em561 lecture notes - part 1 of 3

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IE 360IE 360Probability and Statistics for Probability and Statistics for

EngineersEngineersEngineersEngineers

Gail W. DePuy, PhD, PERoom 311 JB Speed Building

Department of Industrial Engineering

University of LouisvilleUniversity of Louisville

Louisville, KY 40292 USA

Phone: 502-852-0115

Email: [email protected]

Introductions

• My background

M l f thi l

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• My goals for this class• Want students to UNDERSTAND material

• Not just plug-and-chug

• Wide variety of statistical software tools ’available – let’s use them

• Know what button to push

• Know how to interpret results

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Course Objective

This course is designed to give students an introduction to probability theory and

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introduction to probability theory and statistics, and exposure to how the theory may be applied in an engineering environment.

Help students gain an understanding of the effects of variability and the importance of properly considering variability

Course InformationText:Montgomery, D.C., and Runger, G.C., Applied Statistics and Probability for Engineers, 4th edition, John Wiley & Sons, New York, 2007

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Grading:HW#1 (due 7/18) 20%HW#2 (due 7/22) 20%Test 1 (7/20) 30%Test 2 (7/23) 30%

You should consider keeping this book for future reference.

I will follow the book fairly closely

Course Grades:90-100 A80-89 B70-79 C60-69 D<60 F

Note: I try to use different examples than the book so you’ll have many to study

Read the book!!

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Class ScheduleDate Chpt. in

4th edition Course Topics

7/13 1

2 Introduction and Descriptive Statistics Probability

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7/14 2 3

Probability (continued) Discrete Random Variables and Probability Distributions

7/15 3 4

Discrete Random Variables and Probability Distributions (continued) Continuous Random Variables and Probability Distributions

7/16 4 6

Continuous Random Variables and Probability Distributions (continued) Random Sampling and Data Description

7/18 6 7 8

Random Sampling and Data Description (continued) Point Estimation Confidence Intervals Homework #1 reviewHomework #1 review

7/20 Exam 1 (covers chpts. 1,2,3,4) open book 7/21

9 Exam 1 review Test of Hypotheses

7/22 11

Linear Regression Homework #2 review

7/23 Exam 2 (covers chpts. 6,7,8,9,11) open book

Class Policies• Homework Assignments

– Can work in groups of 3. Submit one solution set with all names.

• Test

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– There will be NO communicating of any kind (talking, sharing notes or books, sharing calculators, etc) with anyone else.

– Open book, open notes.

• Extra Credit– None

• BlackboardBlackboard– Check Bb often - Lecture slides will be posted to Bb before they are

covered in class– Check your UofL email account often

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Statistical Software

• Lots of different statistical software packages available

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packages available

• We’ll use:– Excel

– Minitab• See file ‘Minitab Download from IT.ppt’ in ‘Course

Documents’ folder on Bb– Be sure to read last slide of instructions

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?

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Science of dataScience of chance, Science of data Science of chance,uncertainties

what is possible what is probable

collecting, processing,presentation, analyzinginterpretation of data

mathematical formulas numbers with context

Probability Theory

• Branch of mathematics developed to deal with uncertainty

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with uncertainty

• Scientific tool dealing with chance

• Initially developed in 17th century to analyze gambling games then used to analyze mortality tables in medical professionmortality tables in medical profession

• Probability provides the framework for the study and application of statistics.

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Statistics• The field of statistics deals with the collection, presentation,

analysis, and use of data to:M k d i i

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– Make decisions

– Solve problems

– Design products and processes

• Statistical techniques are useful for describing and understanding variability.

• Statistics gives us a framework for describing thisStatistics gives us a framework for describing this variability and for learning about potential sources of variability.– Known source of variation

– Random variation

Random Experiments

• System output

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y paffected by controllable and uncontrollable (noise) variables

• Examples: baseball hitting car brakinghitting, car braking distance, concrete strength. – Controllable variables?

– And random variation!

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Random Experiments

• Definition

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How to mathematically model the system and its variation?

How to use model to draw conclusions or make predictions?

Variation

• Goal: Understand, quantify, and model variation

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variation

• Why? When we incorporate variation into our thinking and analysis, we can make informed judgments that are not invalidated by the variationy

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Effect of Variation• How many call center employees to have so

customers are not put on hold?

Collect c stomer data

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• Collect customer data: • Avg. time between customer calls= 4 min

• Avg. time to serve customer = 3.5 mins

• Based on averages ……..

• 1 employee needed

0 5 10 15 20 25 Time

Effect of Variation• But what about variation?

