WRWP 3-6 Education Initiative NREM 405/505 With Storer Elementary Fall 2002.
Final Exam ESE 405/505 December 15,2008 - Washington … · · 2008-12-07Final Exam ESE 405/505...
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Final Exam ESE 405/505 December 15,2008 Your Name: So\V\-t\OfhS Your ID #: There are 4 problems in this examination.
Transcript of Final Exam ESE 405/505 December 15,2008 - Washington … · · 2008-12-07Final Exam ESE 405/505...
Final Exam ESE 405/505
December 15,2008
Your Name: So\V\-t\OfhS
Your ID #:
There are 4 problems in this examination.
1.(30 pts) Consider the following 25 samples.
Sample XlXlXl
10.06290.06360.0640
20.06300.06310.0622
30.06280.06310.0633
40.06340.06300.0631
50.06190.06280.0630
60.06130.06290.0634
70.06300.06390.0625
80.06280.06270.0622
90.06230.06260.0633
100.06310.06310.0633
110.06350.06300.0638
120.06230.06300.0630
130.06350.06310.0630
140.06450.06400.0631
150.06190.06440.0632
160.06310.06270.0630
170.06160.06230.0631
180.06300.06300.0626
190.06360.06310.0629
200.06400.06350.0629
210.06280.06250.0616
220.06150.06250.0619
230.06300.06320.0630
240.06350.06290.0635
250.06230.06290.0630
(a) Determine X and R charts using all sample points. You should see that bothX and R charts out-of-control.
(b) Remove samples 14, 15, and 22. (The remaining sample points are within the
UCL and LCL of the new X and R charts formed without sample points 14, 15,
and 22.) Using the remaining samples, estimate the standard deviation, ux.
(c) Determine the 6-(J limit using the result of part (b).(d) If the specifications are at 0.0630 ± 0.0015, determine the value of the process
capability ratio, Cpo
(This page is left blank for your work.)
CC\')
0.0640,--'-·_-
.•I UC L;O.063893
~ 0.0635
01III
~ 0.0630-1\ ~\1\/\/ "'.../\r\. !X;0.062952l ~ 0.0625
O.06:W~
."yj)I LCL=0.062011
,
.,.. . ..,.,2
468101214161820
Sampl~,
O.0024-!
WIu c L=O~00;2368.•..:
'" 0.0018'" ... .'!II:i 0.0012 I ___/"\A· I\ I\/ ,'" I R=O.OOO92
E . ,~ 0.0006
o.oo~~
--. y.•..•. VI LCi.=O,
,•,........2
4681012' 141618202224
Sample
/\Ux -=- ~I d~-
( Y\ -::;. 3 ~ Q.
o I 0 00 <32-3, / \ \ 6 <1'3 -:::::. O. 00 t) <t t) 6
W i\-ho--t- S<>WI pIE'> 14- 10;- <L. 22. ::: 0,000 i'2?, )) ,
(()/\
X + 30X :::
::::[0, o6t4-Q 2 ) 0 I o6lf4--0 (:) J
ex w \4 0 J- ~c.lV' p\-e .s. 14- i (S:-) ~ 2-.2- -:.:: 0 I 0628 ~-)
Cd) 0lD030~ -..... ~---
b ~ 0; 0()0 l;.l!-'/)
2.(20 pts) Consider the following 16 samples. Each sample consists of 4 inspections.Determine a control chart for nonconformities per unit. Is the process in control?
Sample Number ofNonconformities1
1
23
32
41
50
62
71
85
92
101
11
012
213
1
141
152
163
1.4
Y\=- 4 m ~·I .()
CL =- AA -=- L M ~IW\ =f:!- X.L/f\ "') 1'tV\
-=- (;2l ( 4 )/ I 6 -=- 0 I 4-~L
(ACL::. AJ: -+ ~ fM: (n "",OIv.:n+3~1.
-= I \~96
I ~ ~ JOI;,~L-(L ~ M _ S ~ M(V\ =- 0\ 4-11.- '3 '+
.::-0 \~~~ ~ 0
UCl=L396
1.2
it:!
:5 1.0•..
l. 0.8.•..c.::I
8 0.6.sl
~ 0.4~ .
0.2
0.0
U=0.422
LCL=O
2 4 8 . 10
SafT1)1e
12 14 16
----- -- --
3.(20 pts) Consider a c-chart for 6-(J control corresponding to 16 nonconformities among30 units inspected.
(a) Determine the a-risk based on this chart.(b) Determine the fi-risk if the average number of defects is actually 2 (i.e., c=2.0).
c~) c ~ 16/30 -=- OIS-~3 ) Do: l6UCL -=- C 't- "3 ~ ~ 2- \ 123>
L C L -:::: C - :5 JC -:::- \ \ is "5 7 -"/ 0
rA -=:.- \7K)~ D< Lc L\ C'. \ -r PJ-of D >vCLI c ')
~ -==- ?r-o~ t> < 0 \ 0 IS-is ~ -t PV6t D > 2. \ 7 2.3 \ o· ~~3 ~
o t\J-. :::.. 0, 0 I 7
c b') r::.. f vot D < veL I c.. \
~ = f' hJ~ D < ~, 1"L \ 2. ")
--
PVO( () <. LCL{ c \
f\-o f D ~ 0 \ ~ '\
4.(30 pts) A process is in statistically control with x = 39.7 and Ii = 2.5. The controlchart uses a sample size of n = 2. Specifications are at 40 ± 5. The quality characteristicis normally distributed.
(a) Determine the potential capability of the process, Cpo
(b) Determine the actual capability of the process, i.e., Cpk = min( Cpu, Cp1 ).
(c) Determine how much improvement could be made in the value of the fractionnonconforming, i.e., p, if the mean could be centered at the nominal value .
f\ -=- 2. )
......
X=-39,7 R=o2,t;)
~ -()~ ~ RId 'L -:::.')~,~ II, l2.8- =. '2.:2.u;
4-(;- - 3 S-
6 (2.. . .Z-/ t )1\
--ft.s - '3q I 1Cb"
C~V\ -:::.
(A~L- Y-
-:: O. 8-D
. ) -...-~ - -3 (2.2/6)3D;
/\ ~ - lS~ _3,'1 \ -, -1S--C p~ -::: ::..0,.130- 3[2,2..1')f\ (' 1\ C PK":-' ~;yj; nee pv I C P..t ) -=- 0 I I I
pvv ~ X < L ~ L '\ -t P VQ ~ x. "/ V\ S L ~
- pv-o ~ l- < L~Li ")+ p"", ~ to 7 ~S:';-Y: ')(; X lIx
- pt.-() ~ ~. < ."~S--]~17_~ -;- PYt? f ~ ....,.t-I-~- 3~~1 "'\2.,2..\6 2.\2..I~
b, D ~~ 34- <6
1\if) P
