.4 0.88490.51510.038
Z k ZZ
k
< = + < −= + − <= −==
Answer is C
13 (0, 1)Z N∼
1
2
2
1
P( ) 0.35P( ) 1 0.35
0.650.385
0.385
Z zZ z
zz
≤ =≤ = −=== −
Answer is B 14 1P( | 0.5) 0.6Z k Z< < =
1
1
1
1
1
1
2
2
2
1
P( 0.5) 0.60.5
P( ) 0.6P( 0.5)
P( ) 0.6 P( 0.5)0.6 0.69150.4149
0.215 (inv norm (0.4149))or 1 P( ) .4149
1 0.4149 P( )0.5851 P( )
0.2150.215
Z k Zk
Z kZZ k Z
kZ k
Z kZ k
kk
< ∩ < =<
< =<
< = × <= ×== −
− < =− = <
= <== −
Answer is A 15 (20, 16), 20, 4
P( ) 0.6X N
X kµ σ= =
< =∼
M B 1 2 Q l d - 1 0 168 T h e n o r m a l d i s t r i b u t i o n
P( ) 0.60.253
200.2534
1.012 2021.012
Z cc
k
kk
< ==
−=
= −=
Answer is D 16 (12, 4), 12, 2
P( ) 0.82Z N
X kµ σ= =
> =∼
1
2
P( ) 0.82P( ) 0.82
Z zZ z
> =< =
2
1
1
0.9150.915
122
0.915 2 121.83 12
10.17
zz
kz
kk
k
== −
−=
− × = −− = −
=
Answer is B 17 160, 8µ σ= =
a P( ) 0.95P( ) 0.95
X kZ z
< =< =
1.6451601.6458
173.16
zk
k
=−=
=
Inv Norm (0.95, 160, 8)=173.16 Theo is 173.16 cm tall. b
1
2
P( ) 0.80P( ) 0.80P( 0.80
X kZ zZ z
> => =< =
2
1
0.8420.842
1600.8428
153.264
zz
k
k
== −
−− =
=
Luisa is 153.26 cm tall. 18 45, 0.5µ σ= =
a 1
1
2
2
1
1
1
P( ) 0.2P( ) 0.2P( ) 0.6 0.2
0.80.842
0.842450.842
0.544.58
X kZ zZ z
zz
k
k
< =< =< = +
=== −
−− =
−
The minimum length of packaged nails is 44.58 mm. b 2
2
2
P( ) 0.2P( ) 0.2P( ) 0.8
X kZ zZ z
> => =< =
2
2
2
0.842450.842
0.545.42
zk
k
=−=
=
The maximum length of packaged nails is 45.42 mm. 19 50, 2µ σ= = 1
2
2
2 1
P( ) 25%P( ) 75%
0.75P( ) 0.75
0.674, 0.674500.674
248.65
X kX k
Z zz z
k
k
=< =
=< =
= = −−− =
=
You would need to run 48.65 sec to qualify. 20 2
1
(20, )P( 19) 0.7P( ) 0.7
X NXZ z
σ≥ =≥ =
∼
1
2
2
1
19 20
P( ) 0.70.524
0.52419 200.524
10.524
1.908
z
Z zzz
σ
σ
σ
σ
−=
≤ === −
−− =
−=−
=
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 169
21 0,µ σ σ= = P( 48) 0.4X < =
1
1
2
2
1
P( ) 0.448 50
P( ) 0.60.2533
0.253320.2533
20.2533
7.896
Z z
z
Z zzz
σ
σ
σ
σ
< =−=
< === −
−− =
−=−
= 22 500µ =
1
2
2
1
P( 485) 0.1P( ) 0.1P( ) 0.9
1.2821.282
485 5001.282
151.282
11.704 grams
XZ zZ z
zz
σ
σ
< =< =< =
== −
−− =
−=−
=
23 ( ,16) 4X N µ σ =∼
1
2
2
1
P( 17) 0.99P( ) 0.99P( ) 0.99
2.3262.326
172.3264
9.304 1726.304
XZ zZ z
zz
µ
µµ
≥ =≥ =≤ =
== −
−− =
− = −=
24 3σ =
P( 27) 0.35P( ) 0.35P( ) 0.65
0.385
XZ zZ z
z
≥ =≥ =≤ =
=
270.3853
1.156 2725.844
µ
µµ
−=
= −=
25 30σ = P( 240) 0.7
P( ) 0.70.5242400.524
3015.72 240
224.28 sec3 min 44 sec
XZ z
zµ
µµ
< =< =
=−=
= −==
Investigation — Sunflower stems 75, 8µ σ= = a A grade is when P( ) 0.1X k> =
P( ) 0.1P( ) 0.9
1.2816751.2816
885.25
X zZ z
zk
k
> =< =
=−=
=
A grade is awarded when stem lengths are greater than 85.25 cm.
