This document contains 5 practice problems related to statistical process control. It provides the problems, calculations for determining control limits, and answers. Problem 1 calculates control limits for a process with a known average and standard deviation. Problem 2 calculates control limits for a sampling process. Problem 3 finds control limits using a table and known average range. Problem 4 calculates control limits for a binomial process. Problem 5 determines if a process is in control using a moving range chart. The document demonstrates how to apply statistical process control methods to different problem types.

1. Practice Problems: Supplement 6 (S6), Statistical Process Control
Problem 1:
Twenty-five engine mounts are sampled each day and found to have an average width of 2 inches,
with a standard deviation of 0.1 inches. What are the control limits that include 99.73% of the
sample means ( Z = 3) ?
Problem 2:
Several samples of size n = 8 have been taken from today’s production of fence posts. The average
post was 3 yards in length and the average sample range was 0.015 yard. Find the 99.73% upper
and lower control limits.
Problem 3:
The average range of a process is 10 lbs. The sample size is 10. Use Table S6.1 to develop upper
and lower control limits on the range.
Problem 4:
Based on samples of 20 IRS auditors, each handling 100 files, we find that the total number of
mistakes in handling files is 220. Find the 95.45% upper and lower control limits.
Problem 5:
There have been complaints that the sports page of the Dubuque Register has lots of typos. The last
6 days have been examined carefully and the number of typos/page is recorded below. Is the
process in control using Z = 2?
Day
Number of Typos
Monday
2
Tuesday
1
Wednesday
5
Thursday
3
Friday
4
Saturday
0
1

2. 2

3. ANSWERS:
Problem 1:
(
= 2 − 3 ( 0.1
UCLX = X + Z σ X = 2 + 3 0.1
LCLX = X − Zσ X
)
25 ) = 2 − 0.06 = 1.94 inches
25 = 2 + 0.06 = 2.06 inches
Problem 2:
X = 3 yds
R = 0.015 yds
A2 = 0.373 from Table S6.1
UCL = X + A2 R = 3 + 0.373(0.015) = 3.006 yds
LCL = X − A2 R = 3 − 0.373(0.015) = 2.994 yds
Problem 3:
From Table S6.1, D4 = 1.78 and D3 = 0.22
UCLR = D4 R = (1.78)(10) = 17.8 lbs
LCLR = D3 R = (0.22)(10) = 2.2 lbs
Problem 4:
P=
Total number of mistakes
220
=
= 0.11
Total number of files
(100)(20)
σP =
(0.11)(1 − 011)
.
= 0.03
100
UCLP = P + zσ P = 0.11 + (2)(0.03) = 0.17
LCLP = P + Zσ P = 0.11 − (2)(0.03) = 0.05
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4. Problem 5:
C = 15 6 = 2.5
UCLC = C + 2 2.5 = 2.5 + 2(1.58) = 5.66
LCLC = C − 2 2.5 = 2.5 − 2(1.58) = −0.66 (or 0)
As none of the observations are outside the control limits, the process is in statistical control.
However, it might be a worthwhile effort to investigate why no typos were made on Saturday. We
would like to better understand this effort so it can be repeated on other days.
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