Statistical Quality Control
Montgomery’s 6th edition
Solutions for Chapter 06
Jan Rohlén
jan.rohlen@hb.se
Question 6.04
Sample No.
Ri
Xi
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
9
7
5
7
6
2
8
6
5
6
8
7
7
6
9
5
4
8
6
4
10
7.75
7.5
9
9.75
10.75
10.75
6.5
9
13.5
12.5
9.75
13.25
10.5
11
12.5
9.75
10.75
8.75
13.25
Table 1: Table 6E.4
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Jan Rohlén
Statistical Quality Control
Chapter 06
(a)
P
P
Ri
R=
= 6.25
m
Sample Size
4
X=
A2
0.729
D3
0
Xi
= 10.325
m
D4
2.282




U CL = X + A2 R = 10.325 + 0.729 × 6.25 = 14.88
X−R Chart  CL = X = 10.325


LCL = X − A2 R = 10.325 − 0.729 × 6.25 = 5.77




U CL = D4 × R = 2.282 × 6.25 = 14.26
R Chart  CL = R = 6.25


LCL = D4 × R = 0 × 6.25 = 0
The process is in-control.
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Statistical Quality Control
Chapter 06
(b)
Specs. (350V ± 5V )
σ=
The real σ can be calculated:
c =
C
p
R
d2
=
6.25
2.059
= 3.0355
U SL − LSL
3550 − 3450
=
= 5.491
6×σ
6 × 3.0355
The minimum capability index for existing processes is 1.33 (i.e., 43 ). Obviously, this process is higher (5.49 1.33).
(c)
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Statistical Quality Control
Chapter 06
Question 6.15
(a)
X=
1000
72
= 20 s =
= 1.44
50
50
Sample Size
4
X −S
A3
1.628
B3
0
B4
2.266




