Sqc irfan bashir kardaash - assistant professor - copy

Statistical Quality
Control
Chapter 20
GOALS
When you have completed this chapter, you will be able to:
• Discuss the role of quality control in production and service
operations
• Define and understand the terms chance cause, assignable cause,
in control, out of control, attribute, and variable
• Construct and interpret a Pareto chart
• Construct and interpret a fishbone diagram
• Construct and interpret a mean and range chart
• Construct and interpret a percent defective and a c-bar chart
• Discuss acceptance sampling
• Construct an operating characteristic curve for various sampling
plans.
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1498
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1948
2000
W
al
ter A. Shewhart (1891–1967)
W
ith the advent of industrial revolution in the 19th
century, mass production replaced manufacturing in
small shops by skilled craftsman and artisans. While
in the small shops the individual worker was completely
responsible for the quality of the work, this was no longer
true in mass production where each individual’s contribution
to the finished product constituted only an insignificant part
in the total process. The quality control by the large compa-
nies was achieved with the help of quality inspectors res-
ponsible for checking a 100 percent inspection of all the
important characteristics.
Dr. Walter A. Shewhart, called the father of quality control
analysis, developed the concepts of statistical quality control.
For the purpose of controlling quality, Shewhart developed charting techniques and sta-
tistical procedures for controlling in-process manufacturing operations. His statistical
procedures are based on the concept of independent and identically distributed random
variables. Based on these concepts, he distinguished between chance causes, producing
random variation intrinsic to the process, and assignable causes, one should look for and
take requiring corrective actions. In order to identify the assignable causes, he recom-
mended 3␴ (standard deviation) limits above and below the mean value of the variable.
Observations beyond these limits were then used to identify the assignable causes.
Shewharts’ other contributions to statistics include his emphasis on providing full
evidence in reporting research results and applications of measurement processes in
science, particularly physics.
Shewhart served as consultant to several organizations including the War Department,
the United Nations, and the Indian government. He also taught at the universities of
Illinois and California, and was involved with various other universities such as Harvard,
Rutgers, and Princeton.
W. Edwards Deming, one of Shewharts’s students—now more popular than his mentor
in the arena of quality control, carried on Shewhart’s work on statistical quality control to
new heights. Deming’s contributions include not only a further development of procedures,
but also a new philosophy, popularly known as Deming’s 14 points in modern statistical
quality control literature.
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Statistical Quality Control 3
Introduction
Throughout this text we have presented many applications of hypothesis testing.
In Chapter 10 we described methods for testing a hypothesis regarding a single popu-
lation value. In Chapter 11 we described methods for testing a hypothesis about two
populations. In this chapter we present another, somewhat different application of
hypothesis testing, called statistical process control or SPC.
Statistical process control is a collection of strategies, techniques, and actions taken
by an organization to ensure they are producing a quality product or providing a qual-
ity service. It begins at the product planning stage, when we specify the attributes of
the product or service. It continues through the production stage. Each attribute
throughout the process contributes to the overall quality of the product. To effectively
use quality control, measurable attributes and specifications must be developed against
which the actual attributes of the product or service can be compared.
20.1 St
a
tistical Quality Control and
Edw
ards Deming’s 14 Points Program
Statistical quality control really came into its own during World War II. The need for
mass-produced war-related items, such as bomb sights, accurate radar, and other
electronic equipment, at the lowest possible cost hastened the use of statistical
sampling and quality control charts. Since World War II these statistical techniques
have been refined and sharpened. The use of computers in the last decade has also
widened the use of these techniques.
World War II virtually destroyed the Japanese production capability. Rather than retool
their old production methods, the Japanese enlisted the aid of the late Dr. W. Edwards
Deming, of the United States Department of Agriculture, to help them develop an over-
all plan. In a series of seminars with Japanese planners he stressed a philosophy that
is known today as Deming’s 14 points. These 14 points are listed on the following page.
He emphasized that quality originates from improving the process, not from inspec-
tion, and that quality is determined by the customers. The manufacturer must be able,
via market research, to anticipate the needs of customers. Upper management has
the responsibility for long-term improvement. Another of his points, and one that the
Japanese strongly endorsed, is that every member of the company must contribute to
the long-term improvement. To achieve this improvement, ongoing education and
training are necessary.
Deming had some ideas that did not mesh with contemporary management philo-
sophies in the West. Two areas where Deming’s ideas differed from Western manage-
ment philosophy were with production quotas and merit ratings. He claimed these two
practices, which are both common in the West, are not productive and should be
eliminated. He also pointed out that Western managers are mostly interested in good
news. Good news, however, does not provide an opportunity for improvement. On the
other hand, bad news opens the door for new products and allows for company
improvement.
Listed below, in a condensed form, are Dr. Deming’s 14 points. He was adamant that
the 14 points needed to be adopted as a package in order to be successful. The under-
lying theme is cooperation, teamwork, and the belief that workers want to do their jobs
in a quality fashion.

1. Create constancy of purpose for the continual improvement of products and
service to society.
2. Adopt a philosophy that we can no longer live with commonly accepted levels
of delays, mistakes, defective materials, and defective workmanship.
3. Eliminate the need for mass inspection as the way to achieve quality. Instead
achieve quality by building the product correctly in the first place.
4. End the practice of awarding business solely on the basis of price. Instead,
require meaningful measures of quality along with the price.
5. Improve constantly and forever every process for planning, production, and
service.
6. Institute modern methods of training on the job for all employees, including
managers. This will lead to better utilization of each employee.
7. Adopt and institute leadership aimed at helping people do a better job.
8. Encourage effective two-way communication and other means to drive out
fear throughout the organization so that everyone may work more effectively
and more productively for the company.
9. Break down barriers between departments and staff areas.
10. Eliminate the use of slogans, posters, and exhortations demanding zero defects
and new levels of productivity without providing methods.
11. Eliminate work standards that prescribe quotas for the workforce and numer-
ical goals for people in management. Substitute aids and helpful leadership in
order to achieve continual improvement in quality and productivity.
12. Remove the barriers that rob hourly workers and the people in management
of their right to pride of workmanship.
13. Institute a vigorous program of education and encourage self-improvement for
everyone. What an organization needs is good people and people who are
improving with education. Advancement to a competitive position will have its
roots in knowledge.
14. Define clearly management’s permanent commitment to ever-improving
quality and productivity to implement all of these principles.
Deming’s 14 points did not ignore statistical quality control, which is often abbre-
viated as SQC, TQC, or just QC. The objective of statistical quality control is to mon-
itor production through many stages of manufacturing. We use the tools of statistical
quality control, such as X-bar and R charts, to monitor the quality of many processes
and services. Control charts allow us to identify when a process or service is “out of
control,” that is, when the point is reached where an excessive number of defective
units are being produced.
Interest in quality has accelerated dramatically since the late 1980s. Turn on the
television and watch the commercials sponsored by Ford, Nissan, and GM to verify
the emphasis on quality control on the assembly line. It is now one of the “in” topics
in all facets of business. V. Daniel Hunt, president of Technology Research Corporation,
wrote in his book Quality in America that in the United States, 20 to 25 percent of
the cost of production is currently spent finding and correcting mistakes. And, he
added, the additional cost incurred in repairing or replacing faulty products in the field
drives the total cost of poor quality to nearly 30 percent. In Japan, he indicates, this cost
is about 3 percent!
In recent years companies have been motivated to improve quality by the chal-
lenge of being recognized for their quality achievements. The Malcolm Baldrige
National Quality Award, established in 1988, is awarded annually to U.S. companies
Chapter Twenty
4

that demonstrate excellence in quality achievement and management. The award
categories include manufacturing, service, and small business. Past winners include
Motorola, Xerox, IBM, Federal Express, and Cadillac. The 2002 award winners
were Motorola Inc., Commercial, Government and Industrial Solutions Sector from
Schaumberg, Illinois (manufacturing category); Branch-Smith Printing Division, Fort
Worth, Texas (small business category); and SSM Health Care, St. Louis, Missouri
(health care category). You can obtain more information on the 2002 winners and
other winners by visiting the Web site: www.quality.nist.gov.
Canada’s National Quality Institute (NQI) similarly awards organizations that
demonstrate sustainable measures of continuous improvement. For the past 19 years,
Canada Awards for Excellence have been awarded for Quality. (An new category,
“Healthy Workplace” has been awarded since 1999.) The 2002 recipients of the
Canada Awards for Excellence—Quality Award Trophy were Dana Canada Inc., Spicer
Driveshaft Group, Magog, Quebec; Canada Post, Saskatoon Operations; Homewood
Health Centre, Guelph Ontario; and Mullen Trucking, Aldersyde, Alberta. To find out
more about these and other winners, visit the Web site www.nqi.ca.
What is quality? There is no commonly agreed upon definition of quality. To cite a
few diverse definitions: From Westinghouse, “Total quality is performance leadership
in meeting the customer requirements by doing the right things right the first time.”
From AT&T, “Quality is meeting customer expectations.” Historian Barbara
W. Tuchman says, “Quality is achieving or reaching the highest standard as against
being satisfied with the sloppy or fraudulent.”
20.2 Causes of V
aria
tion
No two parts are exactly the same. There is always some variation. The weight of each
McDonald’s Quarter Pounder is not exactly 0.25 pounds. Some will weigh more than
0.25 pounds, others less. The standard time for the bus run from downtown Toronto to
airport is 45 minutes. Each run does not take exactly 45 minutes. Some runs take
longer. In some cases there is a reason for the bus being late, an accident on the high-
way or a snowstorm, for example. In other cases the driver may not “hit” the green
lights or the traffic is unusually heavy and slow for no apparent reason. There are two
general causes of variation in a process—chance and assignable.
CHANCE VARIATION Variation that is random in nature. This type of variation
cannot be completely eliminated unless there is a major change in the equip-
ment or material used in the process.
Internal machine friction, slight variations in material or process conditions
(such as the temperature of the mold being used to make glass bottles), atmospheric
conditions (such as temperature, humidity, and the dust content of the air), and
vibrations transmitted to a machine from a passing forklift are a few examples of
sources of chance variation.
If the hole drilled in a piece of steel is too large due to a dull drill, the drill may be
sharpened or a new drill inserted. An operator who continually sets up the machine
incorrectly can be replaced or retained. If the roll of steel to be used in the process does
not have the correct tensile strength, it can be rejected. These are examples of assign-
able variation.
i

