Problem 1 If ν = 141.4 sin (ωt+30°) V and i = 11.31cos (ωt-30°) A, find for each; a) The maximum value, b) The rms value, c) The phasor expression in polar and rectangular form if voltage is the re...
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ME3122E - Tutorial Solution 4
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The solution of all exercise and review question of chapter 1 from the book of Data Communications and Networking By Behrouz A.Forouzan. Actually, I solved those problems for assignment during my ...
Quality Engineering I
Tutorial 3 Control Charts for Attributes
Tutorial 3.1
Determine Determine the trial central line and control limits for a p chart using the following data, which are for the payment of dental insurance claims. claims. Plot the values on graph paper and determine determine if the process is stable. If there are any out-of-control points, assume an assignable cause and determine the revised central line and control limits.
Data for the p chart Subgroup
n
np
p
Subgroup
n
np
p
1
300
3
0.0100
14
300
6
0.0200
2
300
6
0.0200
15
300
7
0.0233
3
300
4
0.0133
16
300
4
0.0133
4
300
6
0.0200
17
300
5
0.0167
5
300
20
0.0667
18
300
7
0.0233
6
300
2
0.0067
19
300
5
0.0167
7
300
6
0.0200
20
300
0
0.0000
8
300
7
0.0233
21
300
2
0.0067
9
300
3
0.0100
22
300
3
0.0100
10
300
0
0.0000
23
300
6
0.0200
11
300
6
0.0200
24
300
1
0.0033
12
300
9
0.0300
25
300
8
0.0267
13
300
5
0.0167
Solution
Plot the p chart;
Determine the trial UCL, CL, and LCL;
Revised control limits if any out-of-control point occurs (assume an assignable cause).
Control limits for the p chart
UCL p 3
p 1 p n
0.04016
CL p 0.01747 LCL p 3
p 1 p n
0.0052 0
The p chart
Revised control limits Discard the 5th point, which has a assignable cause, A new p can be calculated: np npd pnew 0.01542 n nd
Recalculate the control limits using the new fraction nonconforming: UCLnew p 3 LCLnew p 3
p 1 p n p 1 p n
0.03676
0.0059 0
The revised p chart P Chart of C2 0.04 UCL=0.03676 0.03 n o i t r 0.02 o p o r P
_ P=0.01542
0.01
0.00
LCL=0 1
3
5
7
9
11 13 15 Sample
17
19
21
23
Tutorial 3.2
Fifty motor generators are inspected per day from a stable process. The best estimate of the fraction nonconforming is 0.076. Determine the central line and control limits. On a particular day, 5 nonconforming generators were discovered. Is this in control or out of control?
Solution
The best estimate of p=0.076.
UCL p 3
p 1 p n
0.076 3
0.0761 0.076
CL p 0.076 LCL p 3
p=
p 1 p n
0.0364 0
5/50=0.1, in control.
50
0.1884
Tutorial 3.3
Inspection results of video-of-the-month shipments to customers for 25 consecutive days are given in the table. What central line and control limits should be established and posted if it is assumed that any out-of-control points have assignable causes? The number of inspection each day is constant and equals 1750.
The performance of the first shift is reflected in the inspection results of electric carving knives. Determine the trial central line and control limits for each subgroup. Assume that any out-of-control points have assignable causes and determine the standard value for the fraction nonconforming for the next production period.
Tutorial 3.4 Date
Number Inspected
Number
Date
Nonconforming
Number
Number
Inspected Nonconforming
Sept.6
500
5
Sept.23
525
10
7
550
6
24
650
3
8
700
8
27
675
8
9
625
9
28
450
23
10
700
7
29
500
2
13
550
8
30
375
3
14
450
16
Oct. 1
550
8
15
600
6
4
600
7
16
475
9
5
700
4
17
650
6
6
660
9
20
650
7
7
450
8
21
550
8
8
500
6
22
525
7
11
525
1
Solution
Solution P Chart of C2 0.030 UCL
0.025 0.020 n o i t r 0.015 o p o r P
_ P=0.01124
0.010 0.005 0.000
LCL 1
3
5
7
9
11 13 15 Sample
Tests performed with unequal sample sizes
17
19
21
23
Tutorial 3.5
The count of surface nonconformities in 1000 square meters of 20-kg kraft paper is given in the table. Determine the trial central line and control limits and the revised central line and control limits, assuming that out-of-control points have assignable causes.
Tutorial 3.5 Lot Number
Count of Nonconformities
Lot Number
Count of Nonconformities
20
10
36
2
21
8
37
12
22
6
38
0
23
6
39
6
24
2
40
14
25
10
41
10
26
8
42
8
27
10
43
6
28
0
44
2
29
2
45
14
30
8
46
16
31
2
47
10
32
20
48
2
33
10
49
6
34
6
50
3
Tutorial 3.5
Given the count of surface nonconformities in 1000 square meters of 20-kg kraft paper;
Determine the trial UCL, CL, and LCL;
Revised control limits if any out-of-control point occurs.
Solution
UCL c 3 c 16.53; CL c 8.03; LCL c 3 c 0.470 0.
Solution
Solution
Solution
Tutorial 3.6 Construct a control chart for the data in the table for empty bottle inspections of a softdrink manufacturer. Assume assignable causes for any points that are out of control. The subgroup size varies
