1. Statistical process control (SPC) methods are useful for mass production but not for job shop production where each product is unique. Quality in job shops can be ensured by controlling input quality, process conditions, and output quality checks. 2. The quality control manager should be involved in planning quality, not just policing it. However, the quality control function needs separation from production to allow an independent quality assessment. 3. Specification limits define the acceptable performance range for customers, while control limits are set narrower by the producer to monitor the process and ensure it stays within specifications. Process capability affects whether control limits can keep the process within specifications.

1. CHAPTER 9: QUALITY MANAGEMENT – I
RESPONSES TO QUESTIONS:
1. Managing variations is an important aspect of Quality Control. Statistics
is a study of variations. Therefore, statistics can be used in quality
control to manage variations.
However, since statistics involves a large population, these methods
are useful largely in mass production, mass procurement, mass
shipping and any other mass scale operations. When the numbers
involved are very small, statistics cannot be applied.
2. In a Job-shop situation, since each order is unique, statistical process
control method cannot be used. However, quality (i.e. customer’s
requirement) can be ensured by:
a. Controlling the quality of the input (raw materials, components).
b. Ensuring that the process conditions are observed as per design
and planned manufacturing engineering.
c. Checking the quality of the output by non-destructive testing
methods
However, SPC methods can be used in controlling the quality of the
input materials (refer (a) above).
Note that design and manufacturing engineering are vital in controlling
quality in Job-shop situations. One plans for quality and implements the
planned process. Planning quality is as vital as processing.
3. Quality Control manager’s role is not just that of policing quality. He
should be involved in conceptualizing and planning quality.
However many organizations believe in someone in the organization
giving the final verdict on quality of the products andor services. Quality
Control manager has to do that. For this reason, quality control function
needs to be separate from manufacturingoperations.
In cases where productionoperations takes the full responsibility for the
quality of goods going out of the production system, quality control can
be an internal function – internal to productionoperations. Just-in-Time
manufacturing is one such situation where there is no time for
correction.
4. Specification limits are for the productservice performance and are
specified by the customer or by the producing company in the interest

2. 2
of the customer. Any product outside this band of limits should not be
used by the customer.
Control limits are for the production process designed by
the producing company. These limits are there to exercise control over
the processes before the product quality goes beyond the specification
limits. The limits of control are, therefore, narrower than the
specification limits.
5. The assumption is that the process is capable.
When a process is not capable (i.e. capability is not adequate), the
control limits would tend to fall outside the specification limits. In such a
case, just statistical process control will not be of use; the process itself
has to be made less variable.
6. The previous questions have already answered this question. SPC may
not be quite applicable to any production process that produces a
different product in a very small quantity each time it is produced.
7. Sampling assumes some ‘room for error’, however small the error may
be. In crucial space/atomic missions such room may be non-existent or
extremely small, thus drastically reducing the chances to use statistics.
8. Narrower band of Control Limits ≡ More frequent interference in the process.
More interference leads to - More interruptions and delays
in the process - More ‘Finger maintenance’ problems.
3 sigma limits ≡ 3 interruptions per 1000
2 sigma limits ≡ about 5 interruptions per 100
1 sigma limits ≡ about 32 interruptions per 100
When the process mean and variance shift frequently, re-setting may
have to be resorted to more frequently, thus necessitating narrower
limits of ± 2 sigma.
9. p-chart use: In a hospitality industry to control quality of service to the
customers. For example: Are certain service components being
delivered or not? What is the per cent of the time the service is
‘incomplete’ or ‘defective’?
c-chart use: In public health administration e.g. take samples and find
defects in the administration of vaccine.
10. Under the condition that the defects (i.e. non-conformance) are
randomly occurring chance events with low degree of probability, but

3. 3
the number of opportunities for occurrence of non-conformity is very
large.
11. Quality Planning involves:
• Setting quality objectives and targets
• Setting relative importance of quality characteristics.
This would require a larger plan of:
• What the product / service will be?
• Who will be the target customers?
The above is a part of Corporate Planning.
12.The old equipment’s process capability seems poor. Two actions are
possible. One may either:
a. Carry on by frequently resetting the machine, if that helps in
producing acceptable quality product; or
b. Replace the old equipment by new equipment.
The choice depends upon comparative cost considerations:
Quality costs (repair, rework, vs. Cost of investment
lost customers, lost goodwill)
13.The three different categories of the defects are first given different
weightages and then the weighted number of defects is used for
constructing the control chart for defects. The weightages given are: A: 10,
B: 5 and C: 1.
The table for the number of defects can be reconstructed as follows:
Aircraft Number Number of Defects
1 (5X1) = 5
2 (1X5) + (4X1) = 9
3 (1X5) = 5
4 (1X10) = 10
5 (2X5) + (1X1) = 11
6 (3X1) = 3
7 (3X1) = 3
8 (9X1) = 9
9 (1X5) + (6X1) = 11
10 (1X1) = 1
11 (1X10) + (3X1) = 13
12 (1X1) = 1