Assignable Variation Variation that is not random. It can be eliminated or
reduced by investigating the problem and finding the cause.
There are several reasons we should be concerned with variation.
1. It will change the shape, dispersion, and central tendency of the distribution of the
product characteristic being measured.
2. Assignable variation is usually correctable, whereas chance variation usually cannot
be corrected or stabilized economically.
20.3 Diagnostic Charts
There are a variety of diagnostic techniques available to investigate quality problems.
Two of the more prominent of these techniques are Pareto charts and fishbone
diagrams.
PARETO CHARTS
Pareto analysis is a technique for tallying the number and type of defects that happen
within a product or service. The chart is named after a 19th-century Italian scientist,
Vilfredo Pareto. He noted that most of the “activity” in a process is caused
by relatively few of the “factors.” His concept, often called the 80–20 rule, is that
80 percent of the activity is caused by 20 percent of the factors. By concentrating on
20 percent of the factors, managers can attack 80 percent of the problem. For exam-
ple, Emily’s Family Restaurant is investigating “customer complaints.” The five com-
plaints heard most frequently are: discourteous service, cold food, long wait for seating,
few menu choices, and unruly young children. Suppose discourteous service was
mentioned most frequently and cold food second. These two factors total more than
85 percent of the complaints and hence are the two that should be addressed first
because this will yield the largest reduction in complaints.
To develop a Pareto chart, we begin by tallying the type of defects. Next, we rank
the defects in terms of frequency of occurrence from largest to smallest. Finally, we
produce a vertical bar chart, with the height of the bars corresponding to the frequency
of each defect. The following example illustrates these ideas.
Example 20-1 The Mayor of Moncton, NB, is concerned with water usage, particularly in single family
homes. He would like to develop a plan to reduce the water usage in Moncton.
To investigate, he selects a sample of 100 homes and determines the typical daily water
usage for various purposes. These sample results are as follows.
Reasons for Litres per Day Reasons for Litres per Day
Water Usage Water Usage
Laundering 24.9 Swimming pool 28.3
Watering lawn 143.7 Dishwashing 12.3
Personal bathing 106.7 Car washing 10.4
Cooking 5.1 Drinking 7.9
What is the area of greatest usage? Where should he concentrate his efforts to reduce
the water usage?
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6
i

Solution A Pareto chart is useful for identifying the major areas of water usage and focusing on
those areas where the greatest reduction can be achieved. The first step is to convert
each of the activities to a percent and then to order them from largest to smallest. The
total water usage per day is 339.3 litres, found by totaling the litres used in the eight
activities. The activity with the largest use is watering lawns. It accounts for 143.7
litres of water per day, or 42.4 percent of the amount of water used. The next largest
category is personal bathing, which accounts for 31.4 percent of the water used. These
two activities account for 73.8 percent of the water usage.
MINITAB CHART 20-1: Pareto Chart for Water Usage in Moncton, NB
MINITAB INSTRUCTIONS
1. Enter the reasons for water usage in C1 and the litres
used in C2.
2. Click Stat, Quality Tools, Pareto Chart, and Enter.
3. In the dialogue box, select Chart defects table, indicate
the location of the labels and frequencies, type a chart
title, and click OK.
2
3
4
1

Chapter Twenty
8
Methods Materials
Personnel
Equipment
Problem
or
Effects
CHART 20-2: Fishbone Diagram
Reasons for Water Usage Litres per Day Percent
Laundering 24.9 7.3
Watering lawn 143.7 42.4
Personal bathing 106.7 31.4
Cooking 5.1 1.5
Swimming pool usage 28.3 8.3
Dishwashing 12.3 3.6
Car washing 10.4 3.1
Drinking 7.9 2.3
Total 339.3 100.0
To draw the Pareto chart, we begin by scaling the number of litres used on the left
vertical axis and the corresponding percent on the right vertical axis. Next we draw a
vertical bar with the height of the bar corresponding to the activity with the largest
number of occurrences. In the Moncton example, we draw a vertical bar for the activ-
ity watering lawns to a height of 143.7 litres. (We call this the count.) We continue this
procedure for the other activities, as shown in the Minitab Chart 20-1.
Below the chart we list the activities, their frequency of occurrence, and the per-
cent of the time each activity occurs. In the last row we list the cumulative percentage.
This cumulative row will allow us to quickly determine which set of activities account
for most of the activity. These cumulative percents are plotted above the vertical bars.
In the Moncton example, the activities of watering lawn, personal bathing, and pools
account for 82.1 percent of the water usage. The city manager can attain the greatest
gain by looking to reduce the water usage in these three areas.
FISHBONE DIAGRAM
Another diagnostic chart is a cause-and-effect diagram or a fishbone diagram. It is called
a cause-and-effect diagram to emphasize the relationship between an effect and a set of
possible causes that produce the particular effect. This diagram is useful to help organize
ideas and to identify relationships. It is a tool that encourages open “brainstorming”
for ideas. By identifying these relationships we can determine factors that are the cause
of variability in our process. The name fishbone comes from the manner in which the
various causes and effects are organized on the diagram. The effect is usually a partic-
ular problem, or perhaps a goal, and it is shown on the right-hand side of the diagram.
The major causes are listed on the left-hand side of the diagram.
The usual approach to a fishbone diagram is to consider four problem areas,
namely, methods, materials, equipment, and personnel. The problem, or the effect,
is the head of the fish. See Chart 20-2.

Under each of the possible causes are subcauses that are identified and investi-
gated. The subcauses are factors that may be producing the particular effect.
Information is gathered about the problem and used to fill in the fishbone diagram.
Each of the subcauses is investigated and those that are not important eliminated, until
the real cause of the problem is identified.
Chart 20-3 illustrates the details of a fishbone diagram. Suppose a family restaurant,
such as those found along a highway, has recently been experiencing complaints from
customers that the food being served is cold. Notice each of the subcauses are listed as
assumptions. Each of these subcauses must be investigated to find the real problem
regarding the cold food. In a fishbone diagram there is no weighting of the subcauses.
Methods Materials
Personnel
Equipment
Complaints
of
Cold
Food
Food heated to
correct temperature
Food placed under
heating lights
Food at correct
starting temperature
Packaging insulates
enough
Thermostat
working properly
Heating lights at
correct height
Employees operating
equipment correctly
Servers deliver
food quickly
CHART 20-3: Fishbone Diagram for a Restaurant Investigation of
Cold Food Complaints
Source: Adapted from M.A. Vonderembse and G.P. White, Operations Management, 3rd Ed. (South Western College
Publishing, 1996), p. 489.
Self- Review 20-
1
Patients at the Rouse Home have been complaining recently about the conditions at
the home. The administrator would like to use a Pareto chart to investigate. When a
patient or patient’s relative has a complaint, they are asked to complete a complaint
form. Listed below is a summary of the complaint forms received during the last
12 months.
Complaint Number Complaint Number
Nothing to do 45 Dirty conditions 63
Poor care by staff 71 Poor quality food 84
Medication error 2 Lack of respect by staff 35
Develop a Pareto chart. What complaints would you suggest the administrator work on
first to achieve the most significant improvement?

EXERCISES 20-1 TO 20-2
20-1. Tom Sharkey is the owner of Sharkey Chevy. At the start of the year
Tom instituted a customer opinion program to find ways to improve service.
One week after the service is performed, Tom’s administrative assistant
calls the customer to find out whether the service was performed
satisfactorily and how the service might be improved. Listed below is
a summary of the complaints for the first six months. Develop a Pareto chart.
What complaints would you suggest that Tom work on to improve the quality
of service?
Complaint Frequency Complaint Frequency
Problem not corrected 38 Price too high 23
Error on invoice 8 Wait too long for service 10
Unfriendly atmosphere 12
20-2. Out of 110 diesel engines tested, a rework and repair facility found 9 had leaky
water pumps, 15 had faulty cylinders, 4 had ignition problems, 52 had oil
leaks, and 30 had cracked blocks. Draw a Pareto chart to identify the key
problem in the production process.
20.4 Purpose and Types of
Quality Control Charts
Control charts identify when assignable causes of variation or changes have entered
the process. For example, the Wheeling Company makes vinyl-coated aluminum
replacement windows for older homes. The vinyl coating must have a thickness
between certain limits. If the coating becomes too thick, it will cause the windows to
jam. On the other hand, if the coating become too thin, the window will not seal
properly. The mechanism that determines how much coating is put on each window
becomes worn and begins making the coating too thick. Thus, a change has occurred
in the process. Control charts are useful for detecting the change in process condi-
tions. It is important to know when changes have entered the process, so that the
cause may be identified and corrected before a large number of unacceptable items
are produced.
Control charts may be compared to the scoreboard in a baseball game. By looking
at the scoreboard, the fans, coaches, and players can tell which team is winning the
game. However, the scoreboard can do nothing to win or lose the game. Control charts
provide a similar function. These charts indicate to the workers, group leaders, quality
control engineers, production supervisor, and management whether the production of
the part or service is “in control” or “out of control.” If the production is “out of con-
trol,” the control chart will not fix the situation; it is just a piece of paper with figures
and dots on it. Instead, the person responsible will adjust the machine manufacturing
the part or do what is necessary to return production to “in control.”
There are two types of control charts. A variable control chart portrays meas-
urements, such as the amount of cola in a two-litre bottle or the time it takes a
nurse at General Hospital to respond to a patient’s call. A variable control chart
Chapter Twenty
10

requires the interval or the ratio scale of measurement. An attribute control chart
classifies a product or service as either acceptable or unacceptable. It is based on
the nominal scale of measurement. Patients in a hospital are asked to rate the meals
served as acceptable or unacceptable; bank loans are either repaid or they are
defaulted.
CONTROL CHARTS FOR VARIABLES
To develop control charts for variables, we rely on the sampling theory discussed in
connection with the central limit theorem in Chapter 8. Suppose a sample of five
pieces is selected each hour from the production process and the mean of each
sample computed. The sample means are and so on. The mean of these sam-
ple means is denoted as . We use k to indicate the number of sample means.
The overall or grand mean is found by:
Grand Mean 20-1
The standard error of the distribution of the sample means is designated by . It is
found by:
Standard Error of the Mean 20-2
These relationships allow limits to be set up around the sample means to show how
much variation can be expected for a given sample size. These expected limits are
called the upper control limit (UCL) and the lower control limit (LCL). An example
will illustrate the use of control limits and how the limits are determined.
Example 20-2 Statistical Software, Inc., offers a toll-free number where customers can call from
7 A.M. until 11 P.M. daily with problems involving the use of their products. It is
impossible to have every call answered immediately by a technical representative,
but it is important customers do not wait too long for a person to come on the line.
Customers become upset when they hear the message “Your call is important to us.
The next available representative will be with you shortly” too many times. To
understand their process, Statistical Software decides to develop a control chart
describing the total time from when a call is received until the representative
answers the caller’s question. Yesterday, for the 16 hours of operation, five calls were
sampled each hour. This information is reported on the next page, in minutes until
a call was answered.
S
S
n
x =
sx
X
X
= =
Σ Σ
of the means of the subgroups
Number of sample means k
k
X
X X X
1 2 3
, , ,