– Less than 4 min between calls

M th 3 5 i t t

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– More than 3.5 min to serve customer

0 5 10 15 20 25 Time

• Now more than 1 employee needed

• Variation causes disruptions in the system!

• Goal: Understand, quantify, and model variation

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Probability Theory

• Branch of mathematics developed to deal with uncertainty

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with uncertainty

• Scientific tool dealing with chance

• Probability provides the framework for the study and application of statistics.

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Experiments

• In general, an experiment is any process for which more than one outcome is possible

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which more than one outcome is possible

• Probability theory provides mathematical structure for understanding/explaining the chances or likelihoods of the various outcomes actually occurringy g

• First step is to list all possible outcomes– Called Sample Space

2-1.2 Sample Spaces

• Definition

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• Sample space for roll of 1 die?

• Sample space for flip of 2 coins?

• Sample space for # tails in 2 coin flips?

• Sample space for car brake distance?

• Sample space for your grade in this course?

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Sample Space

• Sample Space is either discrete or continuous• Discrete – finite or countable set of outcomes

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Discrete finite or countable set of outcomes

• Continuous – interval (finite or infinite) of real numbers

• Discrete or Continuous sample space?– Sample space for roll of 1 die?

– Sample space for flip of 2 coins?Sample space for flip of 2 coins?

– Sample space for # tails in 2 coin flips?

– Sample space for car brake distance?

– Sample space for your grade in this course?

2-1.2 Sample Spaces• Sample spaces can also be described

graphically with tree diagrams.

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• e.g. digital communication system. 3 messages sent. Each message either received on time or late. Sample space for all 3 messages

OOO OOL8 OutcomesOOO OOL

OLO OLL

LOO LOL

LLO LLL

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2-1.3 Events• Definition

• An event is a subset of the sample space of a random experiment

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experiment

• An event is a collection of related outcomes from a random experiment

• Example: Deck of 52 cards. Draw 1 card.

• Sample space?

• A is the event I draw a heart• A = { }

• B is the event I draw a 4• B = { }

2-1.3 EventsCombinations of Events (a.k.a. Joint Events)• Union of 2 events consists of all outcomes contained in

either of the 2 events

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– Union denoted E1 U E2

– Read “event 1 OR event 2”

• Intersection of 2 events consists of all outcomes contained in both of the 2 events– Intersection denoted E1 ∩ E2

Read “e ent 1 AND e ent 2”– Read “event 1 AND event 2”

• Complement of an event is the set of all outcomes not in the event– Complement denoted E1’

– Read “NOT event 1”

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2-1.3 Events

• Example: Deck of 52 cards. Draw 1 card.• A is the event I draw a heart

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A is the event I draw a heart

• B is the event I draw a 4

• How to denote? Event I draw the 4 of hearts?• Event I draw the 4 of hearts?

• Event I draw any suit other than hearts?

• Event I draw a 4 or a heart?

• Event I draw a 4 of any suit other than hearts?

2-1.3 Events• 3 car manufacturers were asked to submitted 1 car of each

model they produce for emissions testing.

• Each car either conforms or does not conform to current

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Each car either conforms or does not conform to current emissions standards. Conforms

Yes NoFord 40 15GM 50 20Toyota 20 5

• Let A denote the event a car is a Ford• Let B denote the event a car conformed

• Determine the number of car models in A• Determine the number of car models in A ∩ B• Determine the number of car models in A U B• Determine the number of car models in A’ ∩ B

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2-1.3 Events

• Definition

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• Example: Deck of 52 cards Draw 1 card• Example: Deck of 52 cards. Draw 1 card.• A is the event I draw a king

• B is the event I draw a 4

• A ∩ B = ?

Venn Diagrams

Mutually exclusive

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2-2 Probability• Probability - Likelihood of an experimental

outcome or event occurring

Probability of Outcome 1 denoted P(O ) or p

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• Probability of Outcome 1 denoted P(O1) or p1

• All probability values are between 0 and 1 (inclusive)• 0 ≤ P(O1) ≤ 1

• P(O1) = 0 indicates outcome O1 will NOT occur.P(O1) 0 indicates outcome O1 will NOT occur. Outcome O1 is impossible

• P(O1) = 1 indicates outcome O1 WILL occur. Outcome O1 is guaranteed to occur

CANNOT have a probability >1 or <0

2-2 Probability• Remember an event is a subset of the sample

space of a random experiment

P b bili f E d d P(E)

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• Probability of Event denoted P(E)

