GM 533 Week 2 Lecture, Applied Managerial Statistics

1. Welcome! To the Week 2
Live Lecture/Discussion
Applied Managerial Statistics (GM533)
Lecturer – Prof. Brent Heard
Please note that I borrowed these charts from
Joni Bynum and the textbook publisher.
Thanks Joni!
I will put my touch on them (in blue) as we go
along.
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2. Week 2: Probability - Overview
• Week 2 Terminal Course Objectives (TCOs)
• The basic idea of probability
• The four main problem types we’ll cover
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3. Week 2 TCOs
• TCO B Probability Concepts and
Distributions: Given a managerial problem,
utilize basic probability concepts, and standard
probability distributions, e.g., binomial,
normal, as is appropriate, to formulate a
course of action which addresses the problem.
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4. Week 2 TCOs (cont’d)
• TCO F Statistics Software Competency:
Students should be able to perform the
necessary calculations for objectives A through
E using technology, whether that be a
computer statistical package or the TI-83, and
be able to use the output to address a
problem at hand.
(I assume this includes Excel. I show you examples in Excel
because you have access to it in the workplace. If your
instructor requires you to use Minitab, that is their decision. I am
only the lecturer)
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5. Probability: the Basic Idea
“We use the concept of probability to deal with
uncertainty. Intuitively, the probability of an
event is a number (between zero and one)
that measures the chance, or likelihood, that
the event will occur.” The text book, p. 171.
(Probabilities must be between 0 and 1 or 0 and
100%. Zero means it can’t happen, 1 means
it will definitely. Everything else is in
between)
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6. The 4 Main Probability Problem Types
1. Contingency Tables
2. Expected Value
3. The Binomial Distribution
4. The Normal Distribution
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7. 1. Example of Contingency Tables
• Each month a brokerage house studies various
companies and rates each company’s stock as
being either “low risk” or “moderate to high
risk.” In a recent report, the brokerage house
summarized its findings about 15 aerospace
companies and 25 food retailers in the
following table:
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8. Part A. 1. If we randomly select one of the
total of 40 companies, find:
Company Low Moderate Total • The probability that
Type Risk to High
Risk
the company is a
food retailer
Aerospace 6 9 15
company
Food 15 10 25
retailer 25
.625
40
Total 21 19 40
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9. Part A. 2. If we randomly select one of the
total of 40 companies, find:
Company Low Risk Moderate Total • The probability that
Type to High
Risk
the company’s
stock is “low risk”
Aerospace 6 9 15
company
21
.525
40
Food 15 10 25
retailer
Total 21 19 40
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10. Part A. 4. If we randomly select one of the
total of 40 companies, find:
Company Low Risk Moderate Total • The probability that
Type to High
Risk
the company is a
food retailer AND
has a stock that is
Aerospace 6 9 15
“low risk”
company
15
.375
40
Food 15 10 25
retailer
(See that this value comes
Total 21 19 40
from“within” the table.)
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11. Part A. 5. If we randomly select one of the
total of 40 companies, find:
Company Low Risk Moderate Total • The probability that
Type to High
Risk
the company is a
food retailer OR has
a stock that is “low
Aerospace 6 9 15
risk”
company
31
.775
40
Food 15 10 25
retailer
(Mathematically this is
Total 21 19 40 (25+21- 15)/40.
I had to subtract out the “overlap” where
I counted it twice.)
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12. Part B. 2. If we randomly select one of the
total of 40 companies, find:
Company Low Risk Moderate Total • The probability that
Type to High
Risk
the company’s
stock is moderate
to high risk GIVEN
Aerospace 6 9 15
THAT the firm is a
company
food retailer.
10
.40
25
Food 15 10 25
retailer (When I heard “Given That” it should
Alert me that the denominator has
Total 21 19 40 changed. I’m only dealing with a
subset of the data now.)
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13. Part C If we randomly select one of the total
of 40 companies:
Company Low Risk Moderate Total • Determine if the company
Type to High type is independent of the
Risk level of risk of the firm’s
stock.
Aerospace 6 9 15 P(Aero.) =? P(Aero.Low Risk)
company
15 6
.375 .2857
40 21
.357 .2857
Food 15 10 25
retailer
Since they are not equal,
company type is not
Total 21 19 40
independent of level of
risk, i.e., they are
dependent.
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14. 2. Example Expected Value Problem
There are many investment options which offer
prospective gains and losses with their associated
probabilities. For example, an oil company is
considering drilling a number of wells, and, based on
geological data and operating costs, they determine
the following outcomes and probabilities.
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15. Example Expected Value Problem
(cont’d)
x = the outcome in $ p(x) x*p(x)
- $40,000 (no oil) .25 (- $40,000)*.25 = - $10,000
$10,000 (some oil) .70 ($10,000)*.7 = $7,000
$70,000 (much oil) .05 ($70,000)*.05 = $3,500
1.00 -$10,000 + $7,000 + $3,500 = $500
(This is just the total of the probabilities, it
should always be one. Don’t make this
harder than what it is.)
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16. The Binomial Distribution #1
The Binomial Experiment:
1. An experiment consists of n identical trials (e.g., we flip
a coin n=4 times).
2. Each trial (e.g., an individual coin flip) results in either
“success” or “failure” (e.g., a head or a tail).