Sample Number
Time 1 2 3 4 5
A.M. 7 8 9 15 4 11
8 7 10 7 6 8
9 11 12 10 9 10
10 12 8 6 9 12
11 11 10 6 14 11
P.M. 12 7 7 10 4 11
1 10 7 4 10 10
2 8 11 11 7 7
3 8 11 8 14 12
4 12 9 12 17 11
5 7 7 9 17 13
6 9 9 4 4 11
7 10 12 12 12 12
8 8 11 9 6 8
9 10 13 9 4 9
10 9 11 8 5 11
Based on this information, develop a control chart for the mean duration of the call.
Does there appear to be a trend in the calling times? Is there any period in which it
appears that customers wait longer than others?
Solution A mean chart has two limits, an upper control limit (UCL) and a lower control limit
(LCL). These upper and lower control limits are computed by:
Control Limits for the Mean and 20-3
where S is an estimate of the standard deviation of the population, ␴. Notice that in the
calculation of the upper and lower control limits the number 3 appears. It represents
the 99.74 percent confidence limits. The limits are often called the 3-sigma limits.
However, other levels of confidence (such as 90 or 95 percent) can be used.
This application developed before computers were widely available and computing
standard deviations was difficult. Rather than calculate the standard deviation from
each sample as a measure of variation, it is easier to use the range. For fixed sized sam-
ples there is a constant relationship between the range and the standard deviation,
so we can use the following formulas to determine the 99.74 percent control limits for
the mean. It can be demonstrated that the term from Formula 20-3 is equiv-
alent to A2R
–
in the following formula.
Control Limits for the Mean 20-4
UCL X A R LCL X A R
= + = −
2 2
3( / )
s n
LCL X
S
n
= − 3
UCL X
S
n
= + 3
Chapter Twenty
12

where:
A2 is a constant used in computing the upper and the lower control limits. It is based
on the average range, R
–
. The factors for various sample sizes can be found in the
Table of Factors for Control Charts, Appendix 1 on this CD. (Note: n in this table
refers to the number in the sample.) A portion of this table is shown below. To
locate the A2 factor for this problem, find the sample size for n in the left margin.
It is 5. Then move horizontally to the A2 column, and read the factor. It is 0.577.
n A2 D2 D3 D4
2 1.880 1.128 0 3.267
3 1.023 1.693 0 2.575
4 0.729 2.059 0 2.282
5 0.577 2.326 0 2.115
6 0.483 2.534 0 2.004
is the mean of the sample means, computed by , where k is the number of
samples selected. In this problem a sample of 5 observations is taken each hour
for 16 hours, so k = 16.
R
–
is the mean of the ranges of the sample. It is ⌺R/k. Remember the range is the
difference between the largest and the smallest value in each sample. It describes
the variability occurring in that particular sample. (See Table 20-1.)
Σ X k
/
X
TABLE 20-1 Duration of 16 Samples of Five Help Sessions
Time 1 2 3 4 5 Mean Range
A.M. 7 8 9 15 4 11 9.4 11
8 7 10 7 6 8 7.6 4
9 11 12 10 9 10 10.4 3
10 12 8 6 9 12 9.4 6
11 11 10 6 14 11 10.4 8
P.M. 12 7 7 10 4 11 7.8 7
1 10 7 4 10 10 8.2 6
2 8 11 11 7 7 8.8 4
3 8 11 8 14 12 10.6 6
4 12 9 12 17 11 12.2 8
5 7 7 9 17 13 10.6 10
6 9 9 4 4 11 7.4 7
7 10 12 12 12 12 11.6 2
8 8 11 9 6 8 8.4 5
9 10 13 9 4 9 9.0 9
10 9 11 8 5 11 8.8 6
Total 150.60 102
The centreline for the chart is . It is 9.4125 minutes, found by 150.60/16. The
mean of the ranges (R
–
) is 6.375 minutes, found by 102/16. Thus, the upper control
limit of the X bar chart is:
The lower control limit of the X bar chart is:
LCL X A R
= − = − =
2 9 4125 0 577 6 375 5 7341
. . ( . ) .
UCL X A R
= + = + =
2 9 4125 0 577 6 375 13 0909
. . ( . ) .
X

, UCL, and LCL, and the sample means are portrayed in Chart 20-4. The mean, , is
9.4125 minutes, the upper control limit is located at 13.0909 minutes, and the lower
control limit is located at 5.7341. There is some variation in the duration of the calls,
but all sample means are within the control limits. Thus, based on 16 samples of five
calls, we conclude that 99.74 percent of the time the mean length of a sample of 5 calls
will be between 5.7341 minutes and 13.0909 minutes.
X
X
Chapter Twenty
14
7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10
6
7
8
9
10
11
12
13
A.M. P.M.
Time
Call
length
(in
minutes)
CHART 20-4: Control Chart for Mean Length of Customer Calls
to Statistical Software, Inc.
Because the statistical theory is based on the normality of large samples, control
charts should be based on a stable process, that is, a fairly large sample, taken over a
long period of time. One rule of thumb is to design the chart after at least 25 samples
have been selected.
RANGE CHART
In addition to the central tendency in a sample, we must also monitor the amount of
variation from sample to sample. A range chart shows the variation in the sample
ranges. If the points representing the ranges fall between the upper and the lower lim-
its, it is concluded that the operation is in control. According to chance, about 997
times out of 1,000 the range of the samples will fall within the limits. If the range
should fall above the limits, we conclude that an assignable cause affected the opera-
tion and an adjustment to the process is needed. Why are we not as concerned about
the lower control limit of the range? For small samples the lower limit is often zero.
Actually, for any sample of six or less, the lower control limit is 0. If the range is
zero, then logically all the parts are the same and there is not a problem with the vari-
ability of the operation.
The upper and lower control limits of the range chart are determined from the fol-
lowing equations.
Control Chart for Ranges 20-5
The values for D3 and D4, which reflect the usual three ␴ (sigma) limits for various
sample sizes, are found in the Table of Factors for Control Charts (Appendix 1 on this
CD-ROM).
UCL D R LCL D R
= =
4 3
St
a
tistics in
Action
Control charts were
used to help convict a
person who bribed jai
alai players to lose. X
–
and R charts showed
unusual betting
patterns and that
some contestants did
not win as much as
expected when they
made certain bets. A
quality control expert
was able to identify
times when assignable
variation stopped, and
prosecutors were able
to tie those times to
the arrest of the
suspect.
• • •

Example 20-3 The length of time customers of Statistical Software, Inc. waited from the time
their call was answered until a technical representative answered their question or
solved their problem is recorded in Table 20-1. Develop a control chart for the range.
Does it appear that there is any time when there is too much variation in the
operation?
Solution The first step is to find the mean of the sample ranges. The range for the five calls sam-
pled in the 7 A.M. hour is 11 minutes. The longest call selected from that hour was
15 minutes and the shortest 4 minutes; the difference in the lengths is 11 minutes.
In the 8 A.M. hour the range is 4 minutes. The total of the 16 ranges is 102 minutes,
so the average range is 6.375 minutes, found by R
–
= 102/16. Referring to the Table of
Factors for Control Charts in Appendix 1 on this CD or the partial table on page 13, D3 and
D4 are 0 and 2.115, respectively. The lower and upper control limits are 0 and 13.4831.
The range chart with the 16 sample ranges plotted is shown in Chart 20-5. This chart
shows all the ranges are well within the control limits. Hence, we conclude the variation
in the time to service the customer’s calls is within normal limits, that is, “in control.”
Of course, we should be determining the control limits based on one set of data and
then applying them to evaluate future data, not the data we already know.
UCL D R
LCL D R
= = =
= = =
4
3
2 115 6 375 13 4831
0 6 375 0
. ( . ) .
( . )
2
4
6
8
10
12
14
7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10
Range
of
calls
in
minutes
A.M. P.M.
Time
UCL
CHART 20-5: Control Chart for Ranges of Length of Customer Calls
to Statistical Software, Inc.
Minitab will draw a control chart for the mean and the range as shown below for the
Statistical Software example. The data is in Table 20-1. The minor differences in
the control limits are due to rounding.