U CL = X + A3 R = 20 + 1.628 × 1.44 = 22.34
Chart  CL = X = 10.325


LCL = X − A3 R = 205 − 1.628 × 1.44 = 17.66
S



U CL = B4 s = 2.266 × 1.44 = 3.26
Chart CL = s = 1.44

 LCL = B s = 0 × 1.44 = 0
3
(b)
Natural Tolerance Limits
First of all, we need to calculate the real σ: σb =
(
s
C4
=
1.44
0.9213
= 1.563
U N T L = X + 3σ = 20 + 3 × 1.563 = 24.69
LN T L = X − 3σ = 20 − 3 × 1.563 = 15.31
(c)
Specs Limits: 19 ± 4
U SL − LSL
23 − 15
=
= 0.85
6σ
6 × 1.563
=⇒ the process is not capable!
c =
C
p
Cp < 1.33
(d)
Pbrework = P (S > U SL) = 1 − P (X ≤ U SL) = 1 − P (
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U SL − µ
)=
σb
Jan Rohlén
Statistical Quality Control
= 1 − φ(
Chapter 06
23 − 20
) = 1 − φ(1.919) = 0.0275 = 2.75%
1.563
15 − 20
) = φ(−3.199)
1.563
= 1 − 0.99931 = 0.00069 ∼
= 0.069%
Total: Pb = 2.75% + 0.065 ∼
= 2.8%.
Pbscrap = P (X < LSL) = φ(
(e)
µ = 19
PbRework = 1−φ(
23 − 19
) = 1−φ(2.56) = 1−0.9947 = 0.0053 = 0.53%
1.563
PbScrap = 1 − φ(
15 − 19
) = φ(−2.563) = 0.0053 = 53%
1.563
Total: Pb = 0.53% + 0.53% = 1.06%
Centring decreased the Rework but increased the Scrap. Cost analysis
should be done.
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Statistical Quality Control
Chapter 06
Question 6.21
Spec. Limits 100 ± 10, X = 104, R = 9.3 and n=5
d2 of (n=5) = 2.236
R
9.3
=
= 3.998
d2
2.326
Natural Tolerance is: 6 × σ = 6 × 3.998 = 23.99
The Tolerance limit is wider than specification limit (2 × 10 = 20).
Therefore, even adjusting the centre of the process (i.e., change the mean
to 100) wouldn’t meet the specification:
σ=
Cbp =
U SL − LSL
110 − 90
=
= 0.83
6σ
6 × 2.998
Question 6.34
X ∼ N (µx , σx )
Y ∼ N (µy , σy )
)
Independent
each sample n=5.
(a)
We have mx = 20 Samples and my = 10
20
X
10
X
Rxi = 18.608
i=1
Ryi = 6.978
i=1
Estimated σx and σy ?
P
Rxi
18.608
=
= 0.9304
20
20
P
Ryi
6.978
Ry =
=
= 0.6978
20
10
0.9304
0.6978
= 0.4
σby =
= 0.3
σbx =
2.326
2.326
Rx =
(b)
Axis must fit into the hole. Condition: P (X − Y < 0.09) = 0.006 What
difference µ0 = µx − µy should be specified?
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Statistical Quality Control
Chapter 06
D = X − y (if x and y are normally distributed then D would be
normally distributed too).
D ∼ N (µD , σD )
q
√
σb = σbx2 + σby2 = 0.42 + 0.32 = 0.5
D − µ0
0.09 − µ0
P (D < 0.09) = P
<
σ0
σ0
0.09 − µ0
φ
σ0
!
!
= 0.0006
0.09 − µ0
φ(−x) = 1 − φ(x) = φ −
σ0
!
= 1 − 0.0006 = 0.994
By checking the table, the answer is 2.51.
−
0.09 − µ0
= 2.51 =⇒ µ0 = 0.09 + 2.51σD = 1.345
σ0
Question 6.41
Type II Error =⇒ alarm by at least 3rd plot point =⇒
1 − P {no alram by 3rd sample}
|
σx =
{z
}
β3
√σ
n
!
U CL − µ
LCL − µ
φ
−φ
σx
σx
!
104 − 98
96 − 98
√
√
φ
−φ
8/ 5
8/ 5
!
!
= 0.66578 = β
1 − β 3 = 0.7049
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Statistical Quality Control
Question 6.42
ARL1 =
1
1
=
= 2.992
1−β
1 − 0.66578
Question 6.45
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Weight
16.11
16.08
16.12
16.1
16.1
16.11
16.12
16.09
16.12
16.1
16.09
16.07
16.13
16.12
16.1
16.08
16.13
16.15
16.12
16.1
16.08
16.07
16.11
16.13
16.1
P24
MR =
MRi
0.03
0.04
0.02
0
0.01
0.01
0.03
0.03
0.02
0.01
0.02
0.06
0.01
0.02
0.02
0.05
0.02
0.03
0.02
0.02
0.01
0.04
0.02
0.03
M Ri
= 0.02375
24
i=1
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Chapter 06
Jan Rohlén
Statistical Quality Control
d2 = 1.128,
σb =
Chapter 06
MR
0.02375
=
= 0.021055
d2
1.128



U CL = D4 M R = 3.267 × 0.02375 = 0.07759
CL = M R = 0.02375


LCL = D3 M R = 0
LSL − µ
φ
σx
!
16 − 16.1052
=φ
0.021055
!
∼
= 3 × 10−7 ∼
= 3 × 10−5 %




U CL = X + 3 Md2R = 16.1052 + 3 × 0.02375 = 16.1683
CL = X = 16.1052


 LCL = X − 3 M R = 16.1052 + 3 × 0.02375 = 16.04204
d2
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Statistical Quality Control
Chapter 06
Yes, it is normally distributed.
Question 6.49
x chart { U CL = 710, CL = 700 and LCL = 690
y chart { U CL = 18.08, CL = 0 and LCL = 7.979
(a)
µ = x = 700,
σbx =
s
7.979
=
= 8.661
C4
0.9213
(b)
Specs 705 ± 15
!
U SL − µ LSL − µ
=
Pb = P {x < LSL}+p{x > U SL} = φ
+ 1−φ
σbx
σbx
690 − 700 =φ
8.661
720 − 700
+ 1−φ
8.661
φ(−1.15) + 1 − φ(2.31) = 0.1339 ' 13.3%
(c)
!
U CL − µ LCL − µ
P = P {X < LCL}+P {X > U CL} = φ
+ 1−φ
σ
σ
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!
Jan Rohlén
Statistical Quality Control
!
690 − 700
710 − 700
√
√
=φ
+1−φ
8.661/ 4
8.661/ 4
Chapter 06
!
= 0.0208
(d)
Pb = P {X − LCL} + P {X > U CL}
LCL − µ
φ
690 − 693
!
σnew
new
U CL − µnew
+1−φ
σnew
710 − 693
q
√
φ
+1−φ
12/ 4
12 + (4)
!
!
= φ(−0.51)+1−φ(2.83) = 0.3108
(e)
ARL1 =
1
1
1
1
=
=
=
1−β
1 − P {notdetecting}
P {detect}
0.3108
In average; after 3 points, we can expect alarm.
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