b B grade is when P( ) 0.2X k> =
P( ) 0.2P( ) 0.8
0.8416750.8416
881.73
Z zZ z
zk
k
> =< =
=−=
=
B grade is awarded for stem lengths between 81.73 cm and 85.25 cm.
c C grade is when P( ) 0.3X k> =
P( ) 0.3P( ) 0.7
0.5244750.5244
879.19
Z zZ z
Zk
k
> =< =
=−=
=
C grade is awarded for stem lengths between 79.19 cm and 81.73 cm.
Exercise 10C — The normal approximation to the binomial distribution 1 a Bi (500, 0.52)X ∼
500 ( 30)500 0.52260 ( 10)500 0.48240 ( 10)
nnp
nq
= ≥= ×= ≥= ×= ≥
500 0.52 0.48124.8Bi (500, 0.52) (260, 124.8)
npq
X X N
= × ×=
⇒∼ ∼
b Bi (400, 0.48)X ∼
400 ( 30)400 0.48192 ( 10)400 0.52208 ( 10)400 0.48 0.5299.84Bi (400, 0.48) (192, 99.84)
nnp
nq
npq
X X N
= ≥= ×= ≥= ×= ≥= × ×=
⇒∼ ∼
M B 1 2 Q l d - 1 0 170 T h e n o r m a l d i s t r i b u t i o n
c Bi (300, 0.5)X ∼ 300 ( 30)300 0.5150 ( 10)300 0.5150 ( 10)300 0.5 0.575Bi (300, 0.5) (150, 75)
nnp
nq
npq
X X N
= ≥= ×= ≥= ×= ≥= × ×=
⇒∼ ∼
d Bi (250, 0.55)X ∼ 250 ( 30)250 0.55137.5 ( 10)250 0.45112.5 ( 10)250 0.55 0.4561.875Bi (250, 0.55) (137.5, 61.875)
nnp
nq
npq
X X N
= ≥= ×= ≥= ×= ≥= × ×=
⇒∼ ∼
2 a Bi (15, 0.5)X ∼
1515 0.57.515 0.5 0.53.75Bi (15, 0.5) (7.5, 3.75)
nnp
npq
X X N
== ×== × ×=
⇒∼ ∼
i 8 7.5P( 8) P3.75
P( 0.2582)0.6019
X Z
Z
− < = <
= <=
ii 12 7.5P( 12) P3.75
P( 2.324)1 P( 2.324)1 0.98990.0101
X Z
ZZ
− ≥ = ≥
= ≥= − ≤= −=
b Bi (15, 0.5)X ∼ P( 8) binom cdf (15, 0.5, 7)
0.5X < =
=
P( 12) 1 binom cdf (15, 0.5, 11)0.0176
X ≥ = −=
Note: The normal approximation should not be used as the criteria have not been satisfied; the normal does not approximate the binomial in this case.