20.5 Some In-Control and Out
-of
-Control
Situa
tions
Following are three illustrations of in-control and out-of-control processes.
1. The mean chart and the range chart together indicate that the process is in control.
Note the sample means and sample ranges are clustered close to the centrelines.
Some are above and some below the centrelines, indicating the process is quite sta-
ble. That is, there is no visible tendency for the means and ranges to move toward
the “out of control” areas.
Chapter Twenty
16
MINITAB CHART 20-6: Mean and Range Chart for Call Times at Statistical Software
1
2
3
4
1. Enter the information from Table 20-1 into your
worksheet.
2. Click on Stat, Control Charts, Xbar-R, and click OK.
3. Click on the Single Column option and enter the Variable
by name of column. Enter the Subgroup size (5).
4. Click on Options, enter the name of the chart, click OK,
and then OK again.
Everything OK

2. The sample means are in control, but the ranges of the last two samples are out
of control. This indicates there is considerable variation from piece to piece. Some
pieces are large; others are small. An adjustment in the process is probably
necessary.
Mean Chart
Range Chart
UCL
LCL
UCL
LCL
Considerable
variation in
ranges
Mean Chart
Range Chart
UCL
LCL
UCL
LCL
3. The mean was in control for the first samples, but there is an upward trend toward
UCL. The last two sample means were out of control. An adjustment in the process
is indicated.

The above chart for the mean is an example in which the control chart offers some
additional information. Note the direction of the last five observations of the mean.
They are all above and increasing, and, in fact, the last two observations are out of
control. The fact that the sample means were increasing for seven consecutive obser-
vations is very improbable and another indication that the process is out of control.
SELF-REVIEW 20-2
Every hour a quality control inspector measures the outside diameter of four parts.
The results of the measurements are shown below.
Sample Piece
Time 1 2 3 4
9 A.M. 1 4 5 2
10 A.M. 2 3 2 1
11 A.M. 1 7 3 5
(a) Compute the mean outside diameter, the mean range, and determine the control
limits for the mean and the range.
(b) Are the measurements within the control limits? Interpret the chart.
20-3. Describe the difference between assignable variation and chance variation.
20-4. Describe the difference between an attribute control chart and a variable
control chart.
X
Chapter Twenty
18
Mean out of
control UCL
LCL
UCL
LCL
Mean Chart
Range Chart

20-5. Samples of size n = 4 are selected from a production line.
(a) What is the value of the A2 factor used to determine the upper and lower
control limits for the mean?
(b) What are the values of the D3 and D4 factors used to determine the upper
and lower control limits for the range?
20-6. Samples of size 5 are selected from a manufacturing process. The mean of
the sample ranges is .50. What is the estimate of the standard deviation of the
population?
20-7. A new industrial oven has just been installed at the Piatt Bakery. To develop
experience regarding the oven temperature, an inspector reads the tempera-
ture at four different places inside the oven each half hour. The first reading,
taken at 8:00 A.M., was 340 degrees Fahrenheit. (Only the last two digits are
given in the following table to make the computations easier.)
Reading
Time 1 2 3 4
8:00 A.M. 40 50 55 39
8:30 A.M. 44 42 38 38
9:00 A.M. 41 45 47 43
9:30 A.M. 39 39 41 41
10:00 A.M. 37 42 46 41
10:30 A.M. 39 40 39 40
(a) Based on this initial experience, determine the control limits for the mean
temperature. Determine the grand mean. Plot the experience on a chart.
(b) Interpret the chart. Does there seem to be a time when the temperature
is out of control?
20-8. Refer to Exercise 20-7.
(a) Based on this initial experience, determine the control limits for the range.
Plot the experience on a chart.
(b) Does there seem to be a time when there is too much variation in
the temperature?
20.6 A
ttribute Control Charts
Often the data we collect are the result of counting rather than measuring. That is, we
observe the presence or absence of some attribute. For example, the screw top on a
bottle of shampoo either fits onto the bottle and does not leak (an “acceptable” condition)
or does not seal and a leak results (an “unacceptable” condition), or a bank makes a
loan to a customer and the loan is either repaid or it is not repaid. In other cases we
are interested in the number of defects in a sample. Air Canada might count the num-
ber of its flights arriving late per day at Dorval Airport in Montreal. In this section
we discuss two types of attribute charts: the p (percent defective) and the c (number
of defectives).

PERCENT DEFECTIVE CHART
If the item recorded is the fraction of unacceptable parts made in a larger batch of
parts, the appropriate control chart is the percent defective chart. This chart is based
on the binomial distribution, discussed in Chapter 6, and proportions, discussed in
Chapter 9. The centreline is at p, the mean proportion defective. The p replaces the
of the variable control chart. The mean proportion defective is found by:
Mean Proportion Defective 20-6
The variation in the sample proportion is described by the standard error of a proportion.
It is found by:
Standard Error of the Proportion 20-7
Hence, the upper control limit (UCL) and the lower control limit (LCL) are com-
puted as the mean percent defective plus or minus three times the standard error of
the percents (proportions). The formula for the control limits is:
Control Limits for Proportions 20-8
An example will show the details of the calculations and the conclusions.
Example 20-4 The Credit Department at Global National Bank is responsible for entering each trans-
action charged to the customer’s monthly statement. Of course, accuracy is critical
and errors will make the customer very unhappy! To guard against errors, each data
entry clerk rekeys a sample of 1500 of their batch of work a second time and a com-
puter program checks that the numbers match. The program also prints a report of the
number and size of any discrepancy. Seven people were working last hour and here are
their results:
Inspector Number Inspected Number Mismatched
Mullins 1500 4
Rider 1500 6
Gankowski 1500 6
Smith 1500 2
Reed 1500 15
White 1500 4
Reading 1500 4
Construct the percent defective chart for this process. What are the upper and the
lower control limits? Interpret the results. Does it appear any of the data entry clerks
are “out of control”?
Solution The first step is to determine the mean proportion defective p, using Formula 20-6.
It is 0.0039, found by 41/10 500.
LCL, UCL = ±
−
p
p p
n
3
1
( )
S
p p
n
p =
−
( )
1
p =
Total number defective
Total number of items sampled
X
Chapter Twenty
20

Inspector Number Inspected Number Mismatched Proportion Defective
Mullins 1500 4 0.00267
Rider 1500 6 0.00400
Gankowski 1500 6 0.00400
Smith 1500 2 0.00133
Reed 1500 15 0.01000
White 1500 4 0.00267
Reading 1500 4 0.00267
Total 10 500 41
The upper and lower control limits are computed using Formula (20-8).
From the above calculations, the upper control limit is 0.0087, found by 0.0039
+ 0.0048. The lower control limit is 0. Why? The lower limit by formula is determined
by 0.0039 - 0.0048, which is equal to - 0.0009. A negative proportion defective is not
possible, so the smallest value is 0. We set the control limit at 0. Thus, any data entry
clerk whose proportion defective is between 0 and 0.0087 is “in control.” Clerk num-
ber 5, whose name is Reed, is out of control. Her proportion defective is 0.01, or 1.0
percent, which is outside the upper control limit. Perhaps she should receive additional
training or be transferred to another position. This information is summarized in
Chart 20-7, based on Minitab software.
LCL UCL p
p p
n
,
( )
. ( . )
. .
= ±
−
= ±
−
= ±
3
1
41
10500
3
0 0039 1 0 0039
1500
0 0039 0 0
0048
MINITAB CHART 20-7: Control Chart for Proportion of Entries Defective Among Data
Entry Clerks at Global National Bank
1. Enter the data on the number of mismatches from page 20.
2. Click Stat, Control Charts, P, and Enter.
3. Select the Variable indicating the number of mismatches, click on
Subgroup size, and enter 1500.
4. In the lower right, click on Annotation, Title (type the title in the
space provided), click OK, and OK again.
1
2
3
4

c-BAR CHART
The c-bar chart plots the number of defects or failures per unit. It is based on the
Poisson distribution discussed in Chapter 6. The number of bags mishandled on a flight
by Air Canada might be monitored by a c-bar chart. The “unit” under consideration is
the flight. On most flights there are no bags mishandled. On others there may be only
one, on others two, and so on. Canada Customs and Revenue Agency might count and
develop a control chart for the number of errors in arithmetic per tax return. Most
returns will not have any errors, some returns will have a single error, others will have
two, and so on. We let c
– be the mean number of defects per unit. Thus, c
– is the mean
number of bags mishandled by Air Canada per flight or the mean number of arithmetic
errors per tax return. Recall from Chapter 6 that the standard deviation of a Poisson
distribution is the square root of the mean. Thus, we can determine the 3-sigma,
or 99.74 percent limits on a c-bar chart by:
Control Limits for the Number
of Defects per Unit 20-9
Example 20-5 The publisher of the Oak Harbour Daily Telegraph is concerned about the number of
misspelled words in the daily newspaper. They do not print a paper on Saturday or
Sunday. In an effort to control the problem and promote the need for correct spelling,
a control chart is to be instituted. The number of misspelled words found in the final
edition of the paper for the last 10 days is: 5, 6, 3, 0, 4, 5, 1, 2, 7, and 4. Determine
the appropriate control limits and interpret the chart. Were there any days during the
period that the number of misspelled words was out of control?
Solution The sum of the number of misspelled words over the 10-day period is 37. So the mean
number of defects, c
–, is 3.7. The square root of this number is 1.924. So the upper control
limit is:
The computed lower control limit would be 3.7 - 3(1.924) = -2.07. However, the
number of misspelled words cannot be less than 0, so we use 0 as the lower limit. The
lower control limit is 0 and the upper limit is 9.47. When we compare each of the data
points to the value of 9.47, we see they are all less than the upper control limit; the
number of misspelled words is “in control.” Of course, newspapers are going to strive
to eliminate all misspelled words, but control charting techniques offer a means of
tracking daily results and determining whether there has been a change. For example,
if a new proofreader was hired, her work could be compared with others. These results
are summarized in Chart 20-8, based on the Minitab software.
UCL = + = + = + =
c c
3 3 7 3 3 7 3 7 5 77 9 47
. . . . .
LCL UCL c c
, = ± 3
Chapter Twenty
22

