Quality Control Ch 2 Solutions

8/17/2019 Quality Control Ch 2 Solutions

1/31

Chapter 2 Exercise Solutions

Several exercises in this chapter differ from those in the 4 th edition. An “*” following theexercise number indicates that the description has changed (e.g., new values. A second

exercise number in parentheses indicates that the exercise number has changed. !orexample, “"#$%* ("#&” means that exercise "#$% was "#& in the 4th edition, and that thedescription also differs from the 4 th edition (in this case, as'ing for a time series plotinstead of a digidot plot. ew exercises are denoted with an “”.

"#$*.(a

( )$

$%.) $%.)+ $%.) $" $%.)"& o-n

ii

x x n=

= = + + + =∑ L

(b

( )"

"" " "

$ $ ($%.) $%.) ($%.) $%.) $" ).)")" o-$ $" $

n n

i ii i

x x n

sn

= =−∑ ∑ + + − + +

= = =− −

L L

MTB > Stat > Basic Statistics > Display Descriptive Statistics

Descriptive Statistics: Ex2-1Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

Ex2-1 12 0 16.029 0.0053 0.0202 16.000 16.013 16.025 16.0!

Variable Maximum

Ex2-1 16.0"0

"#".(a

( )$

).))$ 4&.&& ).))4 ).))" mmn

ii

x x n=

= = + + + =∑ L

(b

( )"

"" " "

$ $ ().))$ ).))4 ().))$ ).))4 ).))+ mm$ $

n n

i ii i

x x n

sn

= =−∑ ∑ + + − + +

= = =− −

L L

MTB > Stat > Basic Statistics > Display Descriptive StatisticsDescriptive Statistics: Ex2-2Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

Ex2-2 0 50.002 0.00122 0.003!! !9.996 !9.999 50.003 50.005

Variable Maximum

Ex2-2 50.006

2-1


2/31


"#+.(a

( )$

&+ & && & &".& /!n

ii

x x n=

= = + + + =∑ L

(b

( )"

"" " "

$ $ (&*+ &*& (&*+ &*& & +., /!$ & $

n n

i ii i

x x n

sn

= =−∑ ∑ + + − + +

= = =− −

L L

MTB > Stat > Basic Statistics > Display Descriptive Statistics


Ex2-3 9 0 952.9 1.2! 3."2 9!.00 9!9.50 953.00 956.00

Variable Maximum

Ex2-3 959.00

"#4.(a0n ran'ed order, the data are 1&4, &4&, &), &$, 953, &4, &, &, &&2. 3he samplemedian is the middle value.

(bSince the median is the value dividing the ran'ed sample observations in half, it remainsthe same regardless of the si-e of the largest measurement.

"#.MTB > Stat > Basic Statistics > Display Descriptive Statistics


Ex2-5 0 121.25 .00 22.63 96.00 102.50 11".00 1!!.50

Variable Maximum

Ex2-5 156.00

2-2


3/31


"#%.(a, (dMTB > Stat > Basic Statistics > Display Descriptive Statistics

Descriptive Statistics: Ex2-Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

Ex2-6 !0 0 129.9 1.!1 .91 11.00 12!.00 12.00 135.25

Variable MaximumEx2-6 160.00

(b

se √n 5 √4) ≅ binsMTB > !raph > "isto#ra$ > Si$ple

Hours

F r e q u e n c y

160152144136128120112

20

15

10

5

0

Histogram of Time to Failure (Ex2-6)

(cMTB > !raph > Ste$-an%-&ea'

Ste$-an%-&ea' Display: Ex2-Stem-and-lea# $# Ex2-6 N % !0

&ea# 'nit % 1.0

2 11 9

5 12 011

12 233

1" 12 !!!!55555

19 12 6"

(5) 12 999

16 13 0111

12 13 33

10 13

10 13 6""

" 13

" 1! 001

! 1! 22

+ 151, 160

2-3


4/31


"#.

se &) &n = ≅ bins

MTB > !raph > "isto#ra$ > Si$ple

Yield

F r e q u e n c y

96928884

18

16

14

12

10

8

6

4

2

0

Histogram of Process Yield (Ex2-7)

2-(


5/31


"#.(a Stem-and-&ea# l$t

2 12$6

6 13*313!

12 13$""69"

2 1!*3133101332!23!0!

