Chapter 10 Solutions
1.
specs: 24 oz. to 25 oz.
.0062
.0062
=
 = 24.5 oz. [assume  = x ]
 = .2 oz.
-2.5
0
a. [refers to population]
24
24.5
 .5
z
 2.5  2(.0062 )  .0124
.2

.2
b.   2
 24 .5  2
 24 .5  .10 or 24 .40 to 24 .60
n
16
2.
+2.5
z-scale
25
16
 = 1.0 liter
 = .01 liter
n = 25
a.
Control limits :   2

n
b.
1.0043
[z = 2.17 for 97%]
.01
UCL is 1.0  2.17
 1.0043
25
LCL is 1.0  2.17
1.006
.01
 .9957
25
(liters) 1.002
Mean
out
UCL
*


1.000
.998


LCL
.9957
*
.994
out
3. a. n = 20
A2 = 0.18
D3 = 0.41
D4 = 1.59
=
=

X = 3.10 Mean Chart: X ± A2 R = 3.1 ± 0.18(0.45)

= 3.1 ± .081
R = 0.45
Hence, UCL is 3.181
and LCL is 3.019. All means are within these limits.

Range Chart: UCL is D4R = 1.59(0.45) = .7155

LCL is D3R = 0.41(0.45) = .1845
In control since all points are within these limits.
4.
Sample
Mean
Range
1
79.48
2.6
2
80.14
2.3
3
80.14
1.2
4
79.60
1.7
5
80.02
2.0
6
80.38
1.4
=

Mean Chart: X ± A2R = 79.96 ± 0.58(1.87)
= 79.96 ± 1.1
UCL = 81.04, LCL = 78.88

Range Chart: UCL = D4R = 2.11(1.87) = 3.95

LCL = D3R = 0(1.87) = 0
[Both charts suggest the process is in control: Neither has any
points outside the limits.]
Solutions (continued)
5.
n = 200
a.
1
2
3
4
.020 .010 .025 .045
b. (2.0 + 1.0 + 2.5 + 4.5)/4 = 2.5%
c. mean = .025
Std. dev. 
p (1  p )
.025 (.975 )

 .011
n
200
d. z = 2.17
.025 ± 2.17(0.011) = .025% ± .0239% = .0011 to .0489.
e. .025 + z(.011) = .047
Solving, z = 2, leaving .0228 in each tail. Hence, alpha = 2(.0228)
= .0456.
f. Yes.
g. mean = .02
.02 (.98)
 .0099 [round to .01]
200
h. .02 ± 2(.01) = 0 to .04
The last sample is beyond the upper limit.
Std. dev. 
6.
n = 200
p
25
 .0096
13(200 )
Control Limits = p  2
p (1  p )
n
 .0096  2
.0096 (.9904 )
200
 .0096  .0138
Thus, UCL is .0234 and LCL becomes 0.
Since n = 200, the fraction represented by each data point is half
the amount shown. E.g., 1 defective = .005, 2 defectives = .01,
etc.
Sample 10 is too large. Omitting that value and recomputing
limits with
18
p
 .0075 yields
12 (200 )
UCL = .0197 and LCL = 0.
7.
c
110
 7.857
14
Control limits: c  3 c  7.857  8.409
UCL is 16.266, LCL becomes 0.
All values are within the limits.
Solutions (continued)
21
8.
c
 1.5
14
Control limits: c  3 c  1.5  3.67
UCL is 5.17, LCL becomes 0.
All values are within the limits.
9.
p
total number of defectives
87

