1. A. A quality inspector collected five (5) samples each with four (4) observations of the length of
time glue to dry. The analyst computed the mean of each sample and then computed the grand
mean. All values are in minutes. Use this information to obtain three-­‐sigma (i.e. z=3) control
limits for means of future times. It is known from the previous experience that the standard
deviation of the process is 0.02 minutes.
Sample
Observation 1 2 3 4 5
1 12.11 12.15 12.09 12.12 12.09
2 12.10 12.12 12.09 12.10 12.14
3 12.11 12.10 12.11 12.18 12.13
4 12.08 12.11 12.15 12.10 12.12
Mean 12.1 12.12 12.11 12.1 12.12
a. What is the appropriate statistical process chart to use? X-­‐Chart
b. What is the overall mean?
𝑥 =
#$.#&#$.#$&#$.##&#$.#&#$.#$
'
= 12.11
c. What is the upper control limit?
𝑈𝐶𝐿 = 𝑥 + 𝑧
0
1
𝑈𝐶𝐿 = 12.11 + 3
3.3$
4
= 12.14
d. What is the lower control limit?
𝐿𝐶𝐿 = 𝑥 − 𝑧
0
1
𝐿𝐶𝐿 = 12.11 − 3
3.3$
4
= 12.08
e. Is the Process in control? YES
B. A manufacturer of precision machine parts produce round shafts for use in the construction of
drill presses. The average diameter of a shaft is 0.56 inch. Inspection sample contain 6 shafts
each. The average range of these samples is 0.006 inch. Determine the upper and lower control
limit. Construct a mean chart.
Sample Size
n
Mean Factor
A2
Upper Range
D4
Lower Range
D3
2 1.880 3.268 0
3 1.023 2.574 0
4 0.729 2.282 0
5 0.577 2.115 0
6 0.483 2.004 0
a. What is the overall mean?
𝑥 = 0.56
b. What is the upper control limit?
𝑈𝐶𝐿 = 𝑥 + 𝐴$ 𝑅 𝑈𝐶𝐿 = 0.56 + 0.483 ∗ 0.006 = 0.562898
c. What is the lower control limit?
𝐿𝐶𝐿 = 𝑥 − 𝐴$ 𝑅 𝐿𝐶𝐿 = 0.56 − 0.483 ∗ 0.006 = 0.557102
d. Is the process in control? YES

2. C. Processing new accounts at a bank is intended to average 10 minutes each. Five samples of four
(4) observations each have been taken. Use the sample data in conjunction with table 10.2 to
construct upper and lower control limits for both mean chart and a range chart.
Sample
Observation 1 2 3 4 5
1 10.20 10.3 9.7 9.9 9.8
2 9.90 9.8 9.9 10.30 10.2
3 9.80 9.90 9.9 10.1 10.3
4 10.10 10.4 10.1 10.50 9.7
TOTAL 40.00 40.40 39.60 40.80 40.00
mean 10 10.1 9.9 10.2 10
range 0.4 0.6 0.4 0.6 0.6
a. What is the overall mean?
𝑥 =
10 + 10.1 + 9.9 + 10.2 + 10
5
= 10.04
b. What is the upper control limit of mean chart?
𝑈𝐶𝐿 = 𝑥 + 𝐴$ 𝑅 𝑈𝐶𝐿 = 10.04 + 0.729 ∗ 0.52 = 10.42
c. What is the lower control limit of mean chart?
𝐿𝐶𝐿 = 𝑥 − 𝐴$ 𝑅 𝐿𝐶𝐿 = 10.04 − 0.729 ∗ 0.52 = 9.66
d. Is the process in control referring to the mean chart? YES
e. What is the average range?
𝑅 =
0.4 + 0.6 + 0.4 + 0.6 + 0.6
5
= 0.52
f. What is the upper control limit of range chart?
𝑈𝐶𝐿 = 𝐷4 𝑅 𝑈𝐶𝐿 = 2.282 ∗ 0.52 = 1.18664
g. What is the lower control limit of range chart?
𝐿𝐶𝐿 = 𝐷A 𝑅 𝐿𝐶𝐿 = 0 ∗ 0.52 = 0
h. Is the process in control referring to the range chart? YES
D. Altman Distributors, Inc. fills catalog orders have been taken each day over the past six weeks.
The average defective units was 0.05 or 5%. Determin the upper and lower control limits for
this process for 99.73% confidence.
a. What is the appropriate statistical process chart to use? P CHART
b. What is the upper control limit?
𝑈𝐶𝐿 = 0.05 + 3
0.05 1 − 0.05
𝑛
c. What is the lower control limit?
𝐿𝐶𝐿 = 0.05 − 3
0.05 1 − 0.05
𝑛
Comment [Office1]: Sample size was not given in the
problem.