(15) 1!$55669599695

3" 15*332!223!22112232

21 15$569"666

12 16*1!!011

6 16$5996

1 1"*0

Stem /re&ea#

(b

se ) &n = ≅ bins

MTB > !raph > "isto#ra$ > Si$ple

iscosity

F r e q u e n c y

1716151413

20

15

10

5

0

Histogram of iscosity !ata (Ex 2-")

ote that the histogram has $) bins. 3he number of bins can be changed b6 editing the7 scale. 8owever, if & bins are specified, 9003A: generates an #bin histogram.

;onstructing a bin histogram re


6/31


"#(c continued

MTB > )h*ins 12+5 1, +5 c,$ +nterval /reuen4ie er4ent

1 12.25 t$ 12."5 1 1.25

2 12."5 t$ 13.25 2 2.50

3 13.25 t$ 13."5 " ."5 ! 13."5 t$ 1!.25 9 11.25

5 1!.25 t$ 1!."5 16 20.00

6 1!."5 t$ 15.25 1 22.50

" 15.25 t$ 15."5 12 15.00

15."5 t$ 16.25 " ."5

9 16.25 t$ 16."5 ! 5.00

10 16."5 t$ 1".25 ! 5.00

11 $tal 0 100.00

(dMTB > !raph > Ste$-an%-&ea'

Ste$-an%-&ea' Display: Ex2-Stem-and-lea# $# Ex2- N % 0

&ea# 'nit % 0.10 2 12 6

6 13 133!

12 13 6"""9

2 1! 0011122333333!!!

(15) 1! 55556669999

3" 15 11222222223333!!

21 15 56666"9

12 16 0111!!

6 16 5699

1 1" 0

median observation ran' is ().() > ). 5 4). x).) 5 ($4.& > $4.&?" 5 $4.&

@$ observation ran' is ()."() > ). 5 ").@$ 5 ($4.+ > $4.+?" 5 $4.+

@+ observation ran' is ().() > ). 5 %).@+ 5 ($.% > $.?" 5 $.

(d$)th percentile observation ran' 5 ().$)() > ). 5 . x).$) 5 ($+. > $+.?" 5 $+.

&)th percentile observation ran' is ().&)() > ). 5 ". x).&) 5 ($%.4 > $%.$?" 5 $%."

2-


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"#&.

MTB > !raph > .ro*a*ility .lot > Sin#le

Fluid #unces

P e r c e n t

16.0816.0616.0416.0216.0015.98

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.532

16.03StDev 0.02021

N 12

AD 0.297

P-Value

Normal

Pro$a$ility Plot of %iquid !etergent (Ex2-&)

hen plotted on a normal probabilit6 plot, the data points tend to fall along a straightline, indicating that a normal distribution ade


8/31


"#$$.


Hours

P e r c e n t

160150140130120110

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.005

130.0StDev 8.914

N 40

AD 1.259

P-Value

Normal

Pro$a$ility Plot of Failure Times (Ex2-6)

hen plotted on a normal probabilit6 plot, the data points do not fall along a straight line,indicating that the normal distribution does not reasonabl6 describe the failure times.

"#$" .MTB > !raph > .ro*a*ility .lot > Sin#le

Yield

P e r c e n t

10510095908580

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean

0.015

89.48

StDev 4.158

N 90

AD 0.956

P-Value

Normal

Pro$a$ility Plot of Process Yield !ata (Ex2-7)

hen plotted on a normal probabilit6 plot, the data points do not fall along a straight line,indicating that the normal distribution does not reasonabl6 describe process 6ield.

2-


9/31


"#$+ .


(0n the dialog box, select Bistribution to choose the distributions

iscosity

P e r c e n t

18171615141312

99.9

99

95

90

80

7060504030

20

105

1

0.1

Mean

0.740

14.90

StDev 0.9804

N 80

AD 0.249

P-Value

Normal

Pro$a$ility Plot of iscosity !ata (Ex2-")

iscosity

P e r c e n t

1918171615141312

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Lo!

0.841

2.699

S!ale 0.06595

N 80

AD 0.216

P-Value

Lo"normal


2-/


10/31


"#$+ continued

iscosity

P e r c e n t

181716151413121110

99.999

90807060504030

20

10

5

32

1

0.1

S#a$e

0.010

16.10

S!ale 15.36

N 80

AD 1.032

P-Value

%e&'ull


:oth the normal and lognormal distributions appear to be reasonable models for the dataCthe plot points tend to fall along a straight line, with no bends or curves. 8owever, the plot points on the eibull probabilit6 plot are not straightDparticularl6 in the tailsD indicating it is not a reasonable model.