SELF- REVIEW 20-3
The Auto-Lite Company manufactures car batteries. At the end of each shift
the Quality Assurance Department selects a sample of batteries and tests them.
The number of defective batteries found over the last 12 shifts is 2, 1, 0, 2, 1, 1, 7, 1,
1, 2, 6, and 1. Construct a control chart for the process and comment on whether
the process is in control.
20-9. A bicycle manufacturer randomly selects 10 frames each day and tests for
defects. The number of defective frames found over the last 14 days is 3, 2, 1,
3, 2, 2, 8, 2, 0, 3, 5, 2, 0, 4. Construct a control chart for this process
and comment on whether the process is “in control.”
MINITAB CHART 20-8: c-Bar Chart for Number of Misspelled Words Per Edition of
the Oak Harbour Daily Telegraph
1. Enter the data on the number of misspelled words from page 22.
2. Click on Stat, Control Chart, C Chart, and Enter.
3. Select the Variable indicating the number of misspelled words.
In the lower right, click Annotation, and choose Title. Type the
title in the space provided, click OK, and OK again.
1 2
3
4

20-10. Scott Paper tests its toilet paper by subjecting 15 rolls to a wet stress test to
see whether and how often the paper tears during the test. Following are the
number of defectives found over the last 15 days: 2, 3, 1, 2, 2, 1, 3, 2, 2, 1, 2,
2, 1, 0, and 0. Construct a control chart for the process and comment on
whether the process is “in control.”
20-11. Sam’s Supermarkets tests its checkout clerks by randomly examining the
printout receipts for scanning errors. The following numbers are the number
of errors on each receipt for October 27: 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0.
Construct a control chart for this process and comment on whether the
process is “in control.”
20-12. Dave Christi runs a car wash chain scattered throughout Manitoba. He is
concerned that some local managers are giving away free washes to their
friends. He decides to collect data on the number of “voided” sales receipts.
Of course, some of them are legitimate voids. Would the following data
indicate a reasonable number of “voids” at his facilities: 3, 8, 3, 4, 6, 5, 0, 1,
2, 4? Construct a control chart for this process and comment on whether the
process is “in control.”
20.7 Accept
ance Sampling
The previous section was concerned with maintaining the quality of the product as it
is being produced. In many business situations we are also concerned with the qual-
ity of the incoming finished product. What do the following cases have in common?
• Sims Software, Inc., purchases CDs from CD International. The normal purchase
order is for 100 000 CDs, packaged in lots of 1000. Todd Sims, president, does not
expect each CD to be perfect. In fact, he has agreed to accept lots of 1000 with up
to 10 percent defective. He would like to develop a plan to inspect incoming lots, to
ensure that the quality standard is met. The purpose of the inspection procedure is
to separate the acceptable from the unacceptable lots.
• Zenith Electric purchases magnetron tubes from Bono Electronics for use in their
new microwave oven. The tubes are shipped to Zenith in lots of 10 000. Zenith
allows the incoming lots to contain up to 5 percent defective tubes. They would like
to develop a sampling plan to determine which lots meet the criterion and which
do not.
• General Motors purchases windshields from many suppliers. GM insists that the
windshields be in lots of 1000. They are willing to accept 50 or fewer defects in each
lot, that is, 5 percent defective. They would like to develop a sampling procedure to
verify that incoming shipments meet the criterion.
The common thread in these cases is a need to verify that an incoming product
meets the stipulated requirements. The situation can be likened to a screen door,
which allows the warm summer air to enter the room while keeping the bugs out.
Acceptance sampling lets the lots of acceptable quality into the manufacturing area
and screens out lots that are not acceptable.
Of course, the situation in modern business is more complex. The buyer wants pro-
tection against accepting lots that are below the quality standard. The best protection
against inferior quality is 100 percent inspection. Unfortunately, the cost of 100 per-
cent inspection is often prohibitive. Another problem with checking each item is that
Chapter Twenty
24
St
a
tistics in
Action
It was reported during
the late 1980s that
a Canadian firm
ordered some parts
from a Japanese com-
pany with instructions
that there should be
“no more than three
defective parts per one
thousand.” When the
parts arrived, there
was a note attached
that said, “Your three
defective parts are
wrapped separately in
the upper left com-
partment of the ship-
ment.” This is a far cry
from the days when
“Made in Japan” meant
cheap.
• • •

the test may be destructive. If all light bulbs were tested until burning out before they
were shipped, there would be none left to sell. Also, 100 percent inspection may not
lead to the identification of all defects, because boredom might cause a loss of percep-
tion on the part of the inspectors. Thus, complete inspection is rarely employed in
practical situations.
The usual procedure is to screen the quality of incoming parts by using a statistical
sampling plan. According to this plan, a sample of n units is randomly selected from
the lots of N units (the population). This is called acceptance sampling. The inspection
will determine the number of defects in the sample. This number is compared with a
predetermined number called the critical number or the acceptance number. The
acceptance number is usually designated c. If the number of defects in the sample of
size n is less than or equal to c, the lot is accepted. If the number of defects exceeds c,
the lot is rejected and returned to the supplier, or perhaps submitted to 100 percent
inspection.
Acceptance sampling is a decision-making process. There are two possible deci-
sions: accept or reject the lot. In addition, there are two situations under which the
decision is made: the lot is good or the lot is bad. These are the states of nature. If
the lot is good and the sample inspection reveals the lot to be good, or if the lot is
bad and the sample inspection indicates it is bad, then a correct decision is made.
However, there are two other possibilities. The lot may actually contain more
defects than it should, but it is accepted. This is called consumer’s risk. Similarly,
the lot may be within the agreed-upon limits, but it is rejected during the sample
inspection. This is called the producer’s risk. The following summary table for
acceptance decisions shows these possibilities. Notice how this discussion is very
similar to the ideas of Type I and Type II errors presented at the beginning of
Chapter 10. (See text page 418.)
States of Nature
Decision Good Lot Bad Lot
Accept lot Correct Consumer’s risk
Reject lot Producer’s risk Correct
To evaluate a sampling plan and determine that it is fair to both the producer and
the consumer, the usual procedure is to develop an operating characteristic curve, or
an OC curve as it is usually called. An OC curve reports the percent defective along
the horizontal axis and the probability of accepting that percent defective along the
vertical axis. A smooth curve is usually drawn connecting all the possible levels of qual-
ity. The binomial distribution is used to develop the probabilities for an OC curve.
Example 20-6 Sims Software, as mentioned earlier, purchases CDs from CD International. The CDs
are packaged in lots of 1000 each. Todd Sims, president of Sims Software, has agreed
to accept lots with 10 percent or fewer defective CDs. Todd has directed his inspection
department to select a random sample of 20 CDs and examine them carefully. He will
accept the lot if it has two or fewer defectives in the sample. Develop an OC curve
for this inspection plan. What is the probability of accepting a lot that is 10 percent
defective?
OC curve
Acceptance
sampling
Acceptance
number
Consumer’s risk
Producer’s risk

Solution This type of sampling is also called attribute sampling because the sampled item, a CD
in this case, is classified as acceptable or unacceptable. No “reading” or “measurement”
is obtained on the CD. Let’s structure the problem in terms of the states of nature. Let
␲ represent the actual proportion defective in the population.
The lot is good if ␲ £ 0.10.
The lot is bad if ␲ > 0.10.
Let X be the number of defects in the sample. The decision rule is:
Reject the lot if X ³ 3.
Accept the lot if X £ 2.
Here the acceptable lot is one with 10 percent or fewer defective CDs. If the lot is
acceptable when it has exactly 10 percent defectives, it would be even more acceptable
if it contained fewer than 10 percent defectives. Hence, it is the usual practice to work
with the upper limit of the percent of defectives.
The binomial distribution is used to compute the various values on the OC curve.
Recall that for us to use the binomial there are four requirements:
1. There are only two possible outcomes. Here the CD is either acceptable or unac-
ceptable.
2. There is a fixed number of trials. In this instance the number of trials is the sample
size of 20.
3. There is a constant probability of success. A success is the probability of finding a
defective CD. It is assumed to be 0.10.
4. The trials are independent. The probability of obtaining a defective CD on the third
one selected is not related to the likelihood of finding a defect on the fourth CD
selected.
Appendix A in the text gives the various binomial probabilities. We need to convert
the acceptance sampling vocabulary to that used in Chapter 6 for discrete probability
distributions. Let ␲ = 0.10, the probability of a success, and n = 20, the number of
trials, c is the number of defects allowed—two in this case. We will now determine
the probability of accepting an incoming lot that is 10 percent defective using a sam-
ple size of 20 and allowing zero, one, or two defects. First, locate within Appendix A
the case where n = 20 and ␲ = 0.10. Find the row where X, the number of defects, is 0.
The probability is 0.122. Next find the probability of one defect, that is, where X = 1.
It is 0.270. Similarly, the probability of X = 2 is 0.285. To find the probability of two or
fewer defects, we need to add these three probabilities. The total is 0.677. Hence, the
probability of accepting a lot that is 10 percent defective is 0.677. The probability of
rejecting this lot is 0.323, found by 1 - 0.677. This result is usually written in short-
hand notation as follows (the bar, |, means “given that”):
The OC curve in Chart 20-9 shows various values of ␲ and the corresponding prob-
abilities of accepting a lot of that quality. Management of Sims Software will be able to
quickly evaluate the probabilities of various quality levels.
P X n
( | . ) .
≤ = = =
2 0 10 20 0 677
␲ and
Chapter Twenty
26
Decision rule
Attribute
sampling

Self- Review 20-4
Compute the probability of accepting a lot of CDs that is actually 30 percent defective,
using the sampling plan for Sims Software.
20-13. Determine the probability of accepting lots that are 10 percent, 20 percent,
30 percent, and 40 percent defective using a sample of size 12 and
an acceptance number of 2.
20-14. Determine the probability of accepting lots that are 10 percent, 20 percent,
30 percent, and 40 percent defective using a sample of size 14 and
an acceptance number of 3.
20-15. Warren Electric manufactures fuses for many customers. To ensure
the quality of the outgoing product, they test 10 fuses each hour. If no more
than one fuse is defective, they package the fuses and prepare them for
shipment. Develop an OC curve for this sampling plan. Compute the
probabilities of accepting lots that are 10 percent, 20 percent, 30 percent,
and 40 percent defective. Draw the OC curve for this sampling plan using
the four quality levels.
20-16. Grills Radio Products purchases transistors from Mira Electronics. According
to his sampling plan, Art Grills, owner of Grills Radio, will accept a shipment
of transistors if three or fewer are defective in a sample of 25. Develop an OC
curve for these percents defective: 10 percent, 20 percent, 30 percent,
and 40 percent.
CHART 20-9 OC Curve for Sampling Plan (n = 20, c = 2)
0 5 10 15 20 25 30
0.20
0.40
0.60
0.80
1.00
Probability
of
accepting
lot
Incoming lot percent defective
Probability of
accepting a lot
that is 10%
defective
is 0.677
Lot percent Probability of
defective accepting lot
0 1.000
5 0.924
10 0.677
20 0.207
30 0.036

