Numerical problems on Quality
Control
Date: 01-04-2022
Summary of formulas
Source: Stevenson
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?
Solution1:

25  2  0.06  2.06 inches

25  2  0.06  1.94 inches
UCLX  X  Z X  2  3 0.1
LCLX  X  Z X  2  3 0.1


Problem 2:
Several samples of size n = 8 have
been taken from the 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.
Sol. 2:
X  3 yds
R  0.015 yds
A2 = 0.373 from Table 10.3
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:
The average range of a process is 10
lbs. The sample size is 10. Use Table
10.3 to develop upper and lower
control limits on the range.
Sol. 3:
From Table 10.3, 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:
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.
Sol. 4:
Total number of mistakes
220
P

 0.11
Total number of files
(100)(20)
P 
(011
. )(1  011
. )
 0.03
100
UCLP  P  z P  0.11  (2)(0.03)  0.17
LCLP  P  Z  P  0.11  (2)(0.03)  0.05
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
Tuesday
Wednesday
Thursday
Friday
Saturday
2
1
5
3
4
0
Sol. 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.
Determine which of the processes are capable.
Proce
ss
Mea
n
Standard
deviation
Lower
specific
ation
Higher
specific
ation
1
8
0.20
7
9
2
10
0.25
9
12
3
7.5
0.30
6
8
4
9
0.30
8
10
Proce
ss
Mean
1
2
3
4
8
10
7.5
9
Standar
d
deviati
on
0.20
0.25
0.30
0.30
Lower
specifica
tion
7
9
6
8
Me
an
che
ck
8
10.5
7
9
Higher
specifica
tion
9
12
8
10
Process mean is centred between USL & LSL for processes 1 & 4 but
not for processes 2 &3.
Process 1 : Cp =9-76(0.20) = 1.67;
Process 2: Cpk =Smaller of 12-103(0.25) = 2.67 & 10-93(0.25) = 1.33
Process 3: Cpk =Smaller of
3(0.30) = 0.67 &
Process 4 : Cp =
6(0.30) = 1.11
So, processes 1 & 2 only are capable as
in order to be capable, Cp & Cpk must be at least 1.33
3(0.30) = 1.67