2-10


11/31


"#$4 .



ycles to Failure

P e r c e n t

2500020000150001000050000-5000

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.137

8700

StDev 6157

N 20

AD 0.549

P-Value

Normal

Pro$a$ility Plot of ycles to Failure (Ex2-&*)

ycles to Failure

P e r c e n t

100000100001000

99

95

90

80

70

60

50

40

30

20

10

5

1

Lo!

0.163

8.776

S!ale 0.8537

N 20

AD 0.521

P-Value

Lo"normal


2-11


12/31


"#$4 continued

ycles to Failure

P e r c e n t

100001000

99

90

80

706050

40

30

20

10

5

3

2

1

S#a$e

(0.250

1.464

S!ale 9624

N 20

AD 0.336

P-Value

%e&'ull


Elotted points do not tend to fall on a straight line on an6 of the probabilit6 plots, thoughthe eibull distribution appears to best fit the data in the tails.

2-12


13/31


"#$ .



oncentration+ ''m

P e r c e n t

150100500-50

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.005

9.470

StDev 22.56

N 40

AD 8.426

P-Value

Normal

Pro$a$ility Plot of oncentration (Ex2-&,)

oncentration+ ''m

P e r c e n t

100.010.01.00.1

99

95

90

80

70

60

50

40

30

20

10

5

1

Lo!

0.873

0.9347

S!ale 1.651

N 40

AD 0.201

P-Value

Lo"normal


2-13


14/31


"#$ continued

oncentration+ ''m

P e r c e n t

100.00010.0001.0000.1000.0100.001

99

90

80

706050

40

30

20

10

5

3

2

1

S#a$e

0.091

0.6132

S!ale 5.782

N 40

AD 0.637

P-Value

%e&'ull


3he lognormal distribution appears to be a reasonable model for the concentration data.Elotted points on the normal and eibull probabilit6 plots tend to fall off a straight line.

2-1(


15/31


"#$%* ("#&.MTB > !raph > Ti$e Series .lot > Sin#le or Stat > Ti$e Series > Ti$e Series .lot

Time #rder of ollection

E x 2 - "

80726456484032241681

17

16

15

14

13

12

Time eries Plot of iscosity !ata (Ex2-")

!rom visual examination, there are no trends, shifts or obvious patterns in the data,indicating that time is not an important source of variabilit6.

"#$* ("#$).MTB > !raph > Ti$e Series .lot > Sin#le or Stat > Ti$e Series > Ti$e Series .lot

Time #rder of ollection

E x 2 - 7

90817263544536271891

100

95

90

85

Time eries Plot of Yield !ata (Ex2-7)

3ime ma6 be an important source of variabilit6, as evidenced b6 potentiall6 c6clic behavior.

2-15


16/31


17/31


"#") ("#$".MTB > !raph > Ste$-an%-&ea'

Ste$-an%-&ea' Display: Ex2-,Stem-and-lea# $# Ex2-" N % 90

&ea# 'nit % 0.10

2 2 69

6 3 016"

1! ! 01112569

20 5 0111!!

30 6 111!!!!66"

3 " 3333566"

!3 2236

(6) 9 11!66"

!1 90 00113!5666

31 91 12!"

2" 92 1!!

2! 93 1122"

19 9! 11133!6"

11 95 1236

" 96 13!

3 9" 3

1 9 0

either the stem#and#leaf plot nor the fre


18/31


"#"" ("#$4.MTB > !raph > Boxplot > Si$ple

m m

50.0075

50.0050

50.0025

50.0000

49.9975

49.9950

.ox'lot of .earing .ore !iameters (Ex2-2)

"#"+ ("#$. xF 1the sum of two up dice faces2 sample spaceF 1", +, 4, , %, , , &, $), $$, $"2

$ $ $Er1 "2 Er1$,$2% % +%

x = = = × =

( ) ( )$ $ $ $ "Er1 +2 Er1$, "2 Er1",$2 % % % % +% x = = + = × + × =( ) ( ) ( ) +$ $ $ $ $ $Er1 42 Er1$, +2 Er1", "2 Er1+,$2 % % % % % % +% x = = + + = × + × + × =

. . .

