Print Specification: .200 +/- .005 FOLLOW THE SUBGROUP ORDER 1 THROUGH 10 FOR YO
ID: 3244394 • Letter: P
Question
Print Specification: .200 +/- .005
FOLLOW THE SUBGROUP ORDER 1 THROUGH 10 FOR YOUR GRAPH
Subgroup 1
Subgroup 2
Subgroup 3
Subgroup 4
Subgroup 5
.197
.195
.197
.194
.202
.205
.202
.200
.197
.205
.194
.200
.195
.198
.200
Subgroup 6
Subgroup 7
Subgroup 8
Subgroup 9
Subgroup 10
.198
.200
.195
.195
.197
.201
.204
.201
.200
.203
.197
.206
.193
.197
.200
There are 10 subgroups, each with 3 data measurements.
1. What is the sigma of the individuals? Uncoded go to a minimum of 6 decimal places, coded go to a minimum of 3 decimal places.
2. What is the sigma of the subgroups? Uncoded go to a minimum of 6 decimal places, coded go to a minimum of 3 decimal places.
3. Calculate the Cpk; the % Oversized , and the % Undersized (state which is % Oversized and % Undersized within your submission). For all answers go to a minimum of 4 decimal places.
Subgroup 1
Subgroup 2
Subgroup 3
Subgroup 4
Subgroup 5
.197
.195
.197
.194
.202
.205
.202
.200
.197
.205
.194
.200
.195
.198
.200
Subgroup 6
Subgroup 7
Subgroup 8
Subgroup 9
Subgroup 10
.198
.200
.195
.195
.197
.201
.204
.201
.200
.203
.197
.206
.193
.197
.200
Explanation / Answer
Solution
Let X represent the measurement. We assume that X follows Normal Distribution.
Let Xbari = average of the ith sub-group; Ri = range of the ith sub-group
These are computed and presented below:
Subgroup i
Average Xbari
Range Ri
1
0.198666667
0.011
2
0.199
0.007
3
0.197333333
0.005
4
0.196333333
0.004
5
0.202333333
0.005
6
0.198666667
0.004
7
0.203333333
0.006
8
0.196333333
0.008
9
0.197333333
0.005
10
0.2
0.006
Average
0.198933333
0.0061
Q1
The sigma for individuals is given by Rbar/d2, where Rbar = average of Ri’s and d2 is a constant available in standard Control Chart Constants.Table
Rbar = 0.0061 [given at the bottom of the table above]; d2 for sub-group size 3 = 1.693 and hence sigma for individuals = 0.0061/1.693
= 0.003603 ANSWER
Q2
Sigma of sub-groups = sigma of Xbari’s = 0.002255 [using Excel Function] ANSWER
Q3
Given the specification as 0.200 ± 0.005, USL = 0.205 and LSL = 0.195.
Percentage oversize = 100 x P(X > USL) = 100 x P(X > 0.205)
= 100 x P[Z > {(0.205 – 0.198933)/0.003603}, where 0.198933 = overall average [given at the bottom of the table above], 0.003603 = sigma of individuals as obtained under Q1 and Z is the Standard Normal Variate.
So, Percentage oversize = 100 x P(Z > 1.6838) = 100 x 0.04611 or 4.611% ANSWER1
Similarly, Percentage undersize = 100 x P(X < LSL) = 100 x P(X < 0.195)
= 100 x P[Z < {(0.195 – 0.198933)/0.003603}, where 0.198933 = overall average [given at the bottom of the table above], 0.003603 = sigma of individuals as obtained under Q1 and Z is the Standard Normal Variate.
So, Percentage oversize = 100 x P(Z < - 1.0915910 = 100 x 0.137506 or 13.7506% ANSWER2
NOTE: Normal probabilities are obtained using Excel Function
Subgroup i
Average Xbari
Range Ri
1
0.198666667
0.011
2
0.199
0.007
3
0.197333333
0.005
4
0.196333333
0.004
5
0.202333333
0.005
6
0.198666667
0.004
7
0.203333333
0.006
8
0.196333333
0.008
9
0.197333333
0.005
10
0.2
0.006
Average
0.198933333
0.0061
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