3 Bi (120, 0.47)X ∼ 120 ( 30)
120 0.4756.4 ( 10)120 0.5363.6 ( 10)120 0.47 0.5329.892
nnp
nq
npq
= ≥= ×= ≥= ×= ≥= × ×=
Bi (120, 0.47) (56.4, 29.892)50 56.4P( 50) P
29.892P( 1.171)
X X N
X Z
Z
⇒
− < = <
= < −
∼ ∼
1 P( 1.171)0.1208
Z= − <=
4 Bi (200, 0.52)X ∼
200 ( 30)200 0.52104 ( 10)200 0.4896 ( 10)200 0.52 0.4849.92
Bi (200, 0.52) (104, 49.92)110 104P( 110) P
49.92P( 0.8492)1 ( 0.8492)0.1980
nnp
nq
npq
X X N
X Z
ZP Z
= ≥= ×= ≥= ×= ≥= × ×=
⇒
− ≥ = ≥
= ≥= − ≤=
∼ ∼
5 Bi (30, 0.51)X ∼ 30 ( 30)30 0.5115.3 ( 10)30 0.4914.7 ( 10)30 0.51 0.497.497Bi (30, 0.51) (15.3, 7.497)
14 15.3 16 15.3P (14 16) P7.497 7.497
P( 0.475 0.2557)P( 0.2557) P( 0.475)P( 0
nnp
nq
npq
X X N
X Z
ZZ ZZ
= ≥= ×= ≥= ×= ≥= × ×=
⇒
− − = ≤ ≤ = ≤ ≤
= − ≤ ≤= ≤ − ≤ −= ≤
∼ ∼
.2557) [1 P( 0.475)]P( 0.2557) 1 P( 0.475)0.6010 1 0.68260.2835
ZZ Z
− − ≤= ≤ − + ≤= − +=
6 Bi (350, 0.45)X ∼
350 ( 30)350 0.45
nnp
= ≥= ×
157.5 ( 10)350 0.55192.5 ( 10)350 0.45 0.5586.625Bi (350, 0.45) (157.5, 86.625)
nq
npq
X X N
= ≥= ×= ≥= × ×=
⇒∼ ∼
P(150 170)150 157.5 170 157.5P
86.625 86.625P( 0.806 1.343)P( 1.343) [P( 0.806)]P( 1.343) [1 P( 0.806)]0.9104 [1 0.7898]0.7002
X
Z
ZZ XZ X
≤ ≤− − = ≤ ≤
= − ≤ ≤= ≤ − ≤ −= ≤ − − ≤= − −=
7 Bi (400, 0.55)X ∼
400 ( 30)400 0.55220 ( 10)
nnp
= ≥= ×= ≥
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 171
400 0.45180 ( 10)400 0.55 0.4599Bi (400, 0.55) (220, 99)
P( 230 | 200)P( 230 200
P( 200)P( 230)P( 200)
230 220P99
200 220P99
P( 1.005)P( 2.010)1 P( 1.005)
P( 2.