4. 4
13 (2X5) = 10
14 (4X5) + (2X1) = 22
15 = 0
16 (1X5) + (4X1) = 9
17 (1X5) + (6X1) = 11
18 (1X10) + (5X1) + (3X1) = 18
19 (4X1) = 4
20 (2X1) = 2
Total number of defects = 157
Average number of defects per aircraft, c = 157 = 7.85
20
Standard deviation = √ c= √ 7.85 = 2.80
The control limits for the ‘Number of Defects’ chart or c-chart are:
UCLc = c + 3 √ c = 7.85 + 8.40 = 16.25
LCLc =c - 3 √‾c = 7.85 - 8.40 = 0
(Note: Negative values are assigned zero).
Now, we check as to whether any of the values (i.e. number of defects in the
data which has been used to derive these limits) from the earlier table exceed the
derived control limits. We find that the number of defects for the aircrafts number
14 and 18 exceed the Upper Control Limit. Therefore, these two observations
have to be deleted from our analysis, after which, we have:
c = Total no. of defects = 117 = 6.50
no. of observations 18
√‾c = √‾6.50 = 2.55
UCLc = (6.50) + (3) (2.55) = 14.15. LCLc = (6.50) – (3) (2.55) = 0
(Note: Negative values are assigned zero)
Checks performed by us reveal that none of the given data (already screened
once) crosses the control limits. Thus, the last arrived control limits are the stable
control limits and the control chart looks as shown in the figure given below.
14.15 ------------------------------------------ UCLc
6.50 ------------------------------------------ c
0.00 ------------------------------------------- LCLc
Control Chart for Defects (c Chart)

5. 5
14.(a) We compute ‘p’ (fraction defective) for each sample, because the
sample size is varying.
Sample # p Sample # p
1 3/25 = 0.120 9 3/30=0.100
2 5/50 = 0.100 10 2/25=0.080
3 1/45 = 0.022 11 5/55=0.091
4 2/55=0.036 12 4/40=0.100
5 0/35=0.000 13 3/50=0.060
6 1/40=0.025 14 2/25=0.080
7 9/50=0.180 15 2/40 = 0.050
8 2/65=0.031
p = ∑ defective/ ∑n = 44/630 = 0.0698
(b)&(c): Since the sample size is varying, we compute the UCL and LCL for
each of the samples. UCL & LCL, therefore, may not be constant (i.e. the
same).
Sample # 1: n=25 σ = [p (1 -p) / n ]0.5
= [(0.0698) (0.9302)/25]0.5
= 0.05096
UCL = p + 3σ = 0.0698 + 3 (0.05096)
= 0.2227
LCL = p - 3σ = 0.0698 - 3 (0.05096)
= negative = 0
Sample #2: n= 50 σ = [(0.0698) (0.9302)/50] 0.5
= 0.0360
UCL = p + 3σ = 0.0698 + 3 (0.0360) = 0.1779
LCL = p - 3σ = 0.0698 - 3 (0.0360) = negative = 0
Similarly, we can calculate for all samples.
(d) Following table expresses the deviations from the central value (p = 0.0698)
in terms of ’ z ’ number of standard deviations
Sample # p p -p σ z = (p -p)/σ
1 0.120 0.0578 0.05096 1.1342
2 0.100 0.0302 0.03604 0.8398
3 0.022 -0.0478 0.03798 -1.2585

6. 6
4 0.036 -0.0338 0.03436 -0.9837
5 0.000 -0.0698 0.04307 -1.6206
6 0.025 -0.0448 0.04029 1.1119
7 0.180 0.1102 0.03604 3.0600
8 0.031 -0.0388 0.03161 -1.2275
9 0.100 0.0302 0.04652 0.6491
10 0.080 0.0102 0.05096 0.2002
11 0.091 0.0212 0.03436 0.6170
12 0.100 0.0302 0.04029 0.7496
13 0.060 -0.0098 0.03604 -0.2719
14 0.080 0.0102 0.05096 0.2002
15 0.050 -0.0198 0.04029 -0.4914
The control chart is as follows:
UCL ---------------------------------------------------------- p + 3σ
----------------------------------------------------------p
LCL -----------------------------------------------------------p - 3σ
We observe that only sample # 7 has overshot the UCL slightly. All other
values are within the control limits.
15. From the given data, we have:
µ = (∑x ) /n = 497/20 = 24.85 = 25 approximately.
Which means, µ = 0.3025
R = 102/20 = 5.1 = 5 approximately.
Which means, R =0.0005.
Referring to the Appendix (page 9.18 and 9.19):
A2 = 0.577 (since n = 5, given)
Therefore, UCL and LCL are as follows.
UCL = µ + A2R = 0.3025 + (0.557) (0.0005)
= 0.3028
LCL = µ - A2R =0.3025 – (0.577) (0.0005)
= 0.3022
Thus, the x chart is:
µ = 0.3025, UCL & LCL =0.3028 & 0.3022