Chapter Outline
I. The objective of statistical quality control is to control the quality of the product
or service as it is being developed.
II. A Pareto chart is a technique for tallying the number and type of defects that
happen within a product or service.
A. This chart was named after an Italian scientist, Vilfredo Pareto.
B. The concept of the chart is that 80 percent of the activity is caused by
20 percent of the factors.
III. A fishbone diagram emphasizes the relationship between a possible problem
cause that will produce the particular effect.
A. It is also called a cause-and-effect diagram.
B. The usual approach is to consider four problem areas: methods, materials,
equipment, and personnel.
IV. The purpose of a control chart is to monitor graphically the quality of a product
or service.
A. There are two types of control charts.
1. A variable control chart is the result of a measurement.
2. An attribute chart shows whether the product or service is acceptable or
not acceptable.
B. There are two sources of variation in the quality of a product or service.
1. Chance variation is random in nature and cannot be controlled or
eliminated.
2. Assignable variation is not due to random causes and can be eliminated.
C. Four control charts were considered in this chapter.
1. A mean chart shows the mean of a variable, and a range chart shows the
range of the variable.
(a) The upper and lower control limits are set at plus or minus 3 standard
errors from the mean.
(b) The formulas for the upper and lower control limits for the mean are:
20-4
(c) The formulas for the upper and lower control limits for the range are:
20-5
2. A percent defective chart is an attribute chart that shows the proportion
of the product or service that does not conform to the standard.
(a) The mean percent defective is found by
20-6
(b) The control limits for the proportion defective are determined from
the equation
20-8
LCL UCL p
p p
n
,
( )
= ±
−
3
1
p =
Total number defective
Total number of items sampled
UCL D R LCL D R
= =
4 3
UCL X A R LCL X A R
= + = −
2 2
Chapter Twenty
28

3. A c-bar chart refers to the number of defects per unit.
(a) It is based on the Poisson distribution.
(b) The mean number of defects per unit is c
–.
(c) The control limits are determined from the following equation.
20-9
V. Acceptance sampling is a method to determine whether an incoming lot of
a product meets specified standards.
A. It is based on random sampling techniques.
B. A random sample of n units is selected from a population of N units.
C. c is the maximum number of defective units that may be found in the sample
of n and the lot still considered acceptable.
D. An OC (operating characteristic) curve is developed using the binomial
probability distribution to determine the probability of accepting lots of
various quality levels.
Chapter Exercises 20-
17 to 20-36
20-17. The production supervisor at Westburg Electric, Inc., noted an increase in the
number of electric motors rejected at the time of final inspection. Of the last
200 motors rejected, 80 of the defects were due to poor wiring, 60 contained
a short in the coil, 50 involved a defective plug, and 10 involved other
defects. Develop a Pareto chart to show the major problem areas.
20-18. The manufacturer of athletic shoes conducted a study on their newly
developed jogging shoe. Listed below are the type and frequency of the
nonconformities and failures found. Develop a Pareto chart to show the
major problem areas.
Type of Nonconformity Frequency Type of Nonconformity Frequency
Sole separation 34 Lace breakage 14
Heel separation 98 Eyelet failure 10
Sole penetration 62 Other 16
20-19. Wendy’s fills their soft drinks with an automatic machine that operates based
on the weight of the soft drink. When the process is in control, the machine
fills each cup so that the grand mean is 10.0 ounces and the mean range is
0.25 for samples of 5.
(a) Determine the upper and lower control limits for the process for both the
mean and the range.
(b) The manager of a store tested five soft drinks served last hour and found
that the mean was 10.16 ounces and the range was 0.35 ounces.
Is the process in control? Should other action be taken?
LCL UCL c c
, = ± 3

20-20. A new machine has just been installed to cut and rough-shape large slugs.
The slugs are then transferred to a precision grinder. One of the critical meas-
urements is the outside diameter. The quality control inspector randomly
selected five slugs each hour, measured the outside diameter, and recorded
the results. The measurements (in millimetres) for the period 8:00 A.M. to
10:30 A.M. follow.
Outside Diameter (millimeters)
Time 1 2 3 4 5
8:00 87.1 87.3 87.9 87.0 87.0
8:30 86.9 88.5 87.6 87.5 87.4
9:00 87.5 88.4 86.9 87.6 88.2
9:30 86.0 88.0 87.2 87.6 87.1
10:00 87.1 87.1 87.1 87.1 87.1
10:30 88.0 86.2 87.4 87.3 87.8
(a) Determine the control limits for the mean and the range.
(b) Plot the control limits for the mean outside diameter and the range.
(c) Are there any points on the mean or the range chart that are out of
control? Comment on the chart.
20-21. The Long Last Tire Company, as part of its inspection process, tests its tires
for tread wear under simulated road conditions. Twenty samples of three tires
each were selected from different shifts over the last month of operation.
The tread wear is reported below in hundredths of an inch.
Sample Tread Wear Sample Tread Wear
1 44 41 19 11 11 33 34
2 39 31 21 12 51 34 39
3 38 16 25 13 30 16 30
4 20 33 26 14 22 21 35
5 34 33 36 15 11 28 38
6 28 23 39 16 49 25 36
7 40 15 34 17 20 31 33
8 36 36 34 18 26 18 36
9 32 29 30 19 26 47 26
10 29 38 34 20 34 29 32
(a) Determine the control limits for the mean and the range.
(b) Plot the control limits for the mean outside diameter and the range.
(c) Are there any points on the mean or the range chart that are
“out of control”? Comment on the chart.
20-22. The National Bank has a staff of loan officers located in its branch offices
throughout Quebec. The vice president in charge of the loan officers would
like some information on the typical amount of loans and the range in the
amount of the loans. A staff analyst of the vice president selected a sample
Chapter Twenty
30

of 10 loan officers and from each officer selected a sample of five loans he
or she made last month. The data are reported below. Develop a control chart
for the mean and the range. Do any of the officers appear to be
“out of control”? Comment on your findings.
Loan Amount ($000) Loan Amount ($000)
Officer 1 2 3 4 5 Officer 1 2 3 4 5
Weinraub 59 74 53 48 65 Bowyer 66 80 54 68 52
Visser 42 51 70 47 67 Kuhlman 74 43 45 65 49
Moore 52 42 53 87 85 Ludwig 75 53 68 50 31
Brunner 36 70 62 44 79 Longnecker 42 65 70 41 52
Wolf 34 59 39 78 61 Simonetti 43 38 10 19 47
20-23. The producer of a candy bar, called the “A Rod” Bar, reports on the package
that the calorie content is 420 per 50-gram bar. A sample of 5 bars from each
of the last 10 days is sent for a chemical analysis of the calorie content.
The results are shown below. Does it appear that there are any days where
the calorie count is out of control? Develop an appropriate control chart and
analyze your findings.
Calorie Count Calorie Count
Sample 1 2 3 4 5 Sample 1 2 3 4 5
1 426 406 418 431 432 6 427 417 408 418 422
2 421 422 415 412 411 7 422 417 426 435 426
3 425 420 406 409 414 8 419 417 412 415 417
4 424 419 402 400 417 9 417 432 417 416 422
5 421 408 423 410 421 10 420 422 421 415 422
20-24. The Early Morning Delivery Service guarantees delivery of small packages by
10:30 A.M. Of course, some of the packages are not delivered by 10:30 A.M.
For a sample of 200 packages delivered each of the last 15 working days, the
following number of packages were delivered after the deadline: 9, 14, 2, 13,
9, 5, 9, 3, 4, 3, 4, 3, 3, 8, and 4.
(a) Determine the mean proportion of packages delivered after 10:30 A.M.
(b) Determine the control limits for the proportion of packages delivered
after 10:30 A.M. Were any of the sampled days out of control?
(c) If 10 packages out of 200 in the sample were delivered after 10:30 A.M.
today, is this sample within the control limits?
20-25. An automatic machine produces 5.0 mm bolts at a high rate of speed.
A quality control program has been started to control the number of
defectives. The quality control inspector selects 50 bolts at random and
determines how many are defective. The number of defectives in the first
10 samples is 3, 5, 0, 4, 1, 2, 6, 5, 7, and 7.
(a) Design a percent defective chart. Insert the mean percent defective, UCL,
and LCL.
(b) Plot the percent defective for the first 10 samples on the chart.
(c) Interpret the chart.

