Chapter 10 Solutions
1. specs: 24 oz. to 25 oz.
= 24.5 oz. [assume = =
x ]
= .2 oz.
a. [refers to population]
0124.)0062(.25.2
2.
5.
z=→=
+
=
b.
60.24 to 40.24 or 10.5.24
16
2.
25.24
n
2==
2. = 1.0 liter
= .01 liter
n = 25
a.
n
2:limits Control
[z = 2.17 for 97%]
9957.
25
01.
17.20.1 is LCL
0043.1
25
01.
17.20.1 is UCL
=−
=+
3. a.
n = 20
A2 = 0.18
=
X = 3.10 Mean Chart: =
X ± A2 −
R = 3.1 ± 0.18(0.45)
D3 = 0.41
−
R = 0.45
= 3.1 ± .081
D4 = 1.59
Hence, UCL is 3.181
and LCL is 3.019. All means are within these limits.
Range Chart: UCL is D4−
R = 1.59(0.45) = .7155
LCL is D3−
R = 0.41(0.45) = .1845
In control since all points are within these limits.
4.
Sample
Mean
Range
1
79.48
2.6
Mean Chart: =
X ± A2−
R = 79.96 ± 0.58(1.87)
2
80.14
2.3
= 79.96 ± 1.1
3
80.14
1.2
UCL = 81.04, LCL = 78.88
4
79.60
1.7
Range Chart: UCL = D4−
R = 2.11(1.87) = 3.95
5
80.02
2.0
LCL = D3−
R = 0(1.87) = 0
6
80.38
1.4
[Both charts suggest the process is in control: Neither has any
points outside the limits.]
.0062
.0062
24 24.5 25 16
-2.5 0 +2.5 z-scale
1.006
1.0043
1.002
1.000
.998
.9957
.994
UCL
LCL
out
out
(liters)
Mean
•
•
•
•
*
*
b.
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
011.
200
)975(.025.)1(
.dev Std. ==
−
=n
pp
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
.01] to[round 0099.
200
)98(.02.
dev. Std. ==
h. .02 ± 2(.01) = 0 to .04
The last sample is beyond the upper limit.
6. n = 200 Control Limits =
n
pp
p)1(
2−
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
yields 0075.
)200(12
18
p==
UCL = .0197 and LCL = 0.
7.
857.7
14
110
c==
Control limits:
409.8857.7c3c =
UCL is 16.266, LCL becomes 0.
All values are within the limits.
0096.
)200(13
25 ==p
0138.0096.
200
)9904(.0096.
20096.
=
=
Solutions (continued)
8.
5.1
14
21
c==
Control limits:
67.35.1c3c =
UCL is 5.17, LCL becomes 0.
All values are within the limits.
9.
100
)946(.054.
96.1054.
n
)p1(p
zp are limits Control
054.
)100(16
87
nsobservatio ofnumber total
defectives ofnumber total
p
=
−
===
.044.054.=
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%.
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:
cm. 06.15
1
01.
615 =+
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
85.3
39
15.150
39 === x
x
Control limits are
117.85.3
14
146.
3385.3 ==
N
x
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]
Sample
A/B
Mean
U/D
Sample
A/B
Mean
U/D
1
A
3.86
−
21
B
3.84
D
2
A
3.90
U
22
B
3.82
D
3
B
3.83
D
23
A
3.89
U
4
B
3.81
D
24
A
3.86
D
5
B
3.84
U
25
A
3.88
U
6
B
3.83
D
26
A
3.90
U
7
A
3.87
U
27
B
3.81
D
8
A
3.88
U
28
A
3.86
U
9
B
3.84
D
29
A
3.98
U
10
B
3.80
D
30
A
3.96
D
11
A
3.88
U
31
A
3.88
D
12
A
3.86
D
32
B
3.76
D
13
A
3.88
U
33
B
3.83
U
14
B
3.81
D
34
B
3.77
D
15
B
3.83
U
35
A
3.86
U
16
A
3.86
U
36
B
3.80
D
17
B
3.82
D
37
B
3.84
U
18
A
3.86
U
38
B
3.79
D
20
A
3.87
U
obs.
Conclusion
3.08