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.

4-1

2

Week 2: Probability - Overview

• Week 2 Terminal Course Objectives (TCOs)

• The basic idea of probability• The four main problem types we’ll cover

4-2

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.

4-3

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)

4-4

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)

4-5

6

The 4 Main Probability Problem Types

1. Contingency Tables2. Expected Value3. The Binomial Distribution4. The Normal Distribution

4-6

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:

4-7

8

Part A. 1. If we randomly select one of the total of 40 companies, find:

Company Type

Low Risk

Moderate to High Risk

Total

Aerospace company

6 9 15

Food retailer

15 10 25

Total 21 19 40

• The probability that the company is a food retailer

4-8

625.4025

9

Part A. 2. If we randomly select one of the total of 40 companies, find:

Company Type

Low Risk

Moderate to High Risk

Total

Aerospace company

6 9 15

Food retailer

15 10 25

Total 21 19 40

• The probability that the company’s stock is “low risk”

4-9

525.4021

10

Part A. 4. If we randomly select one of the total of 40 companies, find:

Company Type

Low Risk

Moderate to High Risk

Total

Aerospace company

6 9 15

Food retailer

15 10 25

Total 21 19 40

• The probability that the company is a food retailer AND has a stock that is “low risk”

4-10

375.4015

(See that this value comes from“within” the table.)

11

Part A. 5. If we randomly select one of the total of 40 companies, find:

Company Type

Low Risk

Moderate to High Risk

Total

Aerospace company

6 9 15

Food retailer

15 10 25

Total 21 19 40

• The probability that the company is a food retailer OR has a stock that is “low risk”

4-11

775.4031

(Mathematically this is (25+21- 15)/40.

I had to subtract out the “overlap” where I counted it twice.)

12

Part B. 2. If we randomly select one of the total of 40 companies, find:Company Type

Low Risk

Moderate to High Risk

Total

Aerospace company

6 9 15

Food retailer

15 10 25

Total 21 19 40

• The probability that the company’s stock is moderate to high risk GIVEN THAT the firm is a food retailer.

4-12

40.2510

(When I heard “Given That” it shouldAlert me that the denominator has changed. I’m only dealing with a subset of the data now.)

13

Part C If we randomly select one of the total of 40 companies:

Company Type

Low Risk

Moderate to High Risk

Total

Aerospace company

6 9 15

Food retailer

15 10 25

Total 21 19 40

• Determine if the company type is independent of the level of risk of the firm’s stock.

4-13

375.4015

2857.216

2857.357.

P(Aero.) =? P(Aero.\Low Risk)

Since they are not equal, company type is not independent of level of risk, i.e., they are dependent.

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.

4-14

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

4-15

(This is just the total of the probabilities, it should always be one. Don’t make this harder than what it is.)

16

The Binomial Distribution #1

4-16

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 trial4. Trials are independent (e.g., what we get on one flip

does not affect the next flip).

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)

Note: The probability of failure, q, is 1 – p and is constant from trial to trial

17

Binomial Distribution using Excel Template

4-17

• 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.

18

Binomial using Excel Template Example

4-18

• 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)

19

Binomial using Excel Template Example (cont.)

4-19

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.

20

Binomial using Excel Template Example (cont.)

4-20

P(x=2) = 0.4320 P(x≤2) = 0.7840 P(x>1) = 0.6480Note (x>1) is the sameas saying “at least 2”

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.

4-21

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?

4-22

23

Answers using Excel Template (Do in Minitab also)

4-23

At least 3 would be 0.2557

2 or fewer would be 0.7443

At least 1 would be 0.8824

24

Example Binomial Problem – Mini TabSolution looks like this

Probability Density Function

• Binomial with n = 6 and p = 0.3

• x P( X = x )• 0 0.117649• 1 0.302526• 2 0.324135• 3 0.185220• 4 0.059535• 5 0.010206• 6 0.000729

• Cumulative Distribution Function

• Binomial with n = 6 and p = 0.3

• x P( X <= x )• 0 0.11765• 1 0.42017• 2 0.74431• 3 0.92953• 4 0.98906• 5 0.99927• 6 1.00000

4-24

25

Binomial Probabilities Explanations

4-25

)6()5()4()3()3( xPxPxPxPxP00073.001021.005954.018522.0

25569.0

25569.074431.000000.1)2(1)3( xPxP

74431.0)2( xP

88235.011765.000000.1)0(00000.1)1( xPxP

26

The Normal Probability Distribution

5-26

The normal curve is symmetrical and bell-shaped• The normal is symmetrical about

its mean m• The mean is in the middle

under the curve• So m is also the median

• The normal is tallest over its mean m

• The area under the entire normal curve is 1• The area under either half of

the curve is 0.5

27

The Position and Shapeof the Normal Curve

5-27

(a) The mean m positions the peak of the normal curve over the real axis (b) The variance s2 measures the width or spread of the normal curve

28

The Standard Normal Distribution #1

sm

xz

5-28

If x is normally distributed with mean m and standard deviation s, then the random variable z

is normally distributed with mean 0 and standard deviation 1; this normal is called the standard normal distribution

29

The Standard Normal Distribution #2

5-29

z measures the number of standard deviations that x is from the mean m• The algebraic sign on z indicates on which side of m is x• z is positive if x > m (x is to the right of m on the number line)• z is negative if x < m (x is to the left of m on the number line)

30

Cumulative Areas under the Standard Normal Curve

4-30

31

Cumulative Areas under the Standard Normal Curve (cont’d)

4-31

32

Finding Areas under the Standard Normal Curve

4-32

33

Finding Areas under the Standard Normal Curve (cont’d)

4-33

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

4-34

35

Standard Normal Distribution with ExcelTemplate

4-35

36

Finding Areas under the Standard Normal Curve (cont’d)

4-36

37

Finding Z Points ona Standard Normal Curve

4-37

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.

4-38

39

Brent’s Example of the Normal Distribution (cont.)

4-39

I put in mean and std. dev.

Probability of having less than two failures in a year = 0.4338

Probability of having more than three failures in a year = 0.0668

Probability of having between one and three failures in a year = 0.8998

40

Good day/night

• See you next time

4-40