20-26. Steele Breakfast Foods, Inc. produces a popular brand of raisin bran cereal.
The package indicates it contains 25.0 ounces of cereal. To ensure the prod-
uct quality, the Steele Inspection Department makes hourly checks on the
production process. As a part of the hourly check, 4 boxes are selected and
their contents weighed. The results are reported below.
Sample Weights
1 26.1 24.4 25.6 25.2
2 25.2 25.9 25.1 24.8
3 25.6 24.5 25.7 25.1
4 25.5 26.8 25.1 25.0
5 25.2 25.2 26.3 25.7
6 26.6 24.1 25.5 24.0
7 27.6 26.0 24.9 25.3
8 24.5 23.1 23.9 24.7
9 24.1 25.0 23.5 24.9
10 25.8 25.7 24.3 27.3
11 22.5 23.0 23.7 24.0
12 24.5 24.8 23.2 24.2
13 24.4 24.5 25.9 25.5
14 23.1 23.3 24.4 24.7
15 24.6 25.1 24.0 25.3
16 24.4 24.4 22.8 23.4
17 25.1 24.1 23.9 26.2
18 24.5 24.5 26.0 26.2
19 25.3 27.5 24.3 25.5
20 24.6 25.3 25.5 24.3
21 24.9 24.4 25.4 24.8
22 25.7 24.6 26.8 26.9
23 24.8 24.3 25.0 27.2
24 25.4 25.9 26.6 24.8
25 26.2 23.5 23.7 25.0
Develop an appropriate control chart. What are the limits? Is the process out
of control at any time?
20-27. An investor believes there is a 50-50 chance that a stock will increase in price on
a particular day. To investigate this idea, for 30 consecutive trading days the
investor selects a random sample of 50 stocks and counts the number whose
prices increased. The number of stocks in the sample whose prices increased is
reported below.
14 12 13 17 10 18 10 13 13 14
13 10 12 11 9 13 14 11 12 11
15 13 10 16 10 11 12 15 13 10
Develop a percent defective chart and write a brief report summarizing your
findings. Based on these sample results, is it reasonable that the odds are
50-50 that a stock’s price will increase? What percent of the stocks would
need to increase in a day for the process to be “out of control”?
Chapter Twenty
32

20-28. Lahey Motors specializes in selling cars to buyers with a poor credit history.
Listed below is the number of cars that were repossessed from Lahey customers
because they did not meet the payment obligations over the last 36 months.
6 5 8 20 11 10 9 3 9 9
15 12 4 11 9 9 6 18 6 8
9 7 13 7 11 8 11 13 6 14
13 5 5 8 10 11
Develop a c-bar chart for the number repossessed. Were there any months
when the number was out of control? Write a brief report summarizing your
findings.
20-29. A process engineer is considering two sampling plans. In the first a sample
of 10 will be selected and the lot accepted if 3 or fewer are found defective.
In the second the sample size is 20 and the acceptance number is 5. Develop
an OC curve for each. Compare the probability of acceptance for lots that are
5, 10, 20, and 30 percent defective. Which of the plans would you recom-
mend if you were the supplier?
20-30. The Inter Global Moving and Storage Company is setting up a control chart to
monitor the proportion of residential moves that result in written complaints
due to late delivery, lost items, or damaged items. A sample of 50 moves is
selected for each of the last 12 months. The number of written complaints in
each sample is 8, 7, 4, 8, 2, 7, 11, 6, 7, 6, 8, and 12.
(a) Design a percent defective chart. Insert the mean percent defective, UCL,
and LCL.
(b) Plot the proportion of written complaints in the last 12 months.
(c) Interpret the chart. Does it appear that the number of complaints is out
of control for any of the months?
20-31. Eric’s Cookie House sells chocolate chip cookies in shopping malls.
Of concern is the number of chocolate chips in each cookie. Eric, the owner
and president, would like to establish a control chart for the number of
chocolate chips per cookie. He selects a sample of 15 cookies from today’s
production and counts the number of chocolate chips in each. The results
are as follows: 6, 8, 20, 12, 20, 19, 11, 23, 12, 14, 15, 16, 12, 13, and 12.
(a) Determine the centreline and the control limits.
(b) Develop a control chart and plot the number of chocolate chips per cookie.
(c) Interpret the chart. Does it appear that the number of chocolate chips is
out of control in any of the cookies sampled?
20-32. The number of “near misses” recorded for the last 20 months at the Lima
International Airport is 3, 2, 3, 2, 2, 3, 5, 1, 2, 2, 4, 4, 2, 6, 3, 5, 2, 5, 1,
and 3. Develop an appropriate control chart. Determine the mean number of
misses per month and the limits on the number of misses per month.
Are there any months where the number of near misses is out of control?
20-33. The following number of robberies were reported during the last 10 days to
the Robbery Division of the Metro City Police: 10, 8, 8, 7, 8, 5, 8, 5, 4, and 7.
Develop an appropriate control chart. Determine the mean number of
robberies reported per day and determine the control limits. Are there any
days when the number of robberies reported is out of control?

Chapter Twenty
34
20-34. Seiko purchases watch stems for their watches in lots of 10 000. Seiko’s
sampling plan calls for checking 20 stems, and if 3 or fewer stems are
defective, the lot is accepted.
(a) Based on their sampling plan, what is the probability that a lot of
40 percent defective will be accepted?
(b) Design an OC curve for incoming lots that have zero, 10 percent,
20 percent, 30 percent, and 40 percent defective stems.
20-35. Automatic Screen Door Manufacturing Company purchases door latches from
a number of vendors. The purchasing department is responsible for inspect-
ing the incoming latches. Automatic purchases 10 000 door latches per month
and inspects 20 latches selected at random. Develop an OC curve for the
sampling plan if three latches can be defective and the incoming lot is still
accepted.
20-36. At the beginning of each football season Team Sports, the local sporting goods
store, purchases 5000 footballs. A sample of 25 balls is selected, and they are
inflated, tested, and then deflated. If more than two balls are found defective,
the lot of 5000 is returned to the manufacturer. Develop an OC curve for this
sampling plan.
(a) What are the probabilities of accepting lots that are 10 percent, 20 percent,
and 30 percent defective?
(b) Estimate the probability of accepting a lot that is 15 percent defective.
(c) John Brennen, owner of Team Sports, would like the probability
of accepting a lot that is 5 percent defective to be more than 90 percent.
Does this appear to be the case with this sampling plan?

20-1.
Seventy-three percent of the complaints
involve poor food, poor care, or dirty conditions.
These are the factors the administrator should
address.
20-2. (a)
Sample Piece
1 2 3 4 Total Average Range
1 4 5 2 12 3 4
2 3 2 1 8 2 2
1 7 3 5 16 4 6
9 12
(b) Yes. Both the mean chart and the range chart
indicate that the process is in control.
20-3.
20-4. The shift with 7 defects is out of control.
P X n
( | . ) .
≤ = = =
2 0 30 20 0 036
␲ and
UCL
LCL
= + =
= − =
2 083 3 2 083 6 413
2 083 3 2 083 0
. . .
. .
c = =
25
12
2 083
.
9 9:30 10
0
4
9.128 UCL
LCL
Time
Range chart
LCL D R
UCL D R
= = =
= = =
3
4
0 4 0
2 282 4 9 128
( )
. ( ) .
9 9:30 10
0.084
3
5.916 UCL
LCL
Time
Bar Chart
X
UCL LCL
= =
5 916 0 084
. .
UCL LCL X A R
and = ±
= ±
2
3 0 729 4
. ( )
X R
= = = =
9
3
3
12
3
4
50
100
150
200
250
300 100
75
50
25
Percent
Count
Poor
food
Poor
care
Dirty
Nothing
to
do
No
respect
Med.
error
CHAPTER 20 Answers to self-review
Count 84 71 83 45 35 2
Percent 28 24 21 15 12 0
Cum. % 28 52 73 88 100 100

Answers to Chapter 20 Exercises
36
Answers
to Chapter 20 Exercises
20-1.
Count 38 23 12 10 8
Percent 42 25 13 11 9
Cum % 42 67 80 91 100
About 67% of the complaints concern the prob-
lem of not being corrected and the price being
too high.
20-2. Defect Frequency Percent
Oil leak 52 47
Cracked block 30 27
Faulty cylinder 15 14
Leaky radiator 9 8
Ignition problem 4 4
110
Count 52 30 15 9 4
Percent 47 27 14 8 4
Cum % 47 74 88 96 100
Almost 75% of the complaints relate to oil leaks
and cracked blocks.
20-3. Chance variation is random in nature and
because the cause is a variety of factors it can-
not be entirely eliminated. Assignable variation
is not random, it is usually due to a specific
cause and can be eliminated.
20-4. Variable control charts are concerned with
actual measurements (weight, diameter, etc.).
Attribute control charts are based on a
classification of acceptable or not acceptable
(door lock either works or it does not).
20-5. a. The A2 factor is 0.729.
b. The value for D3 is 0, and for D4 is 2.282.
20-6. 0.215, found by [(0.577)(0.50)( )]/3.
20-7. a.
Time X
–
R
Arithmetic range
means
8:00 A.M. 46.0 16
8:30 A.M. 40.5 6
9:00 A.M. 44.0 6
9:30 A.M. 40.0 2
10:00 A.M. 41.5 9
10:30 A.M. 39.5 1
251.5 40
UCL = 41.92 + 0.729(6.67) = 46.78
LCL = 41.92 - 0.729(6.67) = 37.06
b. Interpreting, the mean reading was 341.92
degrees Fahrenheit. If the oven continues oper-
ating as evidenced by the first six hourly read-
ings, about 99.7% of the mean readings will lie
between 337.06 degrees and 346.78 degrees
Fahrenheit.
20-8. a. UCL = D4 R
-
= 2.282(6.67) = 15.22
LCL = D3 R
-
= 0
R = =
40
6
6 67
.
X = =
251 5
6
41 92
.
.
8:00 8:30 9:00 9:30 10:00 10:30
38
40
42
44
46
48
X
L
C
U
L
C
L
5
0
0.1
0.2
0.3
0.4
0.5
0
20
40
60
Oil leak Faulty
cylinder
Ignition
problem
Cracked
block
Leaky
radiator
0
0.1
0.2
0.3
0.4
0.5
0
10
20
30
40
Problem
not
corrected
Unfriendly Invoice
error
Long
wait
Price
high

