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Chapter 9 (last update: May 30, 2006 jvm)
CHAPTER 9: MANAGING FLOW VARIABILITY: PROCESS
CONTROL AND CAPABILITY
9.3 Solutions to the Problem Set
Problem 9.1
a. Given the symmetric shape of normal distribution around its mean, maximum conformance of the
output within the given specifications will be achieved by centering the process at the midpoint of
the specifications, i.e., at  = 32.5 gms. Now if we desire 98% of the output to conform to the
specifications, it means that only 2% can fall outside the specs. Given a centered process and the
symmetry of the normal distribution, this means only 1% below lower spec and 1% above upper
spec. Thus, from the Normal tables (single-sided at 1%), the specification limits should be z = 2.33
standard deviations on either side of the mean, i.e., (32.5 - 30)/ = 2.33, or  = 2.5/2.33 = 1.073
gms. The sigma-capability of this process is 2.33. The corresponding process capability ratio is Cp
= (35 - 30)/6 = (35-30)/(6*1.073) = 0.78.
b. With n = 12, and  = 1.073 as above, we can now determine the ideal control limits on subgroup
averages of 12 bottles as:
Average control chart:  ± 3 / n = 32.5 ± (3)(1.073)/ 12 = (31.57, 33.34)
Problem 9.2
a. In order to produce 98% of the boxes above 15.5 oz., the process mean must be z = 2.055 standard
deviations above 15.5, i.e.,  = 15.5 + (2.055)(0.5) = 16.53 oz.
b. If  = 16.53, then the proportion of overweight boxes will be
P(X > 16) = P[Z > (16-16.53)/0.5] = P(Z > -1.06) = 0.8554.
With  = 16.53,  = 0.5, and n = 9, the control limits on the average weight in a sample of 9 boxes
are:  + 3 / 9 = (16.03, 17.03). The observed average of 15.9 is below the lower control limit of
16.03, which signals that the mean has shifted below 16.53 due to an assignable cause. They
should stop the process and adjust the mean upward. That the average weight is above the
minimum individual weight of 15.5 oz is irrelevant. The specifications are on weights of individual
boxes, not on average weights.
c. The FDA specifications require at least 15.5 ozs per box. To be a 6-sigma process, the filling
process must produce output centered at a mean of μ and a standard deviation of σ, where (μ15.5)/σ = 6. This can be achieved for many different values of μ and σ. In particular, μ = 16.1 and σ
= 0.1 results in a 6-sigma process.
Problem 9.3
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Chapter 9 (last update: May 30, 2006)
From the 26 observations given, we can calculate the average number of errors per thousand transactions
m = 3.3077, which is much better than the industry average = 15. We can then determine the control
limits on the number of errors, assuming Poisson distribution, as m ± 3 m = (0, 8.75). Observe that three
observations out of the 26 given exceed the UCL = 8.75. Hence, the process is not in control, even
though on average it is better than the BAI standard! The process is not stable, and our estimate of m is
not reliable. We first need to stabilize the process by removing assignable causes.
Problem 9.4
a. If the process mean  = 515, standard deviation  = 5 gms, and sample size n = 25, Control limits
are LCL =  25 = 512 gms and UCL =  25 = 518 gms
b. Proportion of underweight output is
Prob(W < 500) = Prob [Z < (500 – 515) / 5] = Prob (Z < –3) = 0.0013 or 0.13%.
c. If Prob (violation) = 0.13% and the mean is  = 503 gms, then  must be decreased so that –3 =
(500 - 503)/ or  = 1.
Problem 9.5
Given: process mean  = 6 cm, standard deviation  = 0.01 cm, and sample size n = 10,
Control limits are LCL =  10 = 5.9905 cm and UCL =  10 = 6.0095 cm
Note: customer specifications are irrelevant.
Problem 9.6
If the specifications do not change, improvement in the sigma capability means the standard deviation
decreases. In that case, the control band should become narrower.
Problem 9.7
Currently, process mean  = 2.2 hours, and standard deviation  = 0.8 hours, so probability of processing
within 4 hours is
P(T < 4) = Prob[Z < (4 – 2.2)/0.8 ] =Prob(Z < 2.25) = 0.9878 < 0.992, the bank’s requirement
With improved process mean  = 2 hours, and standard deviation  = 1.2 hours,
P(T < 4) = Prob[Z < (4 – 2)/1.2 ] =Prob(Z < 1.67) = 0.9616, which is worse than before.
To meet the bank’s requirement, the mean should be lowered to 2 hours without raising variability. In that
case
P(T < 4) = Prob[Z < (4 – 2)/0.8 ] =Prob(Z < 2.5) = 0.9938 > 0.992, the bank’s requirement
Problem 9.8
Chapter 9 (last update: May 30, 2006) 89
If D is the diameter of basketballs produced, we need 95% of diameters to fall between 29.3 and 29.7.
Given a centered process and the symmetry of the normal distribution, this means only 2.5% below lower
spec and 2.5% above upper spec. Thus, from the Normal tables (single-sided at 2.5%), the specification
limits should be z = 1.96 standard deviations on either side of the mean
So z = (29.7 – 29.5) /  = 1.96 or we need  = .2/1.96 = 0.102. inches.
Hence, Cp = (USL – LSL)/6  = (29.7 - 29.3)/6  = 0.65.
Problem 9.9
a. False. Control limits only assure process stability. They have nothing to do with meeting
customer specifications.
b. True. As sample size increases, standard deviation of sample averages decreases, and the control
bands becomes narrower.
c. True. If control limits are 3 standard deviations from the mean, the probability of sample average
falling outside control limits is 0.27%, regardless of the value of the standard deviation.
d. If the process improves sigma capability, its standard deviation goes down, and the width of the
control band decreases.
e. False. A 6-sigma process may be out of control if an assignable cause leads to a mean shift (for
instance). A 6-sigma process, however, is likely to produce very few defectives even when it is
out of control. Thus, a 6-sigma process has the advantage of detecting small shifts even before
they impact the customer.