7. 7
R-chart values from Appendix (pages 9.18 and 9.19) are:
D4 = 2.115 and D3 = 0
Thus, the R-chart is:
R = 0.005, UCL = (2.115)(0.0005)=0.0011
LCL = (0) (0.0005) = 0
Looking at the data given in the problem, we observe that many values of x
and R fall outside the control limits of x & R charts.
These readings are removed and fresh set of values for x chart and R chart
are calculated.
Second Iteration
µ = 0.3026, R = 0.0005
For x chart:
UCL = 0.3026 + (0.577)(0.0005) = 0.3029
LCL = 0.3026 - (0.577)(0.0005) = 0.3023
For R chart:
UCL = (2.115) (0.0005) = 0.0011
LCL = (0) (0.0005) = 0
Reading No.11 and 17 are outside the LCL and are now removed.
Third Iteration
µ = 0.3027, R = 0.0005
x chart: UCL = 0.3030 & LCL = 0.3024
R chart: UCL = 0.0011 & LCL = 0
All the values now fall within the control limits of both the charts. These x
and R charts are stable charts.
The specification limits, as given, are: 0.3035 and 0.3019.
The process, as given, has produced 19 of the 20 results within the
specification limits; that is, the process is able to satisfy the specifications just
about reasonably if it can be called so. It is desirable that the values should
rarely fall outside the “control” limits in order to produce output with very little
defects. This would require the process to have reduced variations in its x
values. The process needs to be set, modified and/or improved accordingly.

8. 8
16. The setting (420g) seen on the machine need not be the actual mean as
observed for the filled containers. We have 20 samples with their readings.
We will first construct X and R charts from the data furnished as:
The mean of X i.e. µ for the given data is:
[414+408+418+415+407+405+411+413+413+419+425+421+416+410+414+419
+422+415+410+405] ÷ 20 = 414 g.
The mean of the range, i.e. R is computed as:
R = [10+12+9+8+9+3+3+10+5+7+12+8+8+6+5+3+11+10+2+15] ÷ 20 = 7.3 g.
The sample size, “n” is 5. The upper and lower control limits for 3σ limits are as
follows.
For the X chart:
UCLX = µ + A2R and LCLX = µ - A2R
The factor A2 for a sample size of 5 is: 0.577.
(This value was obtained by referring to Appendix. Read factor A2 corresponding
to n =5)
Therefore, UCLX = µ + A2R = (414) + (0.577) (7.3) = 418.2 g.
and LCLX = µ - A2R = (414) - (0.577) (7.3) = 409.8 g.
We notice that some of the samples (amongst the 20 furnished) fall beyond the
above control limits. These are: Sample numbers 2, 5, 6, 10, 11, 12, 16, 17 and
20, which will have to be removed in order to construct a stable control chart.
For the latter purpose, we also need to look at the R-chart (with 3σ limits) and
eliminate from our consideration those samples that fall outside the control limits.
For the R-chart: UCLR = D4 .R = (2.115)(7.3) = 15.4 g
LCLR = D3 .R = (0)(7.3) = 0 g
(The values of the factors D4 and D3 were obtained by reference to Appendix.
Read D4 and D3 corresponding to n= 5. Refer to page 9.18).
We observe that none of the R (Range) values, in the 20 samples given, fall
outside the control limits for the R-chart and so no additional samples need to be
eliminated.
Thus, removing the samples numbers 2, 5, 6, 10, 11, 12, 16, 17 and 20 from our
analysis, we compute the control limits once again.
µ = 413.6; R = 8.0 ….Revised values