• All probability values are between 0 and 1 (inclusive)• 0 ≤ P(E) ≤ 1

• P(E) = 0 indicates event will NOT occur Event is• P(E) = 0 indicates event will NOT occur. Event is impossible

• P(E) = 1 indicates event WILL occur. Event is guaranteed to occur

CANNOT have a probability >1 or <0

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2-2 Probability• The sum of all the probability values over all

outcomes in the sample space is 1.• P(O1) + P(O2) + P(O3) + • • • + P(O ) = 1

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P(O1) + P(O2) + P(O3) + + P(On) 1

• Some outcome in S will occur (definition of S)

• P(O1) + P(O2) + P(O3) + • • • + P(On) = 1• P(O) = 1 – P(O’) ← Very useful ( ) ( ) y

• Similarly P(E) = 1 - P(E’)

• Suppose S=(1,2,3,4,5,6,7,8). Find P(Outcome ≤ 6)

• P(Outcome ≤ 6) = P(1)+P(2)+P(3)+P(4)+P(5)+P(6)

= 1 - P(Outcome > 6) = 1 – [P(7)+P(8)]

2-2 Probability• The sum of all the probability values over all

outcomes in the sample space is 1.• P(O ) + P(O ) + P(O ) + • • • + P(O ) = 1

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• P(O1) + P(O2) + P(O3) + • • • + P(On) = 1

• Read “Prob of outcome O1 OR outcome O2 OR outcome O3 • • • OR outcome On occurring is 1”

• Some outcome in sample space WILL occur (definition of S)

Note OR indicated by + Later we’ll see AND• Note OR indicated by + Later we’ll see AND indicated by *• Look at this logically and mathematically…….

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2-2 Probability• OR (Union) indicated by + AND (intersection) indicated by *

• P(outcome O1 OR outcome O2 ) >? <? P(outcome O1)

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P(outcome O1 OR outcome O2 ) >? <? P(outcome O1)– Prob UofL wins Men’s OR Women’s NCAA BB

Championship– Function of sum or product of individual probabilities?

• Remember 0 ≤ P(O) ≤ 1

• P(outcome O1 AND outcome O2 ) >? <? P(outcome O1)– Prob UofL wins Men’s AND Women’s NCAA BB

Championship– Function of sum or product of individual probabilities?

• Remember 0 ≤ P(O) ≤ 1

2-2 Probability

• Used to quantify likelihood or chance of experimental outcome or event occurring

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• The larger the probability, the more likely it is to happen

• Interpreted as the long-term, relative frequency of experimental outcome or event occurring– Repeat experiment many, many times

– Count the number of times each outcome occurs• Probability of an outcome is the # times outcome occurred

relative to the # of experimental trials

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2-2 Probability

• Roll 1 die. Probability I roll a 3?– Total # outcomes in sample space?

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– # outcomes that meet above criteria?

• Choose 1 card from a deck of 52. Probability I choose the ace of spades?

• Choose 1 card from a deck of 52. Probability I choose an ace (of any suit)?

• Choose 1 card from a deck of 52. Probability I choose a spade?

In the above examples, each outcome is equally likely. Not always the case ………

2-2 Probability• 500 computer chips are collected from 2 production lines

for quality inspection. 220 chips are from line 1 and 280 chips are from line 2. A quality inspector chooses a chip at

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random. Probability the chip was produced on line 2?

• Customers can order small, medium, or large drink. Of 1000 observed customers, 300 ordered small, 200 ordered medium, and 500 ordered large. Probability a customer will order a large drink?

R ll 2 di P b bilit th i 5? P b bilit th• Roll 2 dice. Probability the sum is 5? Probability the sum is 7?

• Lottery ticket consists of 6 numbers (1-40) with no repeated numbers. Probability I win if I buy 1 ticket? • # outcomes in sample space? See next discussion……

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2.1-4 Counting number of outcomes in sample space

• Not always convenient (or necessary) to LIST all outcomes in sample space

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outcomes in sample space

• BUT we do need to know HOW MANY outcomes are in the sample space to compute probabilities

• 3 counting techniquesMultiplication rule– Multiplication rule

– Permutations

– Combinations

2.1-4 The Multiplication Rule

• If an experiment consists of a sequence of kt i hi h

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steps in which • there are n1 possible outcomes for the first step,

• n2 possible outcomes for the second step, and so on,

• then the total number of experimental outcomes is given by (n1)*(n2)* . . . *(nk).

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2.1-4 The Multiplication RuleTrue/False test with 5 questions. How many different test

solutions? TTTTT, TFFFF, TFTFT, …..

