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.