9. 9
Therefore, UCLx = (413.6) + (0.577) (8.0) = 418.2 g.
and LCLx = (413.6) - (0.577) (8.0) = 409.0 g.
Also UCLR = (2.115)(8.0) = 16.9 g. and LCLR = (0)(8.0) = 0 g.
From these revised values, we observe that there is no need for any further
elimination of the samples from our analysis. Thus, we find that when the
process is in control, the sample means would be expected to fall within 418.2
and 409.0 g with a probability of 0.997, for ± 3 standard deviation limits.
However, the above statement is true only about the mean of the sample. The
individual values within a sample may vary with an average range of 8.0 and ±
standard deviation limits of 16.9 and 0, which means that there is approximately
a 0.15 per cent chance ( = 0.3 / 2 ) that the X will be undershot by 8.0 g. (=8.0 –
0), within a sample whose mean value is X.
There is a conditional probability here – the probability that the sample falls on or
above the LCL is 0.9985. The probability that the individual container reading
does not deviate from the mean of the sample by 8.0 g on the lower side, is
again 0.9985. Since the probability of the occurrence of the latter event needs to
be conditional to that of the former event, the conditional probability (that an
individual value will not cross (LCLx – 8.0 g) = (490.0 – 8.0) = (401.0 g) is:
0.9985 x 0.9985 = 0.9970, so that 0.9970 is the probability generally expected.
Since the lower specification limit is 405.0 g and the machine setting can only be
adjusted by 5.0 g, we need to re-set the machine upwards by 5.0 g to 425 g.
(Note: We want the (LCLx – 8.0) to be at or above 405.0 g. At present, it is at
401.0 g). This re-setting will result in pushing the (LCLx – 8.0) to 406.0.
The control limits will, therefore, be pushed upwards by 5.0 g. These will now be:
UCLx = 418.2 + 5.0 = 423.2 g. LCLx = 409.0 - 5.0 = 414.0 g.
The R-chart control limits remain the same. The dispersion characteristics of the
process do not get affected (as given in the problem), by resetting the machine.
The reader may try to obtain the same answer in a different way. The distribution
of the population may be assumed to be Normal. The standard deviation of the
distribution of the population can be obtained by the relation: (Standard
deviation) population = ( √ n) x (Std. Deviation) sample-means, where, ‘n’ is the sample
size, which is 5 (five) in our problem. Since the mean of the population
distribution is the same as that of the sample means, viz. 413.6 g., count 3
(standard deviations) population down from 413.6 g in order to give us the lowest
point.
The answers to the problem can be summarized as follows:

10. 10
(a) No; the initial setting is not okay. The machine should be set at 425 g.
(b) The control limits for the process are:
UCLx = 423.2 g, LCLx = 414.0 g. and UCLR = 16.9 g, LCLR = 0 g.
The explanation for the answers has already been provided.

11. 11
CHAPTER 9: Quality Management – I
Objective Questions:
1. Quality is a strategic decision for a firm because:
a. in a competitive market, quality provides the competitive edge.
b. quality is what the customer wants.
√c. the firm has to fit quality in its long term plans drawn up based on
its capabilities and resource availabilities.
d. a & b
2. By ‘good quality’ product, we mean:
a. a product that meets the performance requirements of the
customer.
b. a product that pleasantly ‘surprises’ the customer.
√c. a & b
d. product characteristics that have been decided as per the strategic
plans of the firm.
3. In order to control, quality characteristics or attributes have to be
measurable or countable.
√a. True (b) False
4. Quality is ‘free’, because:
a. Production volumes have increased.
b. Mass customization is possible.
c. a & b
√d. none of the above.
5. Statistical Quality Control is possible in:
√a. mass production or very large batch production.
b. customized unit production.
c. a & b
d. none of the above.
6. Process Control means:
a. inspecting and weeding out unacceptable items from the lot
produced by the production process.
b. correcting the process and thus enhancing the capability of the
process.
√c. bringing the process back to normalcy.
d. a & b

12. 12
7. If the right raw materials are used and right processes operate on them,
then all the produced finished goods will be of good quality.
√a. No; there is still a chance that the process would produce products
of unacceptable quality.
b. Yes; it follows.
c. No; if there is a delay in the operations, the quality may not be
uniform across all products.
8. Uncontrolled ‘variability’ is the bane of quality.
√a. True b. False
9. Entire SQC depends upon ‘sampling’.
a. No; SPC - a part of SQC – does not.
√b. Yes.
c. No; only ‘acceptance sampling’ depends upon sampling.
d. a & c
10. X chart and R chart:
a. are not related; former depicts the ‘mean’ and the latter the ‘range.’
b. are related as they both measure the attribute for quality.
√c. are related as the control limits depend upon the variation in the
quality characteristic and the range of variations.
c. are not related; the former is for averages and latter is for individual
readings.
11. p-chart:
√a. uses Binomial distribution for calculating its variance.
b. uses Poisson distribution for calculating its variance.
c. uses Normal distribution for calculating its variance.
d. none of the above
12.Control limits for the c-chart follow:
a. Binomial distribution.
b. Poisson distribution.
√c. Normal distribution.
d. none of the above.
13.In assuring quality in mass manufacturing:
a. dispersion is important.
b. central tendency is important.
√c. both dispersion and central tendency are important.
d. the applicability of statistical distribution is important.

13. 13
14.Standard deviation used in constructing a c-chart is based upon:
a. Binomial distribution.
√b. Poisson distribution.
c. Normal distribution.
d. None of the above.
15.If a method in manufacturing / operations changes, then:
a. Specification limits and process control limits change.
b. Specification limits and process control limits, both do not change.
c. Specification limits change, but process control limits do not
change.
√d. Specification limits may remain the same, but process control limits
change.