• The 1st question can be answered in 2 ways: n1 = 2

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• The 2nd question can be answered in 2 ways: n2 = 2

• The 3rd question can be answered in 2 ways: n3 = 2

• The 4th question can be answered in 2 ways: n4 = 2

• The 5th question can be answered in 2 ways: n5 = 2

• Then, the number of ways the five questions can be answered is

n1 * n2 * n3 * n4 * n5 = 2*2*2*2*2 = 32

• Probability I get all 5 questions right if I guess on all?

2.1-4 The Multiplication Rule

• The lunch special consists of a soup, sandwich and beverage for $5.00

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g

• There are 3 different soups, 2 different sandwiches, and 4 different beverages

• How many different lunch specials?

• A helpful graphical representation of the sample space is a tree diagram.

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Tree Diagram

Soup (3) Sandwich (2) Drink (4)

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2.1-4 Sampling With or Without Replacement?

• May need to consider whether the experiment is d ith l t ith t l t

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done with replacement or without replacement

• 2 cards drawn from deck of 52.

• With replacement the total number of outcomes in th l i 52*52 2704the sample space is 52*52 = 2704

• Without replacement the total number of outcomes in the sample space is 52*51 = 2652

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2.1-4 Permutations

• A permutation of outcomes is an orderedsequence of the outcomes

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– Suppose S={1,2,3} sequences 123, 132, 213, 231, 312, 321 all different (order matters)

• The number of permutations of n different outcomes is n! (read ‘n factorial’) whereoutcomes is n! (read n factorial ) where

n! = n*(n-1)*(n-2)*….*(2)*(1)

e.g. 6! = 6*5*4*3*2*1 = 720

2.1-4 Permutations

Permutations of subsets

• The number of permutations of subsets of r

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The number of permutations of subsets of r outcomes selected from a set of n different outcomes is

)!(

!rn

nPnr −=

)(

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2.1-4 Permutations• How many sets of three medal winners (gold,

silver, bronze) can there be in a 100m dash with 8 ?

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8 runners ?

• Answer:• P8

3 = 8!/(8-3)! = 8!/5! = (8)(7)(6) = 336

• We have 8 choices for the gold-medal winner, 7 choices for the silver medal winner and 6 choices forchoices for the silver-medal winner and 6 choices for the bronze-medal winner.

We can use the multiplication rule to find the answer:There are 8*7*6 = 336 possible sets of winners.

2.1-4 PermutationsPermutations of Similar Objects• The number of n = n1 + n2 + ••• + nr objects of which n1 are

f t f d t d f th

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of one type, n2 are of a second type, …, and nr are of an rth

type is:

• e.g. Taco Bell gets an order for 6 tacos, 3 taco supremes

!!!!

!

321 rnnnnn

⋅⋅⋅

and 2 burritos. How many different processing sequences exist?• SSTTBTTBTTS, BBSSSTTTTTT, SSSTTTBBTTT, …

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2.1-4 Combinations

• A combination of outcomes is an unorderedsequence of the outcomes

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– Suppose S={1,2,3} sequences 123, 132, 213, 231, 312, 321 all the same (order does not matter)

• The number of combinations of subsets of r outcomes selected from a set of n different outcomes isoutcomes is

)!(!!

rnrn

rn

C nr −

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

2.1-4 Combinations

• Determine the number of ways in which a manufacturer can choose 2 of 7 locations for new

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warehouses.

• Back to my lottery question…..• Lottery ticket consists of 6 numbers (1-40) with no

repeated numbers.

• Probability I win if I buy 1 ticket? • # outcomes in the sample space?

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Permutations vs Combinations

• More permutations (ordered) or combinations (unordered) for n objects taken r at a time?

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)!(

!rn

nPnr −=

)!(!!

rnrn

rn

C nr −

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

• How many different orderings of r objects? r!

End of counting techniques. Now back to probabilities….

2-3 Addition Rules• Can often find probabilities of joint events (union,

intersection, complement) from probabilities of individual outcomes or events

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individual outcomes or events

• Use Addition Rule to find probability of a union

Remember: Union = ‘OR’ → sum of individual probabilities

P(AUB) > P(A) P(AUB) > P(B)

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2-3 Addition Rules• 3 car manufacturers were asked to submitted 1 car of each

model they produce for emissions testing.

• Each car either conforms or does not conform to current

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Each car either conforms or does not conform to current emissions standards. Conforms

Yes NoFord 40 15GM 50 20Toyota 20 5

•Let A denote the event a car is a Ford•Let B denote the event a car conformed

•Probability a randomly selected car is a Ford•Probability a randomly selected car is a Ford and conforms•Probability a randomly selected car is a Ford or conforms•Probability a randomly selected car is not a Ford and conforms

Alternative Interpretations of Probability

• Probability (0.0000 – 1.0000)P b bilit t 1

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– Probability outcome 1 occurs

– What is the probability a randomly selected car is a Ford?