$? +%C " " ? +%C + +? +%C 4 4 ? +%C ? +%C % % ? +%C (

? +%C 4 ? +%C & +? +%C $) " ? +%C $$ $? +%C $" )C otherwise

x x x x x x p x

x x x x x

= = = = = == = = = = =

"#"4 ("#$%.

( ) ( ) ( )$$

$

( " $ +% + " +% $" $ +% i ii

x x p x

=

= = + + + =∑ L

"

"$ $

( ( .&" $$

).+$ $)

n n

i i i ii i

x p x x p x n

S n

= =

−∑ ∑ − = = =−

2-1


19/31


"#" ("#$.

3his is a Eoisson distribution with parameter λ 5 ).)", x G EH0().)".

(a).)" $().)"

Er1 $2 ($ ).)$&%$I

e x p

−

= = = =

(b).)" )().)"

Er1 $2 $ Er1 )2 $ () $ $ ).&)" ).)$&)I

e x x p

−

≥ = − = = − = − = − =

(c

3his is a Eoisson distribution with parameter λ 5 ).)$, x G EH0().)$.).)$ )().)$

Er1 $2 $ Er1 )2 $ () $ $ ).&&)) ).)$)))I

e x x p

−

≥ = − = = − = − = − =

;utting the rate at which defects occur reduces the probabilit6 of one or more defects b6approximatel6 one#half, from ).)$& to ).)$)).

"#"% ("#$.

!or f ( x to be a probabilit6 distribution, ( f x dx+∞

−∞∫ must e


20/31


"#" continued(b

+

$

$ +().) $ "().) ). ().)( $ " + $.%

+ + +i i

i

x p x µ =

+ + + = = × + × + × =∑ +

" " " " " " "

$ ( $ ().++ " ().+% + ().") $.% ).%$i ii x p xσ µ == − = + + − =∑

(c

$.$).++C $

+

$.$ $.$( ).)C "

+

$.$ $.$ ).$.)))C +

+

x

F x x

x

= =

+= = =

+ + = =

"#" ("#").

( )

)

( C ) $C ),$,",

( $ b6 definition

$ $ $

$

x

x

i

p x kr r x

F x kr

k r

k r

∞

=

= < < =

= =∑

− = = −

)

"#"& ("#"$.(a

3his is an exponential distribution with parameter λ 5 ).$"F).$"($Er1 $2 ($ $ ).$$ x F e−≤ = = − =

Approximatel6 $$.M will fail during the first 6ear.

(b9fg. cost 5 N)?calculator Sale profit 5 N"?calculator et profit 5 NJ#)($ > ).$$ > K?calculator 5 N$&.$)?calculator.

3he effect of warrant6 replacements is to decrease profit b6 N.&)?calculator.

2-20


21/31


"#+) ("#"".$"$" $" "

$"

$$.,*

$$.,* $$.,*

4Er1 $"2 ($" ( 4( $$.,* 4, $$.,* $$.,* ).$"*

"

x x F f x dx x dx x

−∞

< = = = − = − = − =∫ ∫

"#+$* ("#"+.3his is a binomial distribution with parameter p 5 ).)$ and n 5 ". 3he process is

stopped if x ≥ $.

) ""

Er1 $2 $ Er1 $2 $ Er1 )2 $ ().)$ ($ ).)$ $ ). )."")

x x x

≥ = − < = − = = − − = − = ÷

3his decision rule means that ""M of the samples will have one or more nonconformingunits, and the process will be stopped to loo' for a cause. 3his is a somewhat difficultoperating situation.

3his exercise ma6 also be solved using Lxcel or 9003A:F

($ Excel unction B46MD4STx7 n7 p7 T89E(" MTB > Calc > .ro*a*ility Distri*utions > Bino$ial

Cu$ulative Distri*ution unctionin$mial it7 n % 25 and 8 % 0.01

x ( :% x )

0 0."""21

"#+"* ("#"4.

x G :0(", ).)4 Stop process if x ≥ $.

) ""

Er1 $2 $ Er1 $2 $ Er1 )2 $ ().)4 ($ ).)4 $ ).+% ).%4)

x x x

≥ = − < = − = = − − = − = ÷

"#++* ("#".3his is a binomial distribution with parameter p 5 ).)" and n 5 ).

4()

)

) ) $ 4& 4 4%

)OEr1 ).)42 Er1 "2 ().)" ($ ).)"