nq
npq
X X NX X
X XX
XX
Z
Z
ZZ
ZZ
= ×= ≥= × ×=
⇒> >
> ∩ >=>
>=>
− > =
− >
>=> −
− <=<
∼ ∼
010)1 0.8425
0.97780.1611
−=
=
8 Bi (650, 0.51)X ∼
650 ( 30)650 0.51331.5 ( 10)650 0.49318.5 ( 10)650 0.51 0.49162.435Bi (650, 0.51) (331.5, 162.435)
P( 310 | 350)P( 310 350)
P( 350)P(310 350)
P( 350)
nnp
nq
npq
X X NX X
X XXX
X
= ≥= ×= ≥= ×= ≥= × ×=
⇒≥ ≤
≥ ∩ ≤=≤
≤ ≤=≤
∼ ∼
310 331.5 350 331.5P162.435 162.435
350 331.5P162.435
( 1.687 1.452)P( 1.452)
Z
Z
ZZ
− − ≤ ≤ =
− ≤
− ≤ ≤=≤
P( 1.452) P( 1.687)0.9268
0.9268 [1 P( 1.687)]0.9268
0.9268 [1 0.9542]0.9268
0.8810.92680.9506
Z Z
Z
≤ − ≤ −=
− − ≤=
− −=
=
=
9 Bi (170, 0.48)170170 0.4881.6170 0.48 0.5242.432
Xn
np
npq
== ×== × ×=
∼
(81.6, 42.432)X N∼ Answer is C
10 Bi (200, 0.5)X ∼
200200 0.5100200 0.5 0.550
(100, 50)90 100P( 90) P
50P( 1.414)1 P( 1.414)1 0.92130.0787
nnp
npq
X N
X Z
ZZ
== ×== × ×=
− < = <
= < −= − <= −=
∼
Answer is B 11 200, 0.5n p= =
200 0.5100200 0.5 0.550
(100, 50)105 100P( 105) P
50P( 0.707)1 P( 0.707)1 0.76010.2399
np
npq
X N
X Z
Z
= ×== × ×=
− > = >
= >= − <= −=
∼
Answer is C
12 A 1200,6
P( 50) 33.327.7
n p
X npnpq
= =
≥ ==
B 1200,2
P( 90) 10050
n p
X npnpq
= =
≥ ==
C 1200,6
P( 20) 33.327.7
n p
X npnpq
= =
≥ ==
D 1200,6
P( 40) 33.327.7
n p
X npnpq
= =
< ==
E 200, 0n p= = B is the correct answer as np is much greater than 10 Answer is B
13 11002
5025
(50, 25)
n p
npnpq
X N
= =
==∼
M B 1 2 Q l d - 1 0 172 T h e n o r m a l d i s t r i b u t i o n
a 48 50P( 48) P25
P( 0.4)( 0.4)0.6554
X Z
ZZ
− > = >
= > −= <=
b 51 50P( 51) P25
P( 0.2)0.5793
X Z
Z
− < = <
= <=
c 45 50 55 50P(45 55) P25 25
P( 1 1)P( 1) ( 1)P( 1) [1 ( 1)]0.8413 (1 0.8413)0.6826
X Z
ZZ P ZZ P Z
− − ≤ ≤ = ≤ ≤
= − ≤ ≤= ≤ − ≤ −= ≤ − − ≤= − −=
14 500, 0.49500 0.49245500 0.49 0.51124.95
(245, 124.95)
n Pnp
npq
X N
= == ×== × ×=∼
a 240 245P( 240) P124.95
P( 0.447)P( 0.447)0.6725
X Z
ZZ
− ≥ = ≥
= ≥ −= ≤=
b 260 245P( 260) P124.95
P( 1.342)0.9102
X Z
Z
− < = <
= <=
15 300, 0.52300 0.52156300 0.52 0.4874.88
n pnp
npq
= == ×== × ×=
a (156, 74.88)160 156P( 160) P
74.88P( 0.4623)1 P( 0.4623)1 0.67790.3221
X N
X Z
ZZ
− ≥ = ≥
= ≥= − ≤= −=
∼
b P( 130)X < don’t last means 0.48p = 300 0.48
14474.88
(144, 74.88)
np
npqX N
= ×==
∼
130 144P( 130) P74.88
P( 1.618)P( 1.618)1 P( 1.618)1 0.94710.0529
X Z
ZZ
Z
− < = <
= < −= >= − <= −=
16 200P (odd numbers) 0.16 3
0.48P (even numbers) 0.52
n == ×==
a Odd numbers 200 0.48
9649.92
95 96P( 95) P49.92
P( 0.142)P( 0.142)0.5565
np
npq
X Z
ZZ
= ×==
− > = >
= > −= <=
b Even numbers 200 0.52
10449.92
100 104P( 100) P49.92
P( 0.566)P( 0.566)1 P( 0.566)1 0.71440.2856
np
npq
X Z
ZZ
Z
= ×==
− < = <
= < −= >= − <= −=
Investigation — Supporting the proposal a 20, 0.75
20 0.751520 0.75 0.253.75
(15, 3.75)
n pnp
npq
X N
= == ×== × ×=∼
i 9 15P( 9) P3.75
P( 3.098)P( 3.098)
X Z
ZZ
− < = <
= < −= >
1 P( 3.098)1 0.99900.0010
Z= − <= −=
ii 15 15P( 15) P3.75
P ( 0)0.5000
X Z
Z
− > = >
= >=
b Bi (20, 0.75)P( 15) P( 16) P( 17) P( 18)P( 19) P( 20)
XX X X XX X
> = = + = + =+ = + =
∼
Use calculator 1 binom cdf (20, 0.75, 15)
0.4148−
=
Difference between normal approximation and binomial is 0.5000 0.4148 0.0852− = . The approximate value compares well with the actual value – slightly higher.
