John M Awbrey
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3/15/2016
IP2
IET 603
Mikey Awbrey
M1014968
Individual Project II (50 points): Please solve and submit your completed project by 225-2014 at 10:00 p.m.
1.
Explain the foundation of Shewhart’s notion of scientific approach and the basic
activities involved in developing means for satisfying the customers (in
approximately 100 words)
PDCA, Plan Do Check Act. Plan – set up experiment or plan for
improvement. Do – Carry out plan from previous step. Check –
Study results. Act – Adopt, abandon, or adapt change based on
results.
2.
Explain the Juran Trilogy, in approximately 50 words:
Quality Planning, Quality Control, Quality Improvement. Quality
planning – identify need and develop plan for process. Quality
control – measure and observe process. Correct significant
differences. Quality Improvement – Identify opportunities for
improvement and provide remedies.
3.
Briefly list and describe the purpose for basic tools for quality improvement
(Basic Process Improvement Toolbox).
Check Sheet – Collect discrete data about a process during use
Cause-and-Effect diagram – quickly see the likely results of a cause
Control Chart – check if process is in control, process capability
Histogram – quickly identify patterns
Pareto Chart – find significant many and insignificant few
Scatter Diagram – graphically represent collected data
Run Chart – easily collect data in simple numeric form
4.
Explain the properties of random variables
A Random variable is the numeric description of the outcome of an
experiment taken from Ding & Yang Powerpoint. Random variables can be discrete or
continuous. They are also able to be dependent or independent. A
discrete random variable is a count, or a pass fail, represented as
integers. Continuous variables can be any number from - to .
Independent variables are variables which can change on their own, such
as X in a linear line equation Y = kX + m. Dependent variables are
numbers whose values change based on another one such as Y in the
example before.
5.
The data shown below are the times in minutes that successive customers had to
wait for service at an oil change facility. Using hand calculation and formula 1)
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find sample mean and standard deviation, 2) construct a histogram, and 3) use
MINITAB to validate your findings and the histogram.
9.93
9.92
9.97
9.96
9.91
9.90
10.09
10.07
10.01
10.10
10.13
9.78
9.97
10.08
10.15
9.93
9.99
9.91
10.05
9.88
9.98
10.07
9.92
10.01
9.94
9.88
10.05
10.06
10.21
10.13
9.92
9.84
10.09
9.84
9.98
9.92
10.01
9.86
9.95
9.83
9.98
10.01
10.09
10.08
10.00
10.02
10.03
10.03
10.02
9.97
I used excel to help make the hand calculations faster. Included next to each
part is the excel formula used.
9.78
=(A1-$D$3)^2 0.043681
9.83
n
SUM
MEAN
0.025281
Variance STD DEVIATION
9.84
50
499.45
9.989
0.022201
0.00843
0.091812809
9.84 =COUNT(A:A) =SUM(A:A) =C3/B3
0.022201
=SQRT(G3)
9.86
0.016641 =SUM(F:F)/(B3-1)
9.88
0.011881
9.88
0.011881
9.9
0.007921
9.91
0.006241
9.91
0.006241
9.92
0.004761
9.92
0.004761
9.92
0.004761
9.92
0.004761
9.93
0.003481
9.93
0.003481
9.94
0.002401
9.95
0.001521
9.96
0.000841
9.97
0.000361
9.97
0.000361
9.97
0.000361
9.98
8.1E-05
9.98
8.1E-05
9.98
8.1E-05
9.99
1E-06
10
0.000121
10.01
0.000441
John M Awbrey
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10.01
10.01
10.01
10.02
10.02
10.03
10.03
10.05
10.05
10.06
10.07
10.07
10.08
10.08
10.09
10.09
10.09
10.1
10.13
10.13
10.15
10.21
3/15/2016
0.000441
0.000441
0.000441
0.000961
0.000961
0.001681
0.001681
0.003721
0.003721
0.005041
0.006561
0.006561
0.008281
0.008281
0.010201
0.010201
0.010201
0.012321
0.019881
0.019881
0.025921
0.048841
The following Minitab results confirm my calculations.
