QUE a) 15 samples of n = 8 have been taken from a cleaning operation. The averag
ID: 3329153 • Letter: Q
Question
QUE
a) 15 samples of n = 8 have been taken from a cleaning operation. The average sample range for the 20 samples was 0.016 minute, and the average mean was 3 minutes. Determine the three- sigma control limits for this process. (3)
(b) 15 samples of n = 10 observations have been taken from a milling process. The average sample range is 0.01 centimetres. Determine upper and lower control limits for sample ranges. (3)
(c) Determine which of these three processes are capable: (9)
Process
Mean
Standard
Deviation
Lower
Specification
Upper
Specification
1
6.0
0.14
5.5
6.7
2
7.5
0.10
7.0
8.0
3
4.6
0.12
4.3
4.9
Table 1: Control Chart parameters Number of Observations in Subgroup
n
_
Factor for X Chart
A2
FACTORS FOR R CHARTS
Lower Control Limit
D3
FACTORS FOR R CHARTS
Upper Control Limit
D4
2
1.88
0
3.27
3
1.02
0
2.57
4
0.73
0
2.28
5
0.58
0
2.11
6
0.48
0
2.00
7
0.42
0.08
1.92
8
0.37
0.14
1.86
9
0.34
0.18
1.82
10
0.31
0.22
1.78
11
0.29
0.26
1.74
12
0.27
0.28
1.72
13
0.25
0.31
1.69
14
0.24
0.33
1.67
15
0.22
0.35
1.65
16
0.21
0.36
1.64
17
0.20
0.38
1.62
18
0.19
0.39
1.61
19
0.19
0.40
1.60
20
0.18
0.41
1.59
STION B3 [15 Marks]
A decision maker faced with four decision alternatives and four sates of nature develops the following profit payoff table.
States of nature Decision alternative
S_1
S_2
S_3
S_4
d_1
14
9
10
5
d_2
11
10
8
7
d_3
9
10
10
11
d_4
8
10
11
13
Process
Mean
Standard
Deviation
Lower
Specification
Upper
Specification
1
6.0
0.14
5.5
6.7
2
7.5
0.10
7.0
8.0
3
4.6
0.12
4.3
4.9
Table 1: Control Chart parameters Number of Observations in Subgroup
n
_
Factor for X Chart
A2
FACTORS FOR R CHARTS
Lower Control Limit
D3
FACTORS FOR R CHARTS
Upper Control Limit
D4
2
1.88
0
3.27
3
1.02
0
2.57
4
0.73
0
2.28
5
0.58
0
2.11
6
0.48
0
2.00
7
0.42
0.08
1.92
8
0.37
0.14
1.86
9
0.34
0.18
1.82
10
0.31
0.22
1.78
11
0.29
0.26
1.74
12
0.27
0.28
1.72
13
0.25
0.31
1.69
14
0.24
0.33
1.67
15
0.22
0.35
1.65
16
0.21
0.36
1.64
17
0.20
0.38
1.62
18
0.19
0.39
1.61
19
0.19
0.40
1.60
20
0.18
0.41
1.59
Explanation / Answer
a. n=20
R= minute
X_bar = 3 minute
For mean:
LCL = X_bar - A2*R, A2=0.37 for n=8
= 3-0.37*0.016 = 2.994 minute
UCL = X_bar + A2*R
= 3-0.37*0.016 = 3.006 minute
For range:
LCL = R*D3 = 0.14*0.016 = 0.0022 minute
UCL = R*D4 = 1.86*0.016 = 0.0298 minute
b.
n=10
R=0.01
LCL = R*D3 = 0.01*0.22 = 0.0022 cm
UCL = R*D4 = 0.01*1.78 = 0.0178 cm
c.
Process
Mean
Standard
Deviation
Lower
Specification
Upper
Specification
Lower Control Limit
1
6.0
0.14
5.5
6.7
6-0.14*3=5.58
2
7.5
0.10
7.0
8.0
7.5-3*0.1=7.2
3
4.6
0.12
4.3
4.9
4.6-3*0.12=4.24
Capable same reason as above
ProcessMean
Standard
Deviation
Lower
Specification
Upper
Specification
Lower Control Limit
Upper Control Limit Result1
6.0
0.14
5.5
6.7
6-0.14*3=5.58
6+3*0.14=6.42 Not capable as USL is beyond UCL2
7.5
0.10
7.0
8.0
7.5-3*0.1=7.2
7.5+3*0.1=7.8 Capable as LSL and USL are within LCL and UCL3
4.6
0.12
4.3
4.9
4.6-3*0.12=4.24
4.6+3*0.12=4.96Capable same reason as above
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.