• Proportion (0.0000 – 1.0000)– Proportion outcome 1 occurs

– What proportion of randomly selected cars are Ford?

55/150 = 0.36667

0.36667What proportion of randomly selected cars are Ford?

• Long-run Percentage (0.0000% - 100.0000%)– Percentage of time outcome 1 occurs

– What percentage of time is a Ford randomly selected?36.667%

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2-3 Mutually Exclusive• Remember

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• Therefore P(A∩B)=0 and the addition rule becomes:becomes:

2-3 Addition Rules

• Rules for probability of 2 events can be extended for 3 (or more) events

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extended for 3 (or more) events

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2-4 Conditional Probability

• Often probabilities need to be reevaluated as additional information becomes available

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• Conditional Probabilities often used

• Written P(B|A)

• Read “probability of event B occurring, given that event A has already occurred”

• Read “given event A has occurred, the probability of event B occurring)

• Read “probability of B given A” or “given A, the probability of B)

2-4 Conditional Probability• Let D denote the event that a part is functionally defective

and let F denote the event that a part has a visible surface flaw.

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• The probability that a part is functionally defective, given that the part has a visible surface flaw denoted P(D|F)

Table 2-3 Parts Classified

Surface Flaws

• Find P(D|F). Find P(F|D). Find P(D’|F). Find P(D|F’).

Yes (event F) NoDefective Yes (event D) 10 18

No 30 342

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2-4 Conditional Probability

• In the previous example, the conditional probabilities were calculated directly.

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• We can also use the formal definition of conditional probability

• Find P(D|F). Find P(F|D). Find P(D’|F). Find P(D|F’).

2-5.1 Multiplication Rule

Probability of intersection

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• Note: Just a rewrite of conditional probability formula

Remember: Intersection = ‘AND’ → product of individual probabilities

P(AUB) < P(A) P(AUB) < P(B)

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2-5.2 Total Probability Rule• Given probability of event under several conditions

(conditional probabilities). How to find probability of event?

• e.g. Given P(B|A) and P(B|A’). How to find P(B)?

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e.g. Given P(B|A) and P(B|A ). How to find P(B)?

• P(B)=P(A∩B) U P(A’∩B)

• Remember multiplication rule:

P(A∩B)=P(B|A)P(A)

• Need P(B|A), P(A),

P(B|A’) and P(A’) Figure 2-15 Partitioning an event into two mutually exclusive subsets.

2-5.2 Total Probability Rule• Total Probability Rule (2 events)

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2-5.2 Total Probability Rule• Extend to multiple events

• Total Probability Rule (multiple events)

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2-5.2 Total Probability Rule• Example

• The surface roughness of a metal part increases

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g pwith blade wear. Only 3% of parts made with a new blade exhibit roughness, 5% of parts made with a moderately used blade exhibit roughness, and 9% of parts made with an old blade exhibit roughness.

C rrentl on shop floor 30% of blades are ne• Currently on shop floor, 30% of blades are new and 25% of blades are old.

• What proportion of parts made will exhibit roughness?

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2-6 Independence

• Events are independent if knowledge of one event occurring (or not occurring) does not effect the

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probability of another event occurring

2-6 Independence

• Batch of 600 manufactured parts. 40 parts are defective.

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• Two parts randomly selected

• Let A = event first part defective

• Let B = event second part defective

Are the events A and B independent?• Are the events A and B independent?• Without replacement

• With replacement

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2-6 Independence

• We often like to assume independence to simplify analysis

• If independent, P(A∩B) = P(A)P(B)

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p , ( ) ( ) ( )

• If not independent, P(A∩B) = P(B|A)P(A) or P(A|B)P(B)

• Extend independence to multiple events

2-6 Independence• This circuit operates only if there is a path of functional

devices from left to right. The probability each devices fails is shown. Assume that devices fail independently. What is

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the probability the circuit operates?

0.08

0.09

0 11

0.07

0.130.11

What if we cannot assume independence?

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2-6 Independence

• Toss a fair coin. The trials are independent (i.e. knowing what I got on the last toss of the coin (or ALL i t ) d NOT ff t b bilit

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ALL previous tosses) does NOT affect probability of next toss)

• Probability on any toss I get a tails? P(T) = ?