Answers to Chapter 20 Exercises 37
b. Oven is operating within bounds except for
the first range (8 A.M.), which is outside the
control limits.
20-9.
The control limits are from 0 to 0.68. The
process is out of control on the seventh day.
20-10.
The control limits are from 0 to 5.39 tears.
Each sample roll is within these limits.
20-11.
The control limits are from 0 to 2.760, so there
are no receipts out of control.
20-12.
The control limits are from 0 to 9.29. There is
no case where the number of voids is greater
than 9, so the process is in control.
20-13. Percent Probability of
10 0.889
20 0.558
30 0.253
40 0.083
20-14. Percent Probability of
10 0.956
20 0.698
30 0.355
40 0.125
20-15. P(X £ 1 | n = 10, p = 0.10) = 0.736
P(X £ 1 | n = 10, p = 0.20) = 0.375
P(X £ 1 | n = 10, p = 0.30) = 0.149
P(X £ 1 | n = 10, p = 0.40) = 0.046
20-16. P(X £ 3 | n = 25, p = 0.10) = 0.763
P(X £ 3 | n = 25, p = 0.20) = 0.235
P(X £ 3 | n = 25, p = 0.30) = 0.032
P(X £ 3 | n = 25, p = 0.40) = 0.002
20-17. Defect Percent Number of
defects
Poor wiring 40.0 80
Short in coil 30.0 60
Defective plug 25.0 50
Other 5.0 10
100.0 200
0 10 20 30 40
Incoming Lot Percent Defective
0
0.2
0.4
0.6
0.8
1
Probability
of
Accepting
0
0.2
0.4
0.6
0.8
1
Probability
of
Acceptance
0 0.10 0.20 0.30 0.40
Percent Defective
3 6 3 3 6 3 6 5 69
. . . .
± = ±
c = =
36
10
3 6
.
0 545 3 0 545 0 545 2 215
. . . .
± = ±
c = =
6
11
0 545
.
c c
± = ± = ±
3 1 6 3 1 6 1 6 3 79
. . . .
c = =
24
15
1 6
.
0 26 3
0 26 0 74
10
0 26 0 42
.
. ( . )
. .
± = ±
p = =
37
140
0 26
.
8:00 8:30 9:00 9:30 10:00 10:30
0
5
10
15
20
2
2
5
1
7
6
6

38
Count 80 60 50 10
Percent 40 30 25 5
Cum % 40 70 95 100
20-18.
Count 98 62 34 16 14 10
Percent 41.9 26.5 14.5 6.8 6.0 4.3
Cum % 41.9 68.4 82.9 89.7 95.7 100.0
A total of 68.4% of the defects relate to heel and
sole separation.
20-19. a. UCL = 10.0 + 0.577(0.25)
= 10.0 + 0.14425 = 10.14425
LCL = 10.0 - 0.577(0.25)
= 10.0 - 0.14425 = 9.85575
UCL = 2.115(0.25) = 0.52875
LCL = 0(0.25) = 0
b. The mean is 10.16, which is above the upper
control limit and is out of control. There is
too much cola in the soft drinks, an adjust-
ment is needed. The process is in control for
variation.
20-20. a. UCL = 88.11 LCL = 86.61
UCL = 2.75 LCL = 0
b.
c.
20-21. a.
UCL = 30.5665 + (1.023)(15.6) = 46.53
LCL = 30.5665 - (1.023)(15.6) = 14.61
UCL = 2.575(15.6) = 40.17
b.
c. The points all seem to be within the control
limits. No adjustments are necessary.
0 10 20
10
20
30
40
50
Means
Subgroup
UCL = 46.53
LCL = 14.61
X= 30.57
0
10
20
30
40
Ranges
UCL = 40.16
R= 15.60
LCL = 0.00
R = =
312
20
15 6
.
X = =
611 23
20
30 5665
.
.
8:00 8:30 9:00 9:30 10:00 10:30
0
0.5
1
1.5
2
2.5
U
C
L
L
C
L
X
10:30
10:00
9:30
9:00
8:30
8:00
87
87.5
88
U
C
L
L
C
L
X
R = 1 3
.
X = 87 36
.
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
100
120
Heel
separation
Sole
penetration
Sole
separation
Other
Lace
breakage
Eyelet
failure
0
20
40
60
80
100
0
0.1
0.2
0.3
0.4
0.5
Poor
wiring
Short
coil
Defective
plug
Other

20-22.
UCL = 55.14 + (0.577)(35.5) = 75.62
LCL = 55.14 - (0.577)(35.5) = 34.66
UCL = 2.115(35.5) = 75.08
The last bank officer, Simonetti, is below the
lower control limit for the group.
20-23.
UCL = 418.3 + (0.577)(16.2) = 427.65
LCL = 418.3 - (0.577)(16.2) = 408.95
UCL = 2.115(16.2) = 34.26
All points are in control for both the mean and
the range.
20-24. a.
b.
UCL = 0.031 + 0.037 = 0.068
LCL = 0.031 - 0.037 = 0
c. Ten packages mean p
– = 0.05, which is well
within the limits.
20-25. a.
UCL = 0.08 + 0.115 = 0.195
LCL = 0.08 - 0.115 = 0
b.
c. There are no points that exceed the limits.
20-26.
The sample 11 mean is 23.3, which is below the
lower control limit of 23.62. The process is “out
of control” at that point.
20-27.
These sample results indicate that the odds
are much less than 50-50 for an increase. The
percent of stocks that increase is “in control”
around 0.25 or 25%. The control limits are
0.06629 and 0.4337.
20-28.
The fourth month (20) was “out of control”
above the upper control limit of 18.62. Other-
wise, the rest of the months appear “in control”
around a mean of 9.417.
20-29. P(X £ 3 | n = 10, p = 0.05) = 0.999
P(X £ 3 | n = 10, p = 0.10) = 0.987
P(X £ 3 | n = 10, p = 0.20) = 0.878
C C
Chart for 1
0 10 20 30 40
Sample Number
0
10
20
Sample
Count
UCL=18.62
C=9.417
LCL=0.2107
1
0.0
0.1
0.2
0.3
0.4
0.5
Proportion
0 10 20 30
Sample Number
P C
Chart for 1
UCL=0.4337
P=0.25
LCL=0.06629
1
23.0
24.0
25.0
26.0
0 5 10 15 20 25
0
2
4
Subgroup
Sample
Mean
Sample
Range
UCL = 26.33
Mean = 24.97
LCL = 23.62
UCL = 4.243
LCL = 0
R = 1.86
Bar / Chart Weights
X R
1 2 3 4 5 6 7 8 9 10
0
0.05
0.1
0.15
0.2
Percent
Defective
Samples
3
0 08 1 0 08
50
0 115
. ( . )
.
−
=
p = =
40
10 50
0 08
( )
.
3
0 031 1 0 031
200
0 037
. ( . )
.
−
=
p = =
93
15 200
0 031
( )
.
R = =
162
10
16 2
.
X = =
4183
10
418 3
.
R = =
355
10
35 5
.
X = =
551 4
10
55 14
.
.

40
P(X £ 3 | n = 10, p = 0.30) = 0.649
P(X £ 5 | n = 20, p = 0.05) = 0.999
P(X £ 5 | n = 20, p = 0.10) = 0.989
P(X £ 5 | n = 20, p = 0.20) = 0.805
P(X £ 5 | n = 20, p = 0.30) = 0.417
The solid line represents the operating charac-
teristic curve for the first plan; the dashed line
represents the second plan. The supplier would
prefer the first because the probability of
acceptance is higher (above). However, if he is
really sure of his quality, the second plan seems
slightly higher at the very low range of defect
percentages and might be preferred.
20-30. a.
UCL = 0.14 + 0.15 = 0.29
LCL = 0.14 - 0.15 = 0
b.
c. None of the moves are out of control in
terms of the number of complaints.
20-31. a.
UCL = 14.2 + 11.3 = 25.5
LCL = 14.2 - 11.3 = 2.9
b.
c. All the points are in control.
20-32.
20-33.
20-34. a. 0.015, found by 0.012 + 0.003
b. P(c £ 3/p = 0.10, n = 20)
= 0.122 + 0.270 + 0.285 + 0.190 = 0.867
P(c £ 3/p = 0.20, n = 20)
= 0.012 + 0.058 + 0.137 + 0.205 = 0.412
1 2 3 4 5 6 7 8 9 10
0
2
4
6
8
10
12
14
16
Robberies
Day
LCL = − = − =
7 0 3 0 7 7 00 7 9 0
. . . .
UCL = + = + =
7 0 3 0 7 7 00 7 9 14 9
. . . . .
c = =
70
10
7 0
.
1 3 5 7 9 11 13 15 17 19
Months
0
2
4
6
8
Near
Misses
LCL = − = − =
3 0 3 0 3 3 00 5 20 0
. . . .
UCL = + = + =
3 0 3 0 3 3 00 5 20 8 20
. . . . .
c = =
60
20
3 0
.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Sample
0
4
8
12
16
20
24
Chocolate
Chips
3 14 2 11 30
. .
=
c = =
213
15
14 2
.
1 2 3 4 5 6 7 8 9 10 11 12
Moves
0
0.05
0.1
0.15
0.2
0.25
Percent
Defective
3
0 1433 1 0 1433
50
0 15
. ( . )
.
−
=
p = =
86
12 50
0 1433
( )
.
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.1 0.2 0.3
C1
Plan
A

P(c £ 3/p = 0.30, n = 20)
= 0.001 + 0.007 + 0.028 + 0.072 = 0.108
P(c £ 3/p = 0.40, n = 20)
= 0.003 + 0.12 + 0.015
20-35. P(X £ 3 | n = 20, p = 0.10) = 0.867
P(X £ 3 | n = 20, p = 0.20) = 0.412
P(X £ 3 | n = 20, p = 0.30) = 0.108
20-36. a. P(X £ 2 | n = 25, p = 0.10) = 0.537
P(X £ 2 | n = 25, p = 0.20) = 0.099
P(X £ 2 | n = 25, p = 0.30) = 0.008
b. P(X £ 2 | n = 25, p = 0.15)
= 0.0172 + 0.0759 + 0.1607 = 0.2538
c. P(X £ 2 | n = 25, p = 0.05)
= 0.2774 + 0.3650 + 0.2305 = 0.8729
The probability (0.8729) is somewhat less
than desired (0.90). (Note: A more extensive
table, or some computer system such as
Minitab, is required for part b.)
0 5 10 15 20 30
0
0.2
0.4
0.6
0.8
1
1.2
Percent Defective
Probability
0 5 10 15 20 30
0
0.2
0.4
0.6
0.8
1
1.2
Probability
Percent Defective
0 10 20 30 40
Incoming Lot Percent Defective
0
0.2
0.4
0.6
0.8
1
1.2
Probability
of
Accepting
Lot

Sqc irfan bashir kardaash - assistant professor - copy

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