) ) )().)" ($ ).)" ().)" ($ ).)" ().)" ($ ).)" ).&"$

) $ 4

x x

x

p x x

−

=

≤ = ≤ = − ÷

= − + − + + − = ÷ ÷ ÷

∑

L

2-21


22/31


"#+4* ("#"%.3his is a binomial distribution with parameter p 5 ).)$ and n 5 $)).

).)$($ ).)$ $)) ).)$))σ = − =

O OEr1 2 $ Er1 2 $ Er1 ( 2 p k p p k p x n k pσ σ σ > + = − ≤ + = − ≤ +

k 5 $

"$))

)

) $)) $ && " &

$ Er1 ( 2 $ Er1 $))($().)$)) ).)$2 $ Er1 "2

$))$ ().)$ ($ ).)$

$)) $)) $))$ ().)$ ().&& ().)$ ().&& ().)$ ().&&

) $ "

$ J).&"$K ).)&

x x

x

x n k p x x

x

σ

−

=

− ≤ + = − ≤ + = − ≤

= − −∑ ÷

= − + + ÷ ÷ ÷

= − =

k 5 "

+$)) + &

)

$ Er1 ( 2 $ Er1 $))("().)$)) ).)$2 $ Er1 +2

$)) $))$ ().)$ ().&& $ ).&"$ ().)$ ().&&

+

$ J).&"K ).)$

x x

x

x n k p x x

x

σ

−

=

− ≤ + = − ≤ + = − ≤

= − = − +∑ ÷ ÷

= − =

k 5 +

4 $)) 4 &%

)

$ Er1 ( 2 $ Er1 $))(+().)$)) ).)$2 $ Er1 42

$)) $))$ ().)$ ().&& $ ).&" ().)$ ().&&4

$ J).&&"K ).))+

x x

x

x n k p x x

x

σ

−

=

− ≤ + = − ≤ + = − ≤

= − = − +∑ ÷ ÷

= − =

2-22


23/31


"#+* ("#".3his is a h6pergeometric distribution with N 5 " and n 5 , without replacement.

(aPiven D 5 " and x 5 )F

" " ") ) ($(++,%4&

Er1Acceptance2 () ).%++" (+,$+)

p

− ÷ ÷− = = = = ÷

3his exercise ma6 also be solved using Lxcel or 9003A:F($ Excel unction ".!E6MD4STx7 n7 D7 (" MTB > Calc > .ro*a*ility Distri*utions > "yper#eo$etric

Cu$ulative Distri*ution unction;8er


24/31


"#+ continued(d

!ind n to satisf6 Er1 x ≥ $ Q D ≥ 2 ≥ ).&, or e


25/31


"#+ ("#"&.3his is a h6pergeometric distribution with N 5 )) pages, n 5 ) pages, and D 5 $)

errors. ;hec'ing n/N 5 )?)) 5 ).$ ≤ ).$, the binomial distribution can be used toapproximate the h6pergeometric, with p 5 D? N 5 $)?)) 5 ).)").

) ) ))

Er1 )2 () ().)") ($ ).)") ($($().+%4 ).+%4)

x p −

= = = − = = ÷

$ ) $

Er1 "2 $ Er1 $2 $ JEr1 )2 Er1 $2K $ () ($

)$ ).+%4 ().)") ($ ).)") $ ).+%4 ).+" )."%4

$

x x x x p p

−

≥ = − ≤ = − = + = = − −

= − − − = − − = ÷

"#+ ("#+).

3his is a Eoisson distribution with λ 5 ).$ defects?unit.).$ )().$

Er1 $2 $ Er1 )2 $ () $ $ ).&) ).)&

)I

e x x p

−

≥ = − = = − = − = − =

3his exercise ma6 also be solved using Lxcel or 9003A:F($ Excel unction .64SS6 7 x7 T89E

(" MTB > Calc > .ro*a*ility Distri*utions > .oisson

Cu$ulative Distri*ution unction$i$n it7 mean % 0.1

x ( :% x )

0 0.90!3"

"#+& ("#+$.3his is a Eoisson distribution with λ 5 ).))))$ stones?bottle.

).))))$ )().))))$Er1 $2 $ Er1 )2 $ $ ).&&&&& ).))))$

)I

e x x

−

≥ = − = = − = − =

"#4) ("#+".

3his is a Eoisson distribution with λ 5 ).)$ errors?bill.