• Suppose I have tossed the coin 99 times and got Tails every time. Probability the 100th time I toss the coin it will be a Tail (or Head) = 0.5

• Probability of getting the sequence HHHHH? TTTTT? HTHTH? THTHT? HTTTH? THHHT?

Independence vs Mutually Exclusive

• NOT the same thing!!!!!

E t i d d t if k l d f t

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• Events are independent if knowledge of one event occurring (or not occurring) does not effect the probability of another event occurring, i.e. P(A|B)=P(A)

• Events are mutually exclusive if they have no ∩ Øoutcomes in common, i.e. A∩B=Ø

• Mutually exclusive based on outcomes. Independence based on probability model.

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Independence vs Mutually Exclusive

• In fact, if 2 events are mutually exclusive they cannot be independent

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• e.g. roll a die.

• Event A = roll an even number {2,4,6}

• Event B = roll an odd number {1,3,5}

• A and B are mutually exclusive.

• P(A)=0.5 and P(B)=0.5.

• If A and B were independent then P(A|B)=P(A) however P(A|B) = 0 ≠ 0.5

• A and B are mutually exclusive but not independent

2-7 Bayes’ Theorem• Often information presented in terms of one conditional

probability, but we are interested in another conditional prob

B ’ Th f 2 t

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• Bayes’ Theorem for 2 events

• Bayes’ Theorem for multiple events

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2-7 Bayes’ Theorem• Cars are assembled in one of 4 possible locations.

• Plant I supplies 20% of the cars; Plant II 24%;

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pp ; ;Plant III 25%, and Plant IV 31%.

• Each new car has a one year warranty. The following data has been collected regarding warranty claims:warranty claims:

P(claim | plant I) = 0.05 P(claim | plant II) = 0.11

P(claim | plant III) = 0.03 P(claim | plant IV) = 0.08

2-7 Bayes’ Theorem

• Suppose a customer walks into a car dealer and buys a car – the customer does not know which

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plant the car came from.

1. What is the probability a customer purchases a car from plant I and that it does not require a claim on its warranty?on its warranty?

•Find ?

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2-7 Bayes’ Theorem2. Probability a claim on the warranty will be made on a

customer’s car ?• Find ?

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Find ?

3. Suppose a customer makes a claim on their warranty, what is the probability the car was made at Plant I?• Find ?

4. Suppose a customer does not make a claim on their warranty, what is the probability the car was made at Plant III?• Find ?

Chapter 2 Probability

• End of new concepts in chapter 2

• So far, not too difficult to know which formula to

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So far, not too difficult to know which formula to use for examples

• HOWEVER in real world (and on test) you will have to decide which formula(s) to use

• Let’s do some ‘random access’ practice problems– Write down what you know (given info)

– Write down what you are looking for

– Then maybe easier to see how to get from what you know to what you are looking for

38

Review of some probability rules

• Addition Rule • P(AUB) = P(A) + P(B) P(A∩B)

EM 561 GW DePuy 75

• P(AUB) = P(A) + P(B) - P(A∩B)

• Multiplication Rule• P(A∩B) = P(B|A) P(A) = P(A|B) P(B)

• Total Probability Rule• P(B) = P(B|A) P(A) + P(B|A’) P(A’)

• Bayes Theorem• P(A|B) = P(B|A) P(A) / P(B)

More Chapter 2 Examples

• The probability an aluminum can has a flaw on its side is 0.02, the probability the can has a flaw on

EM 561 GW DePuy 76

the top is 0.03, and the probability the can has a flaw on both the side and top is 0.01.

1. What is the probability a randomly chosen can has a flaw?

2 What is the probability the can has no flaw?2. What is the probability the can has no flaw?

3. What is the probability the can has a flaw on the top but not the side?

39

More Chapter 2 Examples• A particular model of car can be made with any of 3

engine sizes. Of all cars sold, 45% have the smallest engine 35% have the medium-sized engine and 20%

EM 561 GW DePuy 77

engine, 35% have the medium sized engine, and 20% have the largest engine. Of cars with the smallest engine, 10% fail an emissions test within 2 years of purchase, while 12% of those with the medium and 15% of those with the largest engine fail.

1 What is the probability a randomly chosen car will fail an1. What is the probability a randomly chosen car will fail an emissions test within 2 years?

2. A record for a failed emission test is chosen at random. What is the probability it is for a car with a small engine?

More Chapter 2 Examples• A lot (batch) of 50 spacing washers contains 30 washers

that are thicker than the target dimension. Suppose that three washers are selected at random, without replacement from the lot

EM 561 GW DePuy 78

replacement, from the lot.