3. E. Eighteen rolls of coiled wire have been examined, and number of defects per roll has been
recorded in the table below.
Sample No. of Defects Sample No. of Defects
1 3 10 1
2 2 11 2
3 4 12 3
4 5 13 2
5 1 14 4
6 2 15 2
7 4 16 1
8 1 17 3
9 2 18 1
a. What is the appropriate statistical process chart to use? C-­‐CHART
b. What is the average defect per roll?
𝑐 =
3 + 2 + 4 + 5 + 1 + 2 + 4 + 1 + 2 + 1 + 2 + 3 + 2 + 4 + 2 + 1 + 3 + 1
18
= 2.3889
c. What is the upper control limit?
𝑈𝐶𝐿 = 𝑐 + 3 𝑐 𝑈𝐶𝐿 = 2.3889 + 3 2.3889 = 7.0257
d. What is the lower control limit?
𝐿𝐶𝐿 = 𝑐 − 3 𝑐 𝐿𝐶𝐿 = 2.3889 − 3 2.3889 = −2.24 𝑜𝑟 0
e. Is the process in control? YES
F. A process has a mean of 9.20 grams an a standard deviation of 0.30 gram. The lower
specification limit is 7.50 grams and the upper specification limit is 10.50 grams. Is the process
capable?
a. Compute the ratio for the lower specification.
𝐿𝐶𝐿 =
FGHIH
A3
𝐿𝑆𝐿 =
K.$3GL.'3
A3
= 0.05667
b. Compute the ratio for the upper specification.
𝑈𝑆𝐿 =
MIHGF
A3
𝑈𝑆𝐿 =
#3.'3GK.$3
A3
= 0.04333
c. What is Cpk?
𝐶𝑝𝑘 = 𝑚𝑖𝑛
FGHIH
A3
,
MIHGF
A3
𝐶𝑝𝑘 = 𝑚𝑖𝑛 0.05667 , 0.04333 = 0.04333
d. Is the process capable? YES

4. FORMULAS AND REMINDERS
VARIABLES:
𝑋 − 𝐶𝐻𝐴𝑅𝑇: 𝑈𝐶𝐿 𝐿𝐶𝐿 = 𝑥 ± 𝑧𝜎F 𝑜𝑟 𝑈𝐶𝐿 𝐿𝐶𝐿 = 𝑥 ± 𝑧
𝜎
𝑛
𝑋 − 𝐶𝐻𝐴𝑅𝑇 𝑤/𝑜 𝜎: 𝑈𝐶𝐿 𝐿𝐶𝐿 = 𝑥 ± 𝐴$ 𝑅
𝑅 − 𝐶𝐻𝐴𝑅𝑇: 𝑈𝐶𝐿 = 𝐷4 𝑅 𝐿𝐶𝐿 = 𝐷A 𝑅
ATTRIBUTES:
𝑃 − 𝐶𝐻𝐴𝑅𝑇: 𝑈𝐶𝐿 𝐿𝐶𝐿 = 𝑝 ± 3
𝑝 1 − 𝑝
𝑛
𝐶 − 𝐶𝐻𝐴𝑅𝑇: 𝑈𝐶𝐿 𝐿𝐶𝐿 = 𝑐 ± 3 𝑐
þ Using a c-­‐Chart:
þ Observations are attributes whose defects per unit of output can be counted
þ The number counted is a small part of the possible occurrences
þ Defects such as number of blemishes on a desk, number of typos in a page of text, flaws in a
bolt of cloth
þ Using the p-­‐chart:
þ Observations are attributes that can be categorized in two states
þ We deal with fraction, proportion, or percent defectives
þ Have several samples, each with many observations
SPECIFICATIONS:
𝐶𝑝𝑘: 𝐶𝑝𝑘 = 𝑚𝑖𝑛
𝑥 − 𝐿𝑆𝐿
30
,
𝑈𝑆𝐿 − 𝑥
30
𝐶𝑝: 𝐶𝑝 =
𝑈𝑆𝐿 − 𝐿𝑆𝐿
60
Process capability is a measure of the relationship between the natural variation of the
process and the design specifications
þ A capable process must have a Cp of at least 1.0
þ Does not look at how well the process is centered in the specification range
þ Often a target value of Cp = 1.33 is used to allow for off-­‐center processes
þ Six Sigma quality requires a Cp = 2.0