).)$ $

().)$Er1 $2 ($ ).))&&$

e x p−

= = = =

2-25


26/31


"#4$ ("#++.$

$

$ $

Er( ($ C $,",+,

$($

t

t t

t t

t p p t

d t p p p q

dq p µ

−

∞ ∞−

= =

= − =

= − = =∑ ∑

)

"#4" ("#+4.3his is a Eascal distribution with Er1defective weld2 5 p 5 ).)$, r 5 + welds, and x 5 $ >()))?$)) 5 $.

+ $ +$ $

Er1 $2 ($ ().)$ ($ ).)$ ($""().)))))$().%$"&) ).)))+ $

x p −−

= = = − = = ÷−

) ) $ 4& " 4

Er1 $2 Er1 )2 Er1 $2 Er1 "2

) ) )).)$ ).&& ).)$ ).&& ).)$ ).&& ).&%"

) $ "

x r r r > = = + = + =

= + = ÷ ÷ ÷

"#4+* ("#+. x G N (4), "C n 5 ),)))

8ow man6 fail the minimum specification, S 5 + lb.T

+ 4)Er1 +2 Er Er1 $2 ( $ ).$&

x z z

− ≤ = ≤ = ≤ − = Φ − =

So, the number that fail the minimum specification are (),))) × ().$& 5 &).

3his exercise ma6 also be solved using Lxcel or 9003A:F($ Excel unction 68MD4ST;7 7 σ7 T89E

(" MTB > Calc > .ro*a*ility Distri*utions > or$al

Cu$ulative Distri*ution unctionN$rmal it7 mean % !0 and tandard deviati$n % 5

x ( :% x )

35 0.15655

8ow man6 exceed 4 lb.T

4 4)Er1 42 $ Er1 42 $ Er $ Er1 $.%2

$ ($.% $ ).&4 ).)

x x z z − > = − ≤ = − ≤ = − ≤

= − Φ = − =

So, the number that exceed 4 lb. is (),))) × ().) 5 ").

2-2


27/31


"#44* ("#+%.x G N (, ).)""C S 5 4.& UC S 5 .) U

Er1;onformance2 Er1S S2 Er1 S2 Er1 S2

.) 4.&

(". ( ". ).&&+& ).))%"$ ).&).)" ).)"

x x x= ≤ ≤ = ≤ − ≤

− −

= Φ − Φ = Φ −Φ − = − = ÷ ÷

"#4* ("#+.3he process, with mean U, is currentl6 centered between the specification limits(target 5 U. Shifting the process mean in either direction would increase the numberof nonconformities produced.

Besire Er1;onformance2 5 $ ? $))) 5 ).))$. Assume that the process remains centered

between the specification limits at U. eed Er1 x ≤ S2 5 ).))$ ? " 5 ).))).

$

( ).)))

().))) +."&

z

z −

Φ =

= Φ = −

S S 4.& , so ).)$

+."& z

z

µ µ σ

σ

− − −= = = =

−

Erocess variance must be reduced to ).)$" to have at least &&& of $))) conform tospecification.

"#4% ("#+."G ( , 4 . !ind such that Er1 +"2 ).)"". x N x µ µ < =

$().)"" $.&&&$

+"$.&&&$

4

4( $.&&&$ +" 4).)

µ

µ

−Φ = −−

= −

= − − + =

"#4 ("#+&. x G N (&)), +"Er1 $)))2 $ Er1 $)))2

$))) &))$ Er

+*

$ (".*,$

$ ).&&,&

).))"$

x x

x

> = − ≤

− = − ≤

= − Φ

= −

=

2-2,


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"#4 ("#4). x G N ())), )". !ind S such that Er1 x R S2 5 ).))

$().)) ".

S )))".

)

S )( ". ))) 4$

−Φ = −−

= −

= − + =

"#4& ("#4$.

x$ G N ()), σ$" 5 $)))"C x" G N ()), σ"" 5 ))"C S 5 ,))) hC S 5 $),))) hsales 5 N$)?unit, defect 5 N?unit, profit 5 N$) × Er1good2 > N × Er1bad2 V c

!or Erocess $

$ $ $ $

$ $

proportion defective $ Er1S S2 $ Er1 S2 Er1 S2

$),))) ,)) ,))) ,))$ Er Er $,))) $,)))