 .054
total number of observatio ns 16 (100 )
p(1  p)
.054 (.946 )
 .054  1.96
n
100
 .054  .044 . Hence, UCL = 0.10
LCL = 0.01
Note that observations must be converted to fraction defective, or control limits must be
converted to number of defectives. In the latter case, the upper control limit would be 7.9
defectives and the lower control limit would be .1 defective. Even though all points are
within these limits, the process appears to be out of control because 75% of the values are
above 4%.
Control limits are p  z
10.
There are several slightly different ways to solve this problem. The most straightforward
seems to be the following:
1) Observe that the upper control limit is six standard deviations above the lower control
limit.
2) Compute the value of the upper control limit at the start:
15  6
.01
 15.06 cm.
1
3) Determine how many pieces can be produced before the upper control limit just
touches the upper tolerance, given that the upper limit increases by .004 cm. per
piece:
15.2cm. – 15.06cm.
= 35 pieces.
.004 cm./piece
11.
Out of the 30 observations, only one value exceeds the tolerances, or 3.3%. [This case is
essentially the one portrayed in the text in Figure 10–9A.] Thus, it seems that the
tolerances are being met: approximately 97 percent of the output will be acceptable.
12.
a.  = .146
n = 14
x
 x  150 .15  3.85
39
39

.146
 .385  3
 3.85  .117
Control limits are x  3
N
14
So UCL is 3.97, LCL is 3.73. Sample 29 is outside the UCL, so the
process is not in control.
Solutions (continued)
b. [median is 3.85]
13.
Sample
A/B
Mean
U/D
Sample
A/B
Mean
U/D
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A
A
B
B
B
B
A
A
B
B
A
A
A
B
B
A
B
A
B
A
3.86
3.90
3.83
3.81
3.84
3.83
3.87
3.88
3.84
3.80
3.88
3.86
3.88
3.81
3.83
3.86
3.82
3.86
3.84
3.87

U
D
D
U
D
U
U
D
D
U
D
U
D
U
U
D
U
D
U
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
B
B
A
A
A
A
B
A
A
A
A
B
B
B
A
B
B
B
[B]
3.84
3.82
3.89
3.86
3.88
3.90
3.81
3.86
3.98
3.96
3.88
3.76
3.83
3.77
3.86
3.80
3.84
3.79
3.85
D
D
U
D
U
U
D
U
U
D
D
D
U
D
U
D
U
D
U
Test
Median
Up/down
obs.
18
29

3.08
2.57
exp.
20.5
25.7
Z
–.81
1.28
Conclusion
random
random
a.
A A B B A B A B B A B A A B A A A B B B A B A B A B


























D D D U D U D D U D U U D U U D D D U U D U D U D
b.
A A A A B A A B B B B B A B B B A A A A B B B B B B














 


 






U D U D U D D U D U D U D U D U D U D D U D U D D
Solutions (continued)
14.

z
14
2.50
1.6
17
17
8
14
22
17
2.07
2.50
2.07
Summary:
obs.
exp.
a. median
up/down
b. median
up/down
18
Conclusion
random
0.0 random
–2.40 nonrandom
2.41 nonrandom
a. [Data from Chapter 10, Problem 8]
Median is 1.5 A = Above, B = Below, U = Up, D = Down.
Sample:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Median:
A
A
B
B
B
A
A
B
A
B
A
B
A
B
Data:
2
3
1
0
1
3
2
0
2
1
3
1
2
0
Up/down:
–
U
D
D
U
U
D
D
U
D
U
D
U
D
b. [Data from #11] Median is 7.5.
Day:
Median:
1
B
2
A
3
A
4
A
5
A
6
B
7
B
8
A
9
A
10
B
11
B
12
B
13
B
14
A
Data:
4
10
14
8
9
6
5
12
13
7
6
4
2
10
Up/dow
n
–
U
D
U
D
D
D
D
U
D
U
U
D
U
For part a and b:
N
14
E(r)med =
 1   1  8 runs
2
2
E(r)u/d =
2 N  1 2(14 )  1

 9 runs
3
3
med 
N 1
14  1

 1.803 runs
4
4
16 N  29
224  29

 1.472 runs
90
90
For part a:
10  8
10  9
 1.109
Zup/down 
 .679
Zmed =
1.803
1.472
For part b:
68
79
 1.109
Zup/down 
 1.36
Zmed =
1.803
1.472
Since the absolute values of all Z statistics calculated above are less than 2, all patterns
appear to be random.
u / d 
Solutions (continued)
Summary:
Test
a. median
up/down
b. median
up/down
Exp.
8