Exercise 10D — Hypothesis testing 1 H0: The extra training has no effect
H1: The extra training improves Michael’s times. 2 a H0: The television campaign will have no effect on sales.
b H0: Extra study will have no effect on exam marks. c H0: The life of milk in a bottle is not less than 4 days.
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 173
d H0: The average mass of a packet of chips is not at least 250 g.
3 a H1: The television campaign will increase sales. b H1: Extra study will improve exam marks. c H1: The life of milk in a bottle is less than 4 days. d H1: The average mass of a packet of chips is at least 250 g.
4 0.5, 10P( 7)p n
X= =
≥
H0: the training program has no effect using tables on pages 384
P( 7) 0.1172 0.0439 0.0098 0.00100.1719
X ≥ = + + +=
Since P( 7) 0.1,X ≥ ≥ the null hypothesis is accepted, i.e. this is not a significant result at the 10% level.
5 0.2, 15P( 8)p n
X= =
≥
H0: The new drug will have no effect. Using tables on page 385
P( 8) 0.0035 .0007 0.0001 0
0.0043X ≥ = + + +
=
Since P( 8) 0.05,X ≥ < the null hypothesis is rejected, i.e. this is a significant result at the 5% level.
6 0.7, 20P( 17)p n
X= =
≥
Use tables on page 387 H0: Private tutoring will have no effect on exam results.
P( 17) 0.0716 0.0278 0.0068 0.00080.107
X ≥ = + + +=
a Since P( 17) 0.2,X ≥ ≤ the null hypothesis is rejected. Yes there is a significant result at the 20% level.
b Since P( 17) 0.1,X ≥ ≥ the null hypothesis is accepted. No there is not a significant result at the 10% level.
c Since P( 17) 0.05,X ≥ ≥ the null hypothesis is accepted. No there is not a significant result at the 5% level.
7 H0: the average matchbox contains 50 matches 12, 0.5n p= = 8 boxes contain less than 50, 4 boxes contain ≥ 50
P( 4) 0.1208 0.0537 0.0161 0.0029 0.00020.1937
X ≤ = + + + +=
Since P( 4) 0.1937 0.1,X ≤ = ≥ the null hypothesis is accepted. No there is not a significant result at the 10% level.
8 8, 0.5n p= = H0: Big Ker’s products are the leading brand.
P ( 2) 0.1094 0.0313 0.0039
0.1446X ≤ = + +
=∼
Since P( 2) 0.2,X ≤ ≤ the null hypothesis is rejected. Yes there is a significant result at the 20% level.
9 16, 0.35n p= = H0: The tyre will withstand 20 000 km of motoring 12 tyres are >20,000, 4 tyres are <20 000
P( 4) 0.1553 0.0888 0.0353 0.0087 0.00100.2891
X ≤ = + + + +=
Since P( 4) 0.05,X ≤ > the null hypothesis is rejected. Yes there is a significant result at the 5% level.
10 a H0: Electricity bills will stay the same over time. H1: Electricity bills will increase over time. 8, 0.5n p= = b ( 6)
( 2) 0.1094 0.313 0.00390.1446
P XP X
≥≤ = + +
=
∼
c i.e. the level of significance is 14.46% approximately 15% significance.