$ (". ( ". $ ).&&+ ).))%" ).)$"4

p x x x

z z

= = − ≤ ≤ = − ≤ + ≤

− − = − ≤ + ≤ = − Φ + Φ − = − + =

profit for process $ 5 $) ($ V ).)$"4 > ().)$"4 V c$ 5 &.&+) V c$

!or Erocess "

" " " "

" "

proportion defective $ Er1S S2 $ Er1 S2 Er1 S2

$),))) ,)) ,))) ,))$ Er Er

)) ))

$ ( ( $ $.)))) ).)))) ).))))

p x x x

z z

= = − ≤ ≤ = − ≤ + ≤

− − = − ≤ + ≤

= − Φ + Φ − = − + =

profit for process " 5 $) ($ V ).)))) > ().)))) V c" 5 $) V c"

0f c" W c$ > ).)%"), then choose process $

2-2


29/31


"#) ("#4".Eroportion less than lower specificationF

%Er1 %2 Er (%

$l p x z

µ µ

− = < = ≤ = Φ −

Eroportion greater than upper specificationF

Er1 2 $ Er1 2 $ Er $ (

$u p x x z

µ µ − = > = − ≤ = − ≤ = − Φ −

) within $ "

) $ "

) " ) $ "

Erofit

J ( (% K J (% K J$ ( K

( J ( K ( J (% K

l uC p C p C p

C C C

C C C C C

µ µ µ µ

µ µ

= + − −= Φ − − Φ − − Φ − − − Φ −= + Φ − − + Φ − −

"$J ( K exp( ? "

"

d d t dt

d d

µ

µ µ µ π

−

−∞

Φ − = −∫

Set s 5 V µ and use chain rule

( )" "$ $

J ( K exp( ? " exp $? " ( " "

sd d dst dt

d ds d µ µ

µ µ π π −∞

Φ − = − = − − × −∫

( ) ( )" ") " ) $(Erofit $ $

( exp $? " ( ( exp $? " (% " "

d C C C C

d µ µ

µ π π

= − + − × − + + − × −

Setting e


30/31


"#$ ("#4+.

!or the binomial distribution, ( ($ C ),$,..., x n x

n p x p p x n

x

− = − = ÷

( )$

$ )

( ( ($ $n n x n x

i i

i x

n E x x p x x p p n p p p np

x

µ ∞ −−

= =

= = = − = + − =∑ ∑ ÷

( )

[ ]

" " " "

" " " " "

$ )

"" " "

J( K ( J ( K

( ( $ (

( ($

n n x x

i ii x

E x E x E x

n E x x p x x p p np np np

x

np np np np np p

σ µ

σ

∞ −

= =

= − = −

= = − = + −∑ ∑ ÷

= + − − = −

"#" ("#44.

!or the Eoisson distribution, ( C ),$,I

xe

p x x x

λ λ −

= = )

( )( $

$ ) )

J K ( I ( $I

x x

i ii x x

e E x x p x x e e e

x x

λ λ λ λ λ λ µ λ λ λ

− −∞ ∞ ∞− −

= = =

= = = = = =∑ ∑ ∑ ÷ −

[ ]

" " " "

" " " "

$ )

"" "

J( K ( J ( K

( ( I

(

x

i ii x

E x E x E x

e E x x p x x

x

λ

σ µ

λ λ λ

σ λ λ λ λ

−∞ ∞

= =

= − = −

= = = +∑ ∑ ÷

= + − =

2-30


31/31


"#+ ("#4.

!or the exponential distribution, ( C ) x f x e xλ λ −= ≥

!or the meanF

( )) )( x

xf x dx x e dx

λ

µ λ

+∞ +∞−

= =∫ ∫ 0ntegrate b6 parts, setting u x= and exp( dv xλ λ = −

( ) ( ))

)

$ $exp exp )uv vdu x x x dxλ λ

λ λ

+∞+∞ − = − − + − = + =∫ ∫

!or the varianceF"

" " " " "

" " "

)

$J( K ( J ( K (

( ( exp(

E x E x E x E x

E x x f x dx x x dx

σ µ λ

λ λ +∞ +∞

−∞

= − = − = − ÷

= = −∫ ∫

0ntegrate b6 parts, setting "u x= and exp( dv xλ λ = −

"

"))

"exp( " exp( () )uv vdu x x x x dxλ λ

λ

+∞+∞

− = − + − = − +∫ ∫ "

" " "

" $ $σ

λ λ λ = − =

Quality Control Ch 2 Solutions

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Transcript of Quality Control Ch 2 Solutions