3. Probability of success, p (e.g., p=.5), is constant from
trial to trial
4. Trials are independent (e.g., what we get on one flip
does not affect the next flip).
Note: The probability of failure, q, is 1 – p and is constant from trial to trial
If x is the total number of successes in n trials of a
binomial experiment, then x is a binomial random
variable (e.g., the number of heads out of 4 flips)
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17. Binomial Distribution using Excel Template
• Templates for Binomial
• Found at:
http://highered.mcgraw-
hill.com/sites/0070620164/student_view0/excel_templates.html
Download Binomial Distribution file. When you first open it, go to the
Review tab at the top and click on “Unprotect Sheet.” Save it and
you have your own personal Binomial Calculator to use to check
your Minitab answers and to use at work.
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18. Binomial using Excel Template Example
• We have a binomial experiment with p = .6 and n = 3. Set up the
probability distribution and compute the mean, variance, and
standard deviation.
• Find P(x=2), P(x≤2) and P(x>1)
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19. Binomial using Excel Template Example (cont.)
All I did was enter 3 for n and 0.6 for p. The template does the rest.
It gives me the mean, variance and standard deviation.
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20. Binomial using Excel Template Example (cont.)
P(x=2) = 0.4320 P(x≤2) = 0.7840 P(x>1) = 0.6480
Note (x>1) is the same
as saying “at least 2”
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21. 3.Example of a Binomial Probability
Problem
Consider the following scenario: historically, thirty
percent of all customers who enter a particular store
make a purchase. Suppose that six customers enter
the store in a given time period, and that these
customers make independent purchase decisions.
Each individual customer either does or does not make
a purchase. This is what makes this a binomial
situation.
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22. Typical Questions for the Binomial
Probability Distribution
• What is the probability that at least three customers
make a purchase?
• What is the probability that two or fewer customers
make a purchase?
• What is the probability that at least one customer
makes a purchase?
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23. Answers using Excel Template (Do in Minitab also)
At least 1 would be 0.8824
2 or fewer would be 0.7443
At least 3 would be 0.2557
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24. Example Binomial Problem – Mini Tab
Solution looks like this
Probability Density • Cumulative Distribution
Function Function
• Binomial with n = 6 and p = • Binomial with n = 6 and p =
0.3 0.3
• x P( X = x ) • x P( X <= x )
• 0 0.117649 • 0 0.11765
• 1 0.302526 • 1 0.42017
• 2 0.324135 • 2 0.74431
• 3 0.185220 • 3 0.92953
• 4 0.059535 • 4 0.98906
• 5 0.010206 • 5 0.99927
• 6 0.000729 • 6 1.00000
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25. Binomial Probabilities Explanations
P( x 3) P( x 3) P( x 4) P( x 5) P( x 6)
0.18522 0.05954 0.01021 0.00073
0.25569
P( x 3) 1 P( x 2) 1.00000 0.74431 0.25569
P(x 2) 0.74431
P( x 1) 1.00000 P( x 0) 1.00000 0.11765 0.88235
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26. The Normal Probability Distribution
The normal curve is
symmetrical and bell-shaped
• The normal is symmetrical about
its mean
• The mean is in the middle
under the curve
• So is also the median
• The normal is tallest over its
mean
• The area under the entire normal
curve is 1
• The area under either half of
the curve is 0.5
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27. The Position and Shape
of the Normal Curve
(a) The mean positions the peak of the normal curve over the real axis
(b) The variance 2 measures the width or spread of the normal curve
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28. The Standard Normal Distribution #1
If x is normally distributed with mean and standard
deviation , then the random variable z
x
z
is normally distributed with mean 0 and standard
deviation 1; this normal is called the standard normal
distribution
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29. The Standard Normal Distribution #2
z measures the number of standard deviations that x is from
the mean
• The algebraic sign on z indicates on which side of is x
• z is positive if x > (x is to the right of on the number line)
• z is negative if x < (x is to the left of on the number line)
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30. Cumulative Areas under the Standard
Normal Curve
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31. Cumulative Areas under the Standard
Normal Curve (cont’d)
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32. Finding Areas under the Standard
Normal Curve
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33. Finding Areas under the Standard
Normal Curve (cont’d)
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34. Normal Distribution using an Excel Template
• You can use the Normal Distribution Template from the same site
I noted to do these.
• Just go to the site, download the template, open it, click
“Unprotect Sheet.” Save it to your computer and you have a
“Normal Distribution Calculator” to check your Minitab and hand
calculations.
• When using for Standard Normal Distribution calculations, enter 0
for the mean and 1 for the standard deviation
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35. Standard Normal Distribution with Excel
Template
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36. Finding Areas under the Standard
Normal Curve (cont’d)
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37. Finding Z Points on
a Standard Normal Curve
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38. Brent’s Example of the Normal Distribution
• Let’s say that failures of a particular component are normally
distributed with a mean of 2.1 failures per year and a standard
deviation of .6.
• Find the probability of having less than two failures in a year.
• Find the probability of having more than 3 failures in a year.
• Find the probability of having between 1 and 3 failures in a year.
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39. Brent’s Example of the Normal Distribution (cont.)
I put in mean and std. dev.
Probability of having between
one and three failures in a
year = 0.8998
Probability of having more
Probability of having less than than three failures in a year =
two failures in a year = 0.4338 0.0668
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40. Good day/night
• See you next time
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