Chapter review 1
a P( 2.6) 0.9953Z ≤ = b P( 1.1) 0.8643Z < = c P( 1.34) 0.9099Z ≤ = d P( 0.89) 0.8365Z ≤ =
e P( 1.5) 1 P( 1.5)1 0.93320.0668
Z Z> = − <= −=
f P( 2.3) 1 P( 2.3)1 0.98930.0107
Z Z≥ = − ≤= −=
g P( 1.18) 1 P( 1.18)1 0.88100.1190
Z Z≥ = − ≤= −=
h P( 0.73) 1 P( 0.73)1 0.76730.2327
Z Z> = − <= −=
i P( 1) P( 1)1 P( 1)1 0.84130.1587
Z ZZ
≤ − = ≥= − ≤= −=
j P( 1.8) P( 1.8)1 P( 1.8)1 0.96410.0359
Z ZZ
< − = >= − <= −=
k P( 2.18) P( 2.18)1 P( 2.18)1 0.98540.0146
Z ZZ
< − = >= − <
−=
l P( 2.39) P( 2.39)1 P( 2.39)1 0.99160.0084
Z ZZ
< − = >= − <= −=
M B 1 2 Q l d - 1 0 174 T h e n o r m a l d i s t r i b u t i o n
m P( 2) P( 2)0.9772
Z Z≥ − = ≤=
n P( 2.5) P( 2.5)0.9938
Z Z≥ − = ≤=
o P( 1.61) P( 1.61)0.9463
Z Z> − = <=
p P( 2.03) P( 2.03)0.9788
Z Z> − = <=
2 a P( 1.2 | 1.5)P( 1.2 1.5)
P( 1.5)P( 1.2)P( 1.5)0.88490.93320.9482
Z ZZ Z
ZZZ
< << ∩ <=
<<=<
=
=
b P( 0.53 | 0.265)Z Z≤ − <
P( 0.53)P( 0.265)P( 0.53)
P( 0.265)1 P( 0.53)P( 0.265)
1 0.70190.6045
0.29810.60450.4931
ZZZZ
ZZ
< −=<>=
<− <=
<−=
=
=
c P( 2.19 | 1.2)Z Z≥ ≥
P( 2.19 1.2)P( 1.2)
P( 2.19)P( 1.2)
1 P( 2.19)1 P( 1.2)
1 0.98571 0.88490.01430.11510.1242
Z ZZ
ZZ
ZZ
≥ ∩ ≥=≥
≥=≥
− ≤=− ≤
−=−
=
=
d P( 1.9 | 2.368)Z Z> <
P( 1.9 2.368)P( 2.368)
P(1.9 2.368)P( 2.368)
P( 2.368) P( 1.9)P( 2.368)
0.9911 0.97130.9911
Z ZZZ
ZZ Z
Z
> ∩ <=<
< <=<
< − <=<
−=
0.01980.99110.0200
=
=
3 a 26, (22, 36)
26 226
0.6667
X X NxZ µ
σ
=−=
−=
=
∼
b 18, (16, 16)
18 164
0.5
X X NxZ µ
σ
=−=
−=
=
∼
c 56, (50.9, 100)
56 50.910
5.1100.51
X XxZ µ
σ
=−=
−=
=
=
∼
4 22, 6µ σ= = a P( 23)X >
23 226
0.1667
Z −=
=
P( 23) P( 0.1667)1 P( 0.1667)1 0.56640.4336
X ZZ
> = >= − <= −=
b P( 11)X ≥
11 226
1.833
Z −=
= −
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 175
P( 11) P( 1.833)P( 1.8333)0.9666
X ZZ
≥ = ≥ −= ≤=
c P( 31)X <
31 226
1.5
Z −=
=
P( 31) P( 1.5)0.9332
X Z< = <=
d P( 19)X ≤
19 22
60.5
Z −=
= −
P( 19) P( 0.5)P( 0.5)1 P( 0.5)1 0.69150.3085
X ZZ
Z
≤ = ≤ −= ≥= − ≤= −=
e P(20 26)X≤ ≤
20 226
0.333326 22
60.6667
Z
Z
−=
= −−=
=
P(20 26) P( 0.3333 0.6667)