1. What is the probability all three washers are thicker than the target?

2. What is the probability the third washer selected is thicker than the target if the first two washers selected are notthan the target if the first two washers selected are not thicker than the target?

3. What is the probability the third washer selected is thicker than the target?

40


• Need more Chapter 2 examples?

READ THE BOOK!

EM 561 GW DePuy 79

• READ THE BOOK!

• We covered all sections of Chapter 2

2-8 Random Variables• We often summarize the outcome from an

experiment by a simple number.

EM 561 GW DePuy 80

• Sample Space can be description of outcome OR often useful to associate a number with each outcome.

• Random variable associates a number with the outcome of a random experimentoutcome of a random experiment

41

2-8 Random VariablesEM 561 GW DePuy 81

IE 360 GW DePuy 82

42

3-1 Discrete Random Variables• Remember: Random Variable is a function that assigns a

number to each outcome in the sample space of a random experiment.

EM 561 GW DePuy 83

• Example: Toss a coin 3 times. Let X= random variable that represents the Number of Tails. X = { }

• Example: Office has 10 phone lines. Let X = r.v. that represents the number of lines in use. X= { }

• Example: Batch of 200 parts contain 6 that do not conform to customer specification Parts are selected successivelyto customer specification. Parts are selected successively, without replacement, until the first nonconforming part is obtained. Let X = r.v. that represents the number of parts selected until the first nonconforming part is obtained. X = { }

3-2 Probability Distributions• The probability distribution of a random

variable X is a description of the probabilities associated with the possible values of X

EM 561 GW DePuy 84

associated with the possible values of XProbability Distributions

Discrete Distribution

General

Continuous Distribution

GeneralNegative UniformGeneral

Binomial

Geometric

Negative Binomial

Poisson

Uniform

Normal

Exponential

Uniform

43

3-2 Probability Distributions and Probability Mass Functions

• The probability distribution of a random variable X is a description of the probabilities associated

ith th ibl l f X

EM 561 GW DePuy 85

with the possible values of X

• Distribution can be specified by a list of possible values and their probabilities OR a formula

• Example: Toss a coin 3 times. Let X= random variable that represents the Number of Tails• P(0) = P(HHH) = (.5)(.5)(.5) = 0.125

• P(1) = P(THH)+P(HTH)+P(HHT) = 0.375

• P(2) = P(TTH)+P(THT)+P(HTT) = 0.375

• P(3) = P(TTT) = 0.125


EM 561 GW DePuy 86

• Example: Toss a coin 3 times Let X= random variable thatExample: Toss a coin 3 times. Let X random variable that represents the Number of Tails

X 0 1 2 3f(x) 0.125 0.375 0.375 0.125 Σf(x) = 1?

44


• Batch of 5 parts: 3 good and 2 bad

EM 561 GW DePuy 87

• You keep drawing a part (without replacement) until you get a bad part

• Let X = # of draws to get a bad part

• Possible values of X?

• Probability mass function of X?

3-3 Cumulative Distribution Functions

• Definition

EM 561 GW DePuy 88

• Cumulative Distribution Function, F(x), for previous example?

45

3-4 Mean and Variance of a Discrete Random Variable

• Use Mean and Variance to summarize probability distribution of rv

EM 561 GW DePuy 89

distribution of rv• Mean is measure of center of distribution

• Variance is measure of spread or dispersion of distribution


• Mean = balance point of loading

L i di t hi h b bilit

EM 561 GW DePuy 90

• Longer arrows indicate higher probability

• Equal means but unequal variances

46


• Mean of this distribution? x 2 3 5 10f(x) 0.1 0.1 0.1 0.7

EM 561 GW DePuy 91

NO !! Why? ???0.5

410532)( =

+++=xE

0 1 2 3 4 5 6 7 8 9 10 11 12

0.8)7.0(10)1.0(5)1.0(3)1.0(2)()( =+++== ∑x

xxfxE


• Variance of this distribution? x 2 3 5 10f(x) 0.1 0.1 0.1 0.7

EM 561 GW DePuy 92

0 1 2 3 4 5 6 7 8 9 10 11 12

( ) )()()( 2 xfxExxVx∑ −=

8.9)7.0()810()1.0()85()1.0()83()1.0()82()( 2222 =−+−+−+−=xV

47


Variance of discrete r.v. (alternative formula)

V (X) E(X2) (E(X))2

EM 561 GW DePuy 93

• Var(X) = E(X2) – (E(X))2

• Where E(X2) = ∑x2 P(X=x)

• Remember E(X) = ∑x P(X=x)

Both variance and standard deviation are a measure of spread or dispersion of the data.

