4 - 1 © 1998 Prentice-Hall, Inc. Statistics for Managers Using Microsoft Excel, 1/e Statistics for...

4 - 1

© 1998 Prentice-Hall, Inc.

Statistics for Managers Using Microsoft Excel, 1/e

Statistics for Managers Using Microsoft Excel

Basic Probability & Discrete Probability Distributions

Chapter 4 Define experiment, outcome, event, sample

space, & probability Use a contingency table to find probabilities Describe 4 discrete probability distributions Find the probability of discrete random

variables

4 - 2



Thinking Challenge

What’s the probability of getting a head on the toss of a single fair coin? Use a scale from 0 (no way) to 1 (sure thing).

So toss a coin twice. Do it! Did you get one head & one tail? What’s it all mean?

4 - 3



Many Repetitions!*

0.00

0.25

0.50

0.75

1.00

0 25 50 75 100 125

Number of Tosses

Total Heads Number of Tosses

4 - 4



Introduction to Probability

Experiment Process of obtaining an observation,

outcome or simple event

Outcome Result of an experiment

4 - 5



Sample space depends on experimenter!

Experiment Process of obtaining an observation,

outcome or simple event

Outcome Result of an experiment

Sample space (S) Collection of all possible outcomes Defined by experimenter

Experiments & Outcomes

4 - 6



Outcome Examples

Toss a coin, note face Head, tail

Toss 2 coins, note faces HH, HT, TH, TT

Select 1 card, note kind 2, 2, ..., A (52)

Select 1 card, note color Red, black

Play a football game Win, lose, tie

Inspect a part, note quality Defective, good

Observe gender Male, female

Experiment Sample Space

4 - 7



Outcome Properties

Mutually exclusive 2 outcomes can not occur

at the same time Example: Both male &

female in same person

Collectively exhaustive 1 outcome in sample space

must occur Example: Male or female

Experiment: Observe gender

© 1984-1994 T/Maker Co.

4 - 8



Events

Any collection of outcomes Simple event

Outcome with 1 characteristic

Compound event Collection of outcomes or simple events 2 or more characteristics Joint event: special case

2 events occurring simultaneously

4 - 9



Event Examples

Sample space HH, HT, TH, TT

1 head & 1 tail HT, TH

Heads on 1st coin HH, HT

At least 1 head HH, HT, TH

Heads on both HH

Experiment: Toss 2 coins. Note faces.Experiment: Toss 2 coins. Note faces.

Event Outcomes in Event

4 - 10



Visualizing Sample Space

Listing S = {Head, Tail}

Contingency Table

4 - 11



2nd Coin1st Coin Head Tail Total

Head HH HT HH, HT

Tail TH TT TH, TT

Total HH, TH HT, TT S

Contingency Table

Experiment: Toss 2 coins. Note faces.Experiment: Toss 2 coins. Note faces.

S = {HH, HT, TH, TT} Sample space

Outcome (% or

count) Simpleevent (head on1st coin)

4 - 12



ColorType Red Black Total

Ace Ace &Red

Ace &Black

Ace

Non-Ace Non &Red

Non &Black

Non-Ace

Total Red Black S

Joint Events (Event A and Event B)

Sample space (S):

2R, 2R, 2B,

..., AB

Experiment: Draw 1 card. Note kind, color & suit.

Joint event Ace AND Black:

AB, AB

Simple event Ace:

AR,

AR,

AB,

ABSimple event black: 2B, ..., AB

4 - 13




Ace Ace &Red

Ace &Black

Ace

Non-Ace Non &Red

Non &Black

Non-Ace

Total Red Black S

Either-Or Events (Event A or Event B)

Sample space (S):

2R, 2R,

2B, ..., AB


Joint eventAce ORBlack:

AR, ..., AB,2B, ..., KB

Simple event Ace:

AR,

AR,

AB,

AB

Simple event Black: 2B, ..., AB

4 - 14



Special Events

Null event Club & Diamond on

1 card draw

Complement of event For event A, all

events not in A: A Mutually exclusive event

Events do not occur simultaneously

Null Event

Q

Q

4 - 15



What is Probability?

Numerical measure of likelihood that event will occur P(Event) P(A) Prob(A)

4 - 16



What is Probability?

