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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
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Chapter 2 Exercise Solutions
"#+.(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
Descriptive Statistics: Ex2-3Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
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&, &), &$, 953, &4, &, &, &&2. 3he samplemedian is the middle value.
(bSince the median is the value dividing the ran'ed sample observations in half, it remainsthe same regardless of the si-e of the largest measurement.
"#.MTB > Stat > Basic Statistics > Display Descriptive Statistics
Descriptive Statistics: Ex2-5Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
Ex2-5 0 121.25 .00 22.63 96.00 102.50 11".00 1!!.50
Variable Maximum
Ex2-5 156.00
2-2
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Chapter 2 Exercise Solutions
"#%.(a, (dMTB > Stat > Basic Statistics > Display Descriptive Statistics
Descriptive Statistics: Ex2-Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
Ex2-6 !0 0 129.9 1.!1 .91 11.00 12!.00 12.00 135.25
Variable MaximumEx2-6 160.00
(b
se √n 5 √4) ≅ binsMTB > !raph > "isto#ra$ > 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# $# Ex2-6 N % !0
&ea# 'nit % 1.0
2 11 9
5 12 011
12 233
1" 12 !!!!55555
19 12 6"
(5) 12 999
16 13 0111
12 13 33
10 13
10 13 6""
" 13
" 1! 001
! 1! 22
+ 151, 160
2-3
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Chapter 2 Exercise Solutions
"#.
se &) &n = ≅ bins
MTB > !raph > "isto#ra$ > Si$ple
Yield
F r e q u e n c y
96928884
18
16
14
12
10
8
6
4
2
0
Histogram of Process Yield (Ex2-7)
2-(
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Chapter 2 Exercise Solutions
"#.(a Stem-and-&ea# l$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 $) bins. 3he number of bins can be changed b6 editing the7 scale. 8owever, if & bins are specified, 9003A: generates an #bin histogram.
;onstructing a bin histogram re
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Chapter 2 Exercise Solutions
"#(c continued
MTB > )h*ins 12+5 1, +5 c,$ +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 $%."
2-
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Chapter 2 Exercise Solutions
"#&.
MTB > !raph > .ro*a*ility .lot > Sin#le
Fluid #unces
P e r c e n t
16.0816.0616.0416.0216.0015.98
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean
0.532
16.03StDev 0.02021
N 12
AD 0.297
P-Value
Normal
Pro$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
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Chapter 2 Exercise Solutions
"#$$.
MTB > !raph > .ro*a*ility .lot > Sin#le
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-
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Chapter 2 Exercise Solutions
"#$+ .
MTB > !raph > .ro*a*ility .lot > Sin#le
(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
Pro$a$ility Plot of iscosity !ata (Ex2-")
2-/
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Chapter 2 Exercise Solutions
"#$+ 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
Pro$a$ility Plot of iscosity !ata (Ex2-")
: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
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Chapter 2 Exercise Solutions
"#$4 .
MTB > !raph > .ro*a*ility .lot > Sin#le
(0n the dialog box, select Bistribution to choose the distributions
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
Pro$a$ility Plot of ycles to Failure (Ex2-&*)
2-11
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Chapter 2 Exercise Solutions
"#$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
Pro$a$ility Plot of ycles to Failure (Ex2-&*)
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
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Chapter 2 Exercise Solutions
"#$ .
MTB > !raph > .ro*a*ility .lot > Sin#le
(0n the dialog box, select Bistribution to choose the distributions
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
Pro$a$ility Plot of oncentration (Ex2-&,)
2-13
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Chapter 2 Exercise Solutions
"#$ 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
Pro$a$ility Plot of oncentration (Ex2-&,)
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(
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Chapter 2 Exercise Solutions
"#$%* ("#&.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
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Chapter 2 Exercise Solutions
"#") ("#$".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
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Chapter 2 Exercise Solutions
"#"" ("#$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
"
"$ $
( ( .&" $$
).+$ $)
n n
i i i ii i
x p x x p x n
S n
= =
−∑ ∑ − = = =−
2-1
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Chapter 2 Exercise Solutions
"#" ("#$.
3his is a Eoisson distribution with parameter λ 5 ).)", x G EH0().)".
(a).)" $().)"
Er1 $2 ($ ).)$&%$I
e x p
−
= = = =
(b).)" )().)"
Er1 $2 $ Er1 )2 $ () $ $ ).&)" ).)$&)I
e x x p
−
≥ = − = = − = − = − =
(c
3his is a Eoisson distribution with parameter λ 5 ).)$, x G EH0().)$.).)$ )().)$
Er1 $2 $ Er1 )2 $ () $ $ ).&&)) ).)$)))I
e x x p
−
≥ = − = = − = − = − =
;utting the rate at which defects occur reduces the probabilit6 of one or more defects b6approximatel6 one#half, from ).)$& to ).)$)).
"#"% ("#$.
!or f ( x to be a probabilit6 distribution, ( f x dx+∞
−∞∫ must e
8/17/2019 Quality Control Ch 2 Solutions
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Chapter 2 Exercise Solutions
"#" continued(b
+
$
$ +().) $ "().) ). ().)( $ " + $.%
+ + +i i
i
x p x µ =
+ + + = = × + × + × =∑ +
" " " " " " "
$ ( $ ().++ " ().+% + ().") $.% ).%$i ii x p xσ µ == − = + + − =∑
(c
$.$).++C $
+
$.$ $.$( ).)C "
+
$.$ $.$ ).$.)))C +
+
x
F x x
x
= =
+= = =
+ + = =
"#" ("#").
( )
)
( C ) $C ),$,",
( $ b6 definition
$ $ $
$
x
x
i
p x kr r x
F x kr
k r
k r
∞
=
= < < =
= =∑
− = = −
)
"#"& ("#"$.(a
3his is an exponential distribution with parameter λ 5 ).$"F).$"($Er1 $2 ($ $ ).$$ 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
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Chapter 2 Exercise Solutions
"#+) ("#"".$"$" $" "
$"
$$.,*
$$.,* $$.,*
4Er1 $"2 ($" ( 4( $$.,* 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
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Chapter 2 Exercise Solutions
"#+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
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Chapter 2 Exercise Solutions
"#+* ("#".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
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Chapter 2 Exercise Solutions
"#+ continued(d
!ind n to satisf6 Er1 x ≥ $ Q D ≥ 2 ≥ ).&, or e
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Chapter 2 Exercise Solutions
"#+ ("#"&.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
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Chapter 2 Exercise Solutions
"#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
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Chapter 2 Exercise Solutions
"#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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Chapter 2 Exercise Solutions
"#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
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Chapter 2 Exercise Solutions
"#) ("#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 $? " (% " "
d C C C C
d µ µ
µ π π
= − + − × − + + − × −
Setting e
8/17/2019 Quality Control Ch 2 Solutions
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Chapter 2 Exercise Solutions
"#$ ("#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
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Chapter 2 Exercise Solutions
"#+ ("#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 = − =