## (Solved) Sample Item 1 1 2 3 4 5 6.85 6 6.15 item 2 item 3 Item 4 Item 5 6.01 Xbar R 5.906 6.074 6.942 6 5.9904 Average R Chart D3 D4 UCL = X-double bar +A2*...

3)Â Â Â Â Â For the data in Problem 2, if the true mean and standard deviation of the process are 6.05 and 0.04.Â

a.Â Â Â Â Â Â Is this process capable of meeting its specification?Â Calculate its Cp and CPk.

b.Â Â Â Â Â What % of cylinders are out of specs?

Sample Item 1
1
2
3
4
5 6.1
6.35
5.85
6
6.15 item 2
item 3
Item 4
Item 5
6.15
5.35
5.98
5.95
6.02
6.2
6.1
5.7
5.9
6.2
6.12
6.08
5.8
5.7
6.04
6.17
5.6
6.25
5.99
6.01 Xbar R
5.906
6.074
6.03
5.942
6
5.9904 Average R Chart
D3
D4 UCL = X-double bar +A2* Rbar =
LCL =X doublebar - A2 * R bar =
A2
n= 0.8
0.65
0.35
0.47
0.65
0.584 0.5777
5 R Chart
6.4
1.4 6.2 1.2 6 1
0.8 5.8 0.6 5.6 0.4
5.4 0.2
0 1 2
Column I 3
Column M 4
Column N 5
Column O 5.2 1 2
Column H Xbar
LCL
CL
5.653023
5.653023
5.653023
5.653023
5.653023 5.9904
5.9904
5.9904
5.9904
5.9904 UCL
6.327777
6.327777
6.327777
6.327777
6.327777 R Chart
LCL CL
0
0
0
0
0 0.584
0.584
0.584
0.584
0.584 UCL
1.234576
1.234576
1.234576
1.234576
1.234576 0
2.114 X Bar Chart
6.4
6.2
6
5.8
5.6
5.4
5.2 1 2
Column H 3
Column J 4
Column K 5
Column L Frequency
(number of times)
21
Parts missing
13
Literature not in the box
9
Call went to voice mail
8
Delivery was late
3
Unit not working
2
Type of Problem Cummlative Cummalative%
21
38%
34
61%
43
77%
51
91%
54
96%
56
100% 56 Chart Title
25
20
15
10
5
0 120%
100%
80%
60%
40%
20%
0% Frequency (number of times) Cummalative% 120%
100%
80%
60%
40%
20%
0%

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This question was answered on: Sep 05, 2019

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Sep 05, 2019

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