P( 0.6667) P( 0.3333)P( 0.6667) P( 0.3333)P( 0.6667) [1 ( 0.3333)]0.7477 [1 0.6304]0.3781
X ZZ ZZ ZZ P Z
≤ ≤ = − ≤ ≤= ≤ − ≤ −= ≤ − ≥= ≤ − − ≤= − −=
5 18 mm, 1 mmµ σ= = a P( 20.5)
20.5 181
2.5P( 20.5) P( 2.5)
1 P( 2.5)1 0.99380.0062
X
Z
X ZZ
>−=
=> = >
= − <= −=
b P( 19)19 18
11.0
P( 19) P( 1)0.8413
X
X Z
<−=
=< = <
=
c P( 19 | 20.5)P( 1| 2.5)P( 1 2.5)
P( 2.5)
X XZ ZZ Z
Z
> <= > <
> ∩ <=<
P( 2.5) P( 1)P( 2.5)
0.9938 0.84130.9938
0.1535
Z ZZ
< − <=<
−=
=
6 a P( ) 0.70.524
Z cc
< ==
b P( ) 0.951.645
Z cc
≤ ==
c
1
1
1
P( ) 0.1P( ) 0.1P( ) 0.9
1.2811.281
Z cZ cZ c
cc
≤ =≥ =≤ =
== −
M B 1 2 Q l d - 1 0 176 T h e n o r m a l d i s t r i b u t i o n
d P( ) 0.28Z c< =
1
1
1
P( ) 0.28P( ) 0.72
0.5830.583
Z cZ c
cc
> =< =
== −
e P( ) 0.71Z c≥ =
1
1
P( ) 0.710.553
0.553
Z ccc
≤ === −
f P( ) 0.89Z c> =
1
1
P( ) 0.891.227
1.227
Z ccc
< === −
g P( ) 0.4P( ) 0.6
0.253
Z cZ c
c
> =< =
=
h P( ) 0.27Z c≥ =
P( ) 0.730.613
Z cc
≤ ==
7 a 31, 9P( ) 0.3X kµ σ= =
< =
1
2
2
2
1
P( ) 0.3P( ) 0.3P( ) 0.7
0.5240.524
Z cZ cZ c
cc
xZ µσ
< => =< =
== −
−=
310.5249
4.716 3126.28
x
xx
−− =
− = −=
b P( ) 0.83Z k> =
1
2
P( ) 0.83P( ) 0.83
Z cZ c
> =< =
2
1
0.9540.954
310.9549
8.586 3122.41
cc
x
xx
== −
−− =
− = −=
c P( ) 0.11X k≥ =
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 177
P( ) 0.11Z c≥ =
P( ) 0.891.225
311.2259
11.025 3142.025
Z cc
x
xx
≤ ==
−=
= −=
d P( ) 0.997X k≤ =
P( ) 0.9972.75
312.759
24.75 3155.75
Z cc
x
xx
≤ ==
−=
= −=
8 a 28 cm,P( 25) 0.9Zµ =
> =
1
2
2
1
P( ) 0.9P( ) 0.9
1.2821.282
Z cZ c
cc
xZ µσ
> =< =
== −
−=
25 281.282
31.282
2.34 cm
σ
σ
−− =
−=−
=
b P( 30)30 28
2.340.8547
P( 30) P( 0.8547)
X
Z
X Z
>−=
=> = >
1 P( 0.8547)1 0.80370.1963
Z= − <= −=
c P( ) 0.3X k< =
1
2
2
2
1
P( ) 0.3P( ) 0.3P( ) 0.7
0.5240.524
280.5242.34
1.226 2826.77
Z cZ cZ c
cc
x
xx
< => =< =
== −
−− =
− = −=
maximum fish length is 26.77 cm
9 a
2
Bi (220, 0.5)
220 0.5110
220 0.5 0.555
30, 10, 10Bi (220, 0.5) N(110, 55)
Xnp
npq
n np nqX X
µ
σ
== ×=
== × ×=≥ ≥ ≥
⇒
∼
∼ ∼
M B 1 2 Q l d - 1 0 178 T h e n o r m a l d i s t r i b u t i o n