Numerical measure of likelihood that event will occur P(Event) P(A) Prob(A)

Lies between 0 & 1

Sum of events is 1

1

.5 0

Certain

Impossible© 1984-1994 T/Maker Co.

4 - 17



Assigning Event Probabilities

a priori classical method

Empirical classical method

Subjective method

4 - 18



a priori Classical Method

Prior knowledge of process

Before experiment

P(Event) = X / T X = No. of event outcomes T = Total outcomes in sample space Each of T outcomes is equally likely

P(Outcome) = 1/T

© 1984-1994 T/Maker Co.

4 - 19



Empirical Classical Method

Actual data collected

After experiment

P(Event) = X / T Repeat experiment

T times Event observed X times

Also called relative frequency method

Of 100 parts inspected, only 2 defects!

4 - 20



Subjective Method

Individual knowledge of situation

Before experiment

Unique process Not repeatable

Different probabilities from different people

© 1984-1994 T/Maker Co.

4 - 21



Thinking Challenge

that a box of 24 bolts will be defective?

that a toss of a coin will be a tail?

that Tom will default on his PLUS loan?

that a student will earn an ‘A’ in this class?

that a new store on RTE. 1 will succeed?

Which method should be used to find the probability...

4 - 22



Joint Event Probability

Numerical measure of likelihood that joint event will occur

Can often use contingency table 2 variables only

Formula methods Addition rule Conditional probability formula Multiplication rule

4 - 23



EventEvent B1 B2 Total

A1 P(A1 and B1) P(A1 and B2) P(A1)

A2 P(A2 and B1) P(A2 and B2) P(A2)

Total P(B1) P(B2) 1

Event Probability Using Contingency Table

Joint Probability Marginal (Simple) Probability

4 - 24




Ace 2/52 2/52 4/52

Non-Ace 24/52 24/52 48/52

Total 26/52 26/52 52/52

Contingency Table Example


P(Ace)

P(Ace AND Red)P(Red)

4 - 25



Thinking Challenge

What’s the probability?

P(A) =

P(D) =

P(C and B) =

P(A or D) =

P(B and D) =

EventEvent C D Total

A 4 2 6

B 1 3 4

Total 5 5 10

4 - 26



Solution*

The probabilities are:

P(A) = 6/10

P(D) = 5/10

P(C and B) = 1/10

P(A or D) = 9/10

P(B and D) = 3/10


A 4 2 6

B 1 3 4

Total 5 5 10

4 - 27



Addition Rule

P(A or B) = P(A) + P(B) - P(A and B)

For mutually exclusive events:

P(A or B) = P(A) + P(B)

4 - 28




Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Addition Rule Example


P(Ace OR Black) = P(Ace)+P(Black) - P(Ace AND Black)

4

522652

252

2852

2 2 2452

( )

4 - 29



Thinking Challenge

Using the Addition Rule, what’s the probability?

P(A or D) =

P(B or C) =Event

Event C D Total

A 4 2 6

B 1 3 4

Total 5 5 10

4 - 30



Solution*

Using the Addition Rule, the probabilities are:

P(A or D) = P(A) + P(D) - P(A and D)

610

5

102

109

10

P(B or C) = P(B) + P(C) - P(B and C)

4

105

101

108

10

4 - 31



Conditional Probability

Event probability given that another event occurred

Revise original sample space to account for new information Eliminates certain outcomes

P(A | B) = P(A and B) P(B)

4 - 32




Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Conditional Probability Using Contingency Table

Experiment: Draw 1 card. Note kind, Experiment: Draw 1 card. Note kind, color & suit.color & suit.

P(Ace | Black) = P(Ace and Black)

P(Black)

2 5226 52

226

//

Revised sample

space

4 - 33



Event occurrence does not affect probability of another event e.g., toss 1 coin twice

Causality not implied Tests for independence

P(A | B) = P(A) P(A and B) = P(A)*P(B)

Statistical Independence

4 - 34



Thinking Challenge

Using the table then the formula, what’s the probability?

P(A|D) =

P(C|B) =

Are C & B independent?