1.80
z
1.11
Conclusion
random
10
9
1.47
.68
random
6
8
1.80
–1.11
random
7
9
1.47
–1.36
random
obs.
10
15.
Day
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B
B
A
B
A
A
A
A
A
A
B
A
A
B
B
B
A
A
B
B
Amount
27.69
28.13
33.02
30.31
31.59
33.64
34.73
35.09
33.39
32.51
27.98
31.25
33.98
25.56
24.46
29.65
31.08
33.03
29.10
25.19
Day
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

U
U
D
U
U
U
U
D
D
D
U
U
D
D
U
U
U
D
D
Amount
B 28.60
B 20.02
B 26.67
A 36.40
A 32.07
A 44.10
A 41.44
B 29.62
B 30.12
B 26.39
A 40.54
A 36.31
B 27.14
B 30.38
A 31.96
A 32.03
A 34.40
B 25.67
A 35.80
A 32.23
U
D
U
U
D
U
D
D
U
D
U
D
D
U
U
U
U
D
U
D
Day
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Amount
B 26.76
B 30.51
B 29.35
B 24.09
B 22.45
B 25.16
B 26.11
B 29.84
A 31.75
B 29.14
A 37.78
A 34.16
A 38.28
B 29.49
B 30.81
B 30.60
A 34.46
A 35.10
A 31.76
A 34.90
D
U
D
D
D
U
U
U
U
D
U
D
U
D
U
D
U
U
D
U
Summary:
Test
median
up/down
obs.
22
35
exp.

z
Conclusion
31
3.84 –2.34 non-random
39.67 3.22 –1.45 random
Since one of the tests suggests non-randomness, the conclusion must be that the process
is not in control. In other words, the variation in daily expenses is not random. Further
investigation would be necessary in order to determine what sort of pattern is present.
Solutions (continued)
16.
3.5
Cm
UCL
n=1
 = 0.01 cm
3.44
Mean
LCL
3.0 Cm
3 sigma
3 sigma
(i) The upper control limit is 6 standard deviations above the lower control limits.
0.01
(ii) When UCL = 3.5 Cm, the LCL = 3.5 – 6
 3.5  0.06  3.44 Cm
1
(iii) Determine how many pieces can be produced before the LCL just crosses the lower
tolerance of 3 Cm.
3.44 – 3.00
0.44
440
=
=
= 440 pieces
0.001
0.001
1
17.
It is necessary to see if the process variability is within 9.96 and 10.35. Two observations
have values above the specified limits, i.e., 10% of the 20 observations fall outside the
limits. Perhaps the process mean should be set a bit lower.
18.
1 Step 10% scrap, 2nd 6%, and 3rd 6%.
a. Let x be the number of units started initially at Step 1. With a scrap rate of 10% in
Step 1 the input to Step 2 is 0.9x. The input to Step 3 is (1 – 0.06) (1 – 0.10)x. With a
scrap rate of 6% at Step 3 the number of good units after Step 3 = (1 – 0.06) (1 –
0.06) (1 – 0.10)x = (0.94)2(0.90)x.
The required output is 450 units
(0.94)2(0.90)x = 450 units
x = 565.87  566 units
b. (1 – 0.03)2(1 – 0.05)x = 450 units
(0.97)2(0.95)x = 450 units
x = 503.44  504 units
Savings of 566 – 504 = 62 units
c. From (a)
The scrap = 566 – 450 = 116 units
@ $10 per unit, The Total Cost = $1,160.00
Solutions (continued)
19.
Sample #
Median = 2.5
A/B
Up/Down
N = 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
3 2 4 5 1 2 4 1 2 1 3 4 2 4 2 1 3 1 3 4
A B A A B B A B B B A A B A B B A B A A
– D U U D U U D U D U U D U D D U D U U
Observed Expected
Median
13
11
Up/down
14
13
20.
a.
1
4.3
2
4.5
3
4.5
Standard
Deviation
2.1794
1.7981
Z
Conclude
0.9177 Random
0.5561 Random
4
4.7
b. =
x = (4.3 + 4.5 + 4.5 + 4.7)/4 = 4.5
std. dev. (of data set) = .192
21.
c. mean = 4.5, std. dev. = .192 / 5  .086
d. 4.5 ± 3(.086) = 4.5 ± .258 = 4.242 to 4.758
The risk is 2(.0013) = .0026
e. 4.5 + z(.086) = 4.86
Solving, z = 4.19, so the risk is close to zero
f. None