b
2
Bi (600, 0.49)
600 0.49294
600 0.49 0.51149.9430, 10, 10Bi (600, 0.49) N(294, 149.94)
Xnp
npq
n np nqX X
µ
σ
== ×=
== × ×=≥ ≥ ≥
⇒
∼
∼ ∼
c
2
Bi (325, 0.51)
325 0.51165.75
325 0.51 0.4981.217530, 10, 10Bi (325, 0.51) N(165.75, 81.2175)
Xnp
npq
n np nqX X
µ
σ
== ×=
== × ×=≥ ≥ ≥
⇒
∼
∼ ∼
d
2
Bi (1000, 0.47)
1000 0.47470
1000 0.47 0.53249.130, 10, 10Bi (1000, 0.47) N(470, 249.1)
Xnp
npq
n np ngX X
µ
σ
== ×=
== × ×=≥ ≥ ≥
⇒
∼
∼ ∼
10
2
Bi (250, 0.52)250 0.52130
250 0.52 0.4862.4N(130, 62.4)
P( 120)
X
XX
µ
σ
= ×=
= × ×=
<
∼
∼
P( 120) P( 1.266)120 130
62.41.266
P( 1.266)1 P( 1.266)1 0.89720.1028
X Z
Z
ZZ
< = < −−=
= −= >= − <= −=
11
2
Bi (100, 0.49)100 0.4949
100 0.49 0.5125
P(47 50)
X
X
µ
σ
= ×=
= × ×=≤ ≤
∼
1
2
47 4925
0.450 49
250.2
P( 0.4 0.2)
z
z
Z
−
−=
=−=
=− ≤ ≤
[ ]
P( 0.2) P( 0.4)P( 0.2) P( 0.4)P( 0.2) 1 P( 0.4)0.5793 (1 0.6554)0.2347
Z ZZ ZZ Z
= ≤ − ≤ −= ≤ − ≥= ≤ − − ≤= − −=
12
2
300, 0.5P( 160)
300 0.5150
300 0.5 0.575N(150, 75)
n pX
X
µ
σ
= =≥= ×=
= × ×=∼
1160 150
751.1547
P( 1.547)
z
Z
−=
=≥
T h e n o r m a l d i s t r i b u t i o n M B 1 2 Q l d - 1 0 179
1 P( 1.547)1 0.87590.1241
Z= − ≤= −=
13 a 0 :H The alternative route makes no difference to travelling time. 1:H The alternative route shortens travelling time. b 0 :H Not many overweight vehicles use Fiona’s street. 1:H Too many overweight vehicles use Fiona’s street. c 0 :H Using the software will make no impact on
mathematics marks. 1:H Using the software will increase mathematics
marks. 14 a 0 :H The coin is not biased.
1:H The coin is biased.
b 20, 0.5P( 16) 0.0046 0.0011 0.0002
0.0059
n pX
= =≥ = + +
=
or binom cdf (20, 0.5, 16, 20) c Significant at the 0.6% level of confidence
15 There are 5 numbers less than or equal to 9 hours. 0 :H Average shift time stays the same. 1:H Average shift time is becoming longer. 16, 0.5
P( 5) 0.0667 0.0278 0.0085 0.0018 0.00020.105
n pX
= =≤ = + + + +
=
Since P( 5) 0.1 (10% level)X ≤ > we accept 0H and acknowledge that the shift times aren’t getting longer.