A 4 2 6

B 1 3 4

Total 5 5 10

4 - 35



Solution*

Using the formula, the probabilities are:

P(A | D) = P(A and D)P(D)

2 105 10

25

//

P(C | B) = P(C and B)P(B)

P(C) = 510

1 104 10

14

14

//

Dependent

4 - 36



Multiplication Rule

P(A and B) = P(A)*P(B|A)

= P(B)*P(A| B)

For independent events:

P(A and B) = P(A)*P(B)

4 - 37




Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Multiplication Rule Example


P(Ace and Black) = P(Ace) P(Black |Ace)4

52

2

4

2

52

4 - 38



Thinking Challenge

Using the Multiplication Rule, what’s the probability?

P(C and B) =

P(B and D) =

P(A and B) =


A 4 2 6

B 1 3 4

Total 5 5 10

4 - 39



Solution*

Using the Multiplication Rule, the probabilities are:

P(A and B) = P(A) P(B|A) 0

P(C and B) = P(C) P(B|C) =

P(B and D) = P(B) P(D|B) =

510

15

110

410

34

310

4 - 40



Discrete Random Variables

4 - 41



Thinking Challenge

You’re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What’s the chance you’ll get it right?

If you guessed on all 33 questions, what would be your grade? Pass?

4 - 42



Data Types

Data

Numerical(Quantitative)

Categorical(Qualitative)

Discrete Continuous

4 - 43



Random Variable

A numerical outcome of an experiment Number of tails in 2 coin tosses

Observe 0, 1, or 2 tails

Discrete random variable Whole number (0, 1, 2, 3 etc.) Obtained by counting Usually finite number of values

Poisson random variable is exception ()

4 - 44



Discrete Random Variable Examples

Experiment RandomVariable

PossibleValues

Make 100 sales calls # Sales 0, 1, 2, ..., 100

Inspect 70 radios # Defective 0, 1, 2, ..., 70

Answer 33 questions # Correct 0, 1, 2, ..., 33

Count cars at tollbetween 11:00 & 1:00

# Carsarriving

0, 1, 2, ...,

4 - 45



Discrete Probability Distribution

List of all possible [ Xi, P(Xi) ] pairs Xi = Value of random variable (outcome)

P(Xi) = Probability associated with value

Mutually exclusive (no overlap) Collectively exhaustive (nothing left out)

0 P(Xi) 1

P(Xi) = 1

4 - 46



Discrete Probability Distribution Example

Probability Distribution

Values, Xi Probabilities, P(Xi)

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

Experiment: Toss 2 coins. Count # tails.

© 1984-1994 T/Maker Co.

4 - 47



Visualizing Discrete Probability Distributions

.00

.25

.50

0 1 2

X

P(X)

# Tails f(Xi)Count

P(Xi)

0 1 .251 2 .502 1 .25

P Xn

x n xp px n x( )

!!( )!

( )

1

{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }

ListingTable

Graph Equation

4 - 48



Summary MeasuresNotation for a Population

Expected value Mean of probability distribution Weighted average of all possible values = E(X) = Xi P(Xi)

Variance Weighted average squared deviation about

mean 2 = E[ (Xi (Xi P(Xi)

4 - 49



Summary Measures Calculation Table

Xi P(Xi) XiP(Xi) Xi - (Xi-)2 (Xi-)2 P(Xi)

Total XiP(Xi) (Xi-)2 P(Xi)

4 - 50



Thinking Challenge

You toss 2 coins. You’re interested in the number of tails. What are the expected value & standard deviation of this random variable, number of tails?

© 1984-1994 T/Maker Co.

4 - 51



Expected Value & Variance Solution*

Xi P(Xi) XiP(Xi) Xi - (Xi-)2 (Xi-)2 P(Xi)

0 .25 0 -1.00 1.00 .25

1 .50 .50 0 0 0

2 .25 .50 1.00 1.00 .25

= 1.0 2 = .50

4 - 52



Discrete Probability Distribution Function

Type of model Representation of some

underlying phenomenon

Mathematical formula Represents discrete

random variable Used to get exact

probabilities

P X x

x

x

( )

!

e-

4 - 53



Discrete Probability Distribution Models

DiscreteProbabilityDistribution

BinomialHyper-

GeometricNegativeBinomial

Poisson

4 - 54





BinomialHyper-


Poisson

4 - 55



Binomial Distribution

Number of ‘successes’ in a sample of n observations (trials)

# reds in 15 spins of roulette wheel # defective items in a batch of 5 items # correct on a 33 question exam # customers who purchase out of 100

customers who enter store

4 - 56



Binomial Distribution Properties

Two different sampling methods Infinite population without replacement Finite population with replacement

Sequence of n identical trials Each trial has 2 outcomes

‘Success’ (desired outcome) or ‘failure’

Constant trial probability Trials are independent

4 - 57



Binomial Probability Distribution Function

P Xn

x n xp px n x( )

!!( )!