g. R = (.3 + .4 + .2 + .4)/4 = .325
n=5

Means: A2 = 0.58 =
x ± A2R = 4.5 ± 0.58(.325) = 4.3115 to 4.6885.
The last mean is above the upper limit.
Ranges: D3 = 0
0 to 2.11(.325) = 0 to .6875
All ranges are within the limits.
h. Two different measures of dispersion are being used, the standard deviation and the
range.
0.18
i. 4.4  3
 4.4  .241  4.16 to 4.64. The last value is above the upper limit.
5
Solution
specificat ion width
.02
.02


 1.11
a. C p 
process width
6(.003 ) .018
b. In order to be capable, the process capability ratio must be at least 1.33. In this
instance, the index is 1.11, so the process is not capable.
Solutions (continued)
22.
23.
Machine Standard Deviation (in.) Job Specification (in.) Cp
Capable ?
001
0.02
0.05
0.833
No
002
0.04
0.07
0.583
No
003
004
0.10
0.05
0.18
0.15
0.600
1.000
No
No
005
0.01
0.04
1.333
Yes
Machine Cost per unit ($) Standard Deviation (mm.) Cp
A
20
0.059
1.355
B
12
0.060
1.333
C
11
0.063
1.27
D
10
0.061
1.311
You can narrow the choice to machines A and B because they are the only ones with a
capability ratio of at least 1.33. You would need to know if the slight additional
capability of machine A is worth an extra cost of $8 per unit.
24.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation
For process H:
X  LSL 15  14 .1

 .93
3
(3)(.32 )
USL  X 16  15

 1.04
3
(3)(.32 )
C pk  min .938 , 1.04  .93
.93  1.0, not capable
For process K:
X  LSL 33  30

 1.0
3
(3)(1)
USL  X 36 .5  33

 1.17
3
(3)(1)
C pk  min{ 1.0, 1.17}  1.0
Since 1.0 < 1.33, the process is not capable.
Solutions (continued)
For process T:
X  LSL 18 .5  16 .5

 1.67
3
(3)(0.4)
USL  X 20 .1  18 .5

 1.33
3
(3)(0.4)
C pk  min{ 1.67 , 1.33}  1.33
Since 1.33 = 1.33, the process is capable.
25.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation.
USL = 90 minutes,
X 1 = 74 minutes,
LSL = 50 minutes,
 1 = 4.0 minutes
X 2 = 72 minutes,
 2 = 5.1 minutes
For the first repair firm:
X  LSL 74  50

 2.0
3
(3)(4.0)
USL  X 90  74

 1.333
3
(3)(4.0)
C pk  min{ 2.0, 1.33}  1.333
Since 1.333 = 1.333, the firm 1 is capable.
For the second repair firm:
X  LSL 72  50

 1.44
3
(3)(5.1)
USL  X 90  72

 1.18
3
(3)(5.1)
C pk  min{ 1.44,1.88}  1.18
Since 1.18 < 1.33, the firm 2 is not capable.
Solutions (continued)
26.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation.
USL = 30 minutes,
LSL = 45 minutes,
X Armand = 38 minutes,  Armand = 3 minutes
X Jerry = 37 minutes,
 Jerry = 2.5 minutes
X Melissa = 37.5 minutes,
 Melissa = 2.5 minutes
For Armand:
X  LSL 38  30

 .89
3
(3)(3)
USL  X 45  38

 .78
3
(3)(3)
C pk  min{. 89, .78}  .78
Since .78 < 1.33, Armand is not capable.
For Jerry:
X  LSL 37  30

 .93
3
(3)(2.5)
USL  X 45  37

 1.07
3
(3)(2.5)
C pk  min{. 93, 1.07}  .93
Since .93 < 1.33, Jerry is not capable.
For Melissa, since USL  X  X  LSL  7.5 , the process is centered, therefore we
will use Cp to measure process capability.
Cp 
USL  LSL 45  30

 1.39
6
(6)(1.8)
Since 1.39 > 1.33, Melissa is capable.