Standard Deviation of C1
Standard deviation of C1 = 0.0918128
Mean of C1
Mean of C1 = 9.989
Sum of C1
Sum of C1 = 499.45
Number of Rows in C1
Total number of observations in C1 = 50
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Histogram of C1
Normal
12
Mean
StDev
N
10
9.989
0.09181
50
Frequency
8
6
4
2
0
6.
9.8
9.9
10.1
10.2
If the probability that any individual will react positively to a drug is 0.8, what is
the probability that 4 individuals will react positively from a sample of 10
individuals?
P=0.8
X=4
N=10
𝑛
p(x) = ( ) 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥
n!
p(x) =
𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥! (𝑛 − 𝑥)!
10!
p(x) =
0.84 (1 − 0.8)10−4
4! (10 − 4)!
5040
p(x) =
0.4096(0.000064)
24
p(x) = 0.005505
0.55%
The following is Minitab result
Probability Density Function
Binomial with n = 10 and p = 0.8
x
4
10.0
C1
P( X = x )
0.0055050
John M Awbrey
7.
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Suppose the average number of customers arriving at ATM during the lunch hour
is 12 customers per hour. The probability of exactly two arrivals during the lunch
hour is:
𝜆 =12
X=2
𝑒 −𝜆 𝜆𝑥
p(X = x) =
𝑥!
𝑒 −12 122
p(X = x) =
2!
p(X = x) = 0.00044238
0.044%
The following is the minitab result
Probability Density Function
Poisson with mean = 12
x
2
8.
P( X = x )
0.0004424
In a sample of 100 items produced by a machine that produces 2% defective
items, what is the probability that 5 items are defective? (Calculate with binomial
distribution formula and verify your response using MINITAB).
Solve the question #8 using Poisson distribution formula and verify your
response using MINITAB.
N = 100
P = 0.02
X=5
𝑛
p(x) = ( ) 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥
n!
p(x) =
𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥! (𝑛 − 𝑥)!
100!
p(x) =
0.025 (1 − 0.02)100−5
5! (100 − 5)!
9034502400
p(x) =
0.0000000032(0.1467159029466343)
120
p(x) = 0.0353468
3.53%
The following is the minitab result
Probability Density Function
Binomial with n = 100 and p = 0.02
x
5
P( X = x )
0.0353468
John M Awbrey
9.
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It is assumed that the inductance of particular inductors produced by ABC
Company is normally distributed. The  of inductors is = 20,000 mH, and  of
90 my. If acceptable inductance range is from 19,750 mH to 20,200 mH. Using
both formula and MINITAB, determine the expected number of rejected inductors
in a production run of 10,000 inductors.


LSL = 19750
USL = 20200
N = 10000
𝐿𝑆𝐿 − µ
𝑍𝐿𝑆𝐿 =
𝜎
19750 − 20000
𝑍𝐿𝑆𝐿 =
90
𝒁𝑳𝑺𝑳 = 𝟎. 𝟎𝟎𝟐𝟖 = 𝑷𝑳𝑺𝑳
𝑈𝑆𝐿 − µ
𝑍𝑈𝑆𝐿 =
𝜎
20200 − 20000
𝑍𝑈𝑆𝐿 =
90
𝑍𝑈𝑆𝐿 = 0.9868
𝐏𝐔𝐒𝐋 = 𝟏 − 𝐙𝐔𝐒𝐋 = 𝟎. 𝟎𝟏𝟑𝟐
Pfailure = PUSL + PLSL
Pfailure = 0.0028 + 0.0132
𝐏𝐟𝐚𝐢𝐥𝐮𝐫𝐞 = 𝟎. 𝟎𝟏𝟔
n𝐟𝐚𝐢𝐥 = 𝐧(𝐏𝐟𝐚𝐢𝐥𝐮𝐫𝐞 )
n𝐟𝐚𝐢𝐥 = 𝟏𝟎𝟎𝟎𝟎(𝟎. 𝟎𝟏𝟔)
𝒏𝒇𝒂𝒊𝒍 = 𝟏𝟔𝟎
The following is the minitab result
Cumulative Distribution Function
Normal with mean = 20000 and standard deviation = 90
x
19750
20200
P( X <= x )
0.002737
0.986866
10.