( )

1

P(X) = Probability of X ‘successes’

n = Sample size

p = Probability of ‘success’

x = Number of ‘successes’ in sample (X = 0, 1, 2, ..., n)

4 - 58



Binomial Probability Distribution Example

P Xn

x n xp p

P X

x n x( )!

!( )!( )

( )!

!( )!. ( . )

1

34

3 4 35 1 53 4 3

.25

Experiment: Toss 1 coin 4 times in a row. Note # tails.

What’s the probability of 3 tails?

4 - 59



.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

Binomial Distribution Characteristics

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

n = 5 p = 0.1

n = 5 p = 0.5

E X np

np p

( )

( )1

Mean

Standard Deviation

4 - 60



Binomial Distribution Thinking Challenge

You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls (p = .20). If you call 12 people tonight, what’s the probability ofA. No sales?

B. Exactly 2 sales?

C. At most 2 sales?

D. At least 2 sales?

4 - 61



Binomial Distribution Solution*

A. P(0) = .0687

B. P(2) = .2835

C. P(at most 2) = P(0) + P(1) + P(2)= .0687 + .2062 + .2835= .5584

D. P(at least 2) = P(2) + P(3)...+ P(12)= 1 - [P(0) + P(1)] = 1 - .0687 - .2062= .7251

4 - 62





BinomialHyper-


Poisson

4 - 63



Poisson Distribution

Number of events that occur in an area of opportunity Events per unit

Example: Time, length, area, space

Examples # customers arriving in 20 minutes # strikes per year in the U.S. # defects per lot (group) of VCR's

4 - 64



Poisson Process

Constant event probability Average of 60/hr. is 1/min.

for 60 1-minute intervals

One event per interval Don’t arrive together

Independent events Arrival of 1 person does not

affect another’s arrival

© 1984-1994 T/Maker Co.

4 - 65



Poisson Probability Distribution Function

P Xx

x

( )!

e-

P(X) = Probability of X ‘successes’

= Expected (mean) number of ‘successes’

e = 2.71828 (base of natural logs)

x = Number of ‘successes’ per unit

4 - 66



Poisson Distribution Characteristics

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 2 4 6 8 10

X

P(X)

.0

.2

.4

.6

0 2 4 6 8 10

X

P(X)

= 0.5

= 6

E X

X P Xii

N

i

( )

( )1

Mean

Standard Deviation

4 - 67



Poisson Distribution Example

Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes?

© 1995 Corel Corp.

4 - 68



Poisson Distribution Solution

72 per hr. = 1.2 per min.

= 3.6 per 3 min.

interval

P Xx

P X

x

( )!

( ).

!

e

e

= .1912

-

- .6

436

4

3 4

4 - 69



Thinking Challenge

You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction?

© 1984-1994 T/Maker Co.

4 - 70



Poisson Distribution Solution: Finding *

75 words/min = (75 words/min)(60 min/hr)

= 4500 words/hr

6 errors/hr= 6 errors/4500 words

= .00133 errors/word

In a 255-word transaction (interval):

= (.00133 errors/word )(255 words)

= .34 errors/255-word transaction

4 - 71



Poisson Distribution Solution: Finding P(0)*

P Xx

P X

x

( )!

( ).

!

.

e

e

= .7118

-

- 34

034

0

0

4 - 72



Conclusion

Defined experiment, outcome, event, sample space, & probability

Used a contingency table to find probabilities

Described 4 discrete probability distributions

Found the probability of discrete random variables

4 - 1 © 1998 Prentice-Hall, Inc. Statistics for Managers Using Microsoft Excel, 1/e Statistics for...

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Transcript of 4 - 1 © 1998 Prentice-Hall, Inc. Statistics for Managers Using Microsoft Excel, 1/e Statistics for...