Explain Null Hypothesis, Alternative Hypothesis, Types of error, Significance
Level, Risk Level in hypothesis testing
Null Hypothesis – the result that you are trying to disprove
Alternative Hypothesis – what to except is the null hypothesis is false
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Types of error – Type 1 occurs when null hypothesis is rejected when it
should have been rejected. Type 2 is when the null hypothesis is
accepted when it should have been rejected.
Significance level – The probability of wrongly rejecting the null
hypothesis. Probability of a type 1 error.
Risk Level – The probability of wrongly accepting the null hypothesis.
Probability of a type 2 error.
11.
Formulate the appropriate null hypothesis and alternative hypothesis for testing
that the starting salary for graduates with a B.S. degree in electrical engineering
is greater than $38,000 per year. The significance, or risk, is: What does  =
0.05 mean?
H0 ≤ 38,000
H1 > 38,000
 = 0.05
CI = 100(1-)% = 95%
 is the significance level of the test and the probability of the type 1 error.
This is the probability that our null hypothesis is true. Confidence Interval
(CI) is the probability that our alternative hypothesis is true.
12.
In a New York Times/CBS poll, 56 percent of 2,000 randomly selected voters in
New York City said that they would vote for the incumbent in a certain twocandidate race. Calculate a 95 percent confidence interval for the population
proportion. Discuss its implication. Carefully discuss what is meant by the
population, how you would carry out the random sampling, and what other factors
could lead to differences between the responses to the surveys and the actual
votes on the day of the election.
The poll implies that the majority of the voters would vote for the
incumbent in question. The population that we are estimating for is
all of the voters, not only the ones surveyed. The best way to
randomly sample the voters is to select people at random from
different areas in the voting district. This should ensure the correct
diversity to estimate the population. Potential problems with the
sampling could be that only a certain demographic of people were
polled. There could be outside factors such as political scandal that
could change the minds of voters before election day.
CI = 95%
P = 56%
N = 2000
X = P*N = 1120
Multiplier for 95% CI = 1.96 = z
𝑝(1 − 𝑝)
𝐶𝐼 = 𝑃 ± 𝑍 ∗ √
𝑛
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0.56(1 − 0.56)
2000
𝑪𝑰 = 𝟎. 𝟓𝟔 ± 𝟎. 𝟎𝟐𝟏𝟕𝟓
𝑪𝑰+= 𝟎. 𝟓𝟖𝟐 = 𝟓𝟖. 𝟐%
𝑪𝑰−= 𝟎. 𝟓𝟑𝟖 = 𝟓𝟑. 𝟖%
𝐶𝐼 = 0.56 ± 1.96 ∗ √
The following is the minitab result
Test and CI for One Proportion
Sample
1
13.
X
1120
N
2000
Sample p
0.560000
95% CI
(0.537921, 0.581903)
A random sample of 50 teaching assistants at the University of Iowa in the fall of
1996 indicated that 30 of them were planning to join the union for teaching
assistants. Calculate a 95 percent confidence interval for the proportion of
University of Iowa teaching assistants who are in favor of joining a union.
CI = 95% = 1.96 = Z
N = 50
X = 30
P = X/N = 0.6
𝑝(1 − 𝑝)
𝐶𝐼 = 𝑃 ± 𝑍 ∗ √
𝑛
0.6(1 − 0.6)
50
𝑪𝑰 = 𝟎. 𝟔 ± 𝟎. 𝟏𝟑𝟓𝟕𝟗
𝑪𝑰+= 𝟎. 𝟕𝟑𝟔 = 𝟕𝟑. 𝟔%
𝑪𝑰−= 𝟎. 𝟒𝟔𝟒 = 𝟒𝟔. 𝟒%
𝐶𝐼 = 0.6 ± 1.96 ∗ √
The following is the minitab result
Test and CI for One Proportion
Sample
1
14.
X
30
N
50
Sample p
0.600000
95% CI
(0.451794, 0.735922)
Suppose that the number of wire-bonding defects per unit that occur in a
semiconductor device is Poisson distributed with parameter lambda (Mu or
variance) of 4. Then, the probability that a randomly selected semiconductor
device will contain 2 or fewer wire-bonding defects is:
λ=4
x=2
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𝑒 −𝜆 𝜆𝑥
p(X = x) =
𝑥!
𝑒 −4 42
p(X = x) =
2!
p(X = x) = 0.14652511
14.65 %
The following is the minitab result
Probability Density Function
Poisson with mean = 4
x
2
P( X = x )
0.146525
15.
List and explain Deming’s 7 deadly diseases.
I.
Lack of constancy of purpose – this relates to Demings 14 points and
ensures stakeholders that the business will grow
II. Too much emphasis on short term profits – this commonly involves
lowered quality which hurts the business long term
III. Performance evaluation - Encourages short term performance by focusing
on performance, employees can be left unhappy and actually perform
worse.
IV. Job hoping – managers who do not spend a long in a company and are
constantly thinking about their next career move rather than their current
work.
V. Managing by visible figures alone – much of what makes a business
successful is unknowable and managing based only on known facts can
be a detriment.
VI. Excessive medical expenses – Health care is important but extravagant
spending on employees is unnecessary and costly.
VII. Liability and excessive damage awards – minimize cost by reducing
lawyer needs and relying on government intervention in large scale
issues.
16.
Explain who, when, and why six sigma was developed. What is the focus of six
sigma?
Motorola developed six sigma to meet the customer needs of quality on a
level where defects are extremely unlikely. The old three sigma quality
performance only offered results up to parts per thousand, with six sigma
they can now find a low number of defects per million (almost 1). This is
necessary for such large scale production. Especially when producing
assemblies from parts built in many different plants, three sigma is just not
acceptable enough for a final product.
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17.
Explain the DMAIC process and its relevance to quality.
Define, Measure, Analyze, Improve, Control. First you define the problem
to be tested, then measure the results of the process. The results are
analyzed and possible methods for improvement are made. The last step
is to setup a plan for continuing to control the improvement. This is used
in quality as a standard process for improving a process.
18.
What is HISTOGRAM and why it is being used?
Histograms are a representation of data into cells, intervals, or bins. This is a
useful way to quickly note patterns and get a general idea about the distribution
of the data. Commonly, they will have the bellcurve of the data placed over
them to accentuate the results.
19.
Define Probability Distribution. What is the difference between discrete and
continuous distributions? Provide Examples.
Probability Distribution – the relationship of a variable with the probability of that
variable taking on a certain value. Discrete distributions are used for counts and
Bernoulli trials, such as a binomial or geometric distribution. A Continuous
distribution is commonly used for measurements and can range from - to .
These are used for distributions such as chi-square or normal distributions.
20.
Define Bernoulli trials and Poisson distributions and explain their uses in Quality.
Give example.
Bernoulli trials are simple pass-fail tests of a sample taken from a population,
such as the number of computers that work in the computer lab. A Poisson
distribution is used for when there are multiple possible results of a Bernoulli trial,
such as a part having an average number of defects, or an average number of
events happening at a particular time. Bernoulli trials are useful for finding
simple pass-fail results of a sample taken from a population of parts. Poisson
distributions are useful for predicting the possible number of events given the
previous averages of that event occurring, such as the number of defects within a
part when an average number of defects are known.