Standard deviation is in the same units as the data whereas variance is in units2


• Mean and variance DO NOT uniquely describe a distribution.

EM 561 GW DePuy 94

– Different distributions can have the same mean and variance

• However they are simple, useful summaries of the probability distribution.

48


• Batch of 5 parts: 3 good and 2 bad

EM 561 GW DePuy 95

• You keep drawing a part (without replacement) until you get a bad part

• Let X = # of draws to get a bad part

• Find Mean and Variance of X

3-2 Probability Distributions• The probability distribution of a random


EM 561 GW DePuy 96



General



Binomial

Geometric

Negative Binomial

Poisson

Uniform

Normal

Exponential

Uniform

49

3-2 Probability Distributions

• Probability Mass Functions for Common Discrete Probability Distributions (uniform discrete,

EM 561 GW DePuy 97

binomial, geometric, negative binomial, poisson) exhibit same basic properties as general p.m.f.

• 0 ≤ f(x) ≤ 1

• Σf(x) = 1

3-5 Discrete Uniform Distribution• Definition and Probability Mass Function

EM 561 GW DePuy 98

• Each outcome equally likely

50

3-5 Discrete Uniform DistributionEM 561 GW DePuy 99

3-5 Discrete Uniform Distribution

• Example: Thickness of plate glass parts measured to nearest mm. The thicknesses

EM 561 GW DePuy 100

are uniformly distributed between 42mm and 49mm.

1. Probability a part measures 45mm?

2. Probability a part is less than 47mm thick?

3. Find the mean and variance of the plate thickness

51

3-6 Binomial Distribution

Bernoulli Trial

E i t ith l 2 ibl t

EM 561 GW DePuy 101

• Experiment with only 2 possible outcomes– Yes/No, Go/No Go, Pass/Fail, Defective/Not

Defective, Heads/Tails, Win/Lose, etc.

– We label outcomes ‘success’ and ‘failure’

• Trials are independent

• Probability of a success, p, is constant for each trial

3-6 Binomial Distribution• Describes the number of successes, X, in n

Bernoulli trials

EM 561 GW DePuy 102

• Definition and Probability Mass Function

52


• Mean and Variance of Binomial Distribution

EM 561 GW DePuy 103

3-6 Binomial DistributionEM 561 GW DePuy 104

Figure 3-8 Binomial distributions for selected values of the parameters n and p.

Binomial Distribution described by 2 parameters, n and p

53


• Example: The quality control department inspects parts. Historically 10% of parts are defective.

EM 561 GW DePuy 105

Each day 12 parts are randomly selected for inspection.

1. What is the probability 4 parts are defective?

2. What is the probability more than 2 parts are defective?defective?

3. What is the expected number of defective parts?

3-7.1 Geometric Distribution• Describes the number of Bernoulli trials, X, until

the first success


EM 561 GW DePuy 106


• Mean and Variance of Geometric Distribution

54

3-7.1 Geometric DistributionEM 561 GW DePuy 107

Figure 3-9. Geometric distributions for selected values of the parameter p.

3-7.1 Geometric Distribution

• My car’s engine starts successfully on a given attempt with probability of 0.75.

EM 561 GW DePuy 108

g y

1. Probability the engine starts in 3 or fewer attempts?

2 Expected number of attempts required to2. Expected number of attempts required to start the engine?

55

3-7.2 Negative Binomial Distribution

• Describes the number of Bernoulli trials, X, until r success occur

EM 561 GW DePuy 109



• Mean and Variance of Negative Binomial Distribution

EM 561 GW DePuy 110

56

3-7.2 Negative Binomial DistributionEM 561 GW DePuy 111

Figure 3-10. Negative binomial distributions for selected values of the parameters r and p.


• Example: Suppose a company wishes to hire 3 new workers, and that each applicant

EM 561 GW DePuy 112

, ppinterviewed has a probability of 0.6 of being found acceptable.

1. Probability more than 6 applicants will need to be interviewed?

2. Expected number of interviews required?

57

3-9 Poisson Distribution

• Not related to Bernoulli trials

• Poisson distribution describes the number of

EM 561 GW DePuy 113

Poisson distribution describes the number of events that occur in an interval

• Interval could be time, distance, area, volume, etc.

• Count the number of events in an interval• # particles contaminating a semiconductor wafer

• # blemishes in 1 ft3 concrete

• # logons to computer site in 1 minute

• # customers to arrive to a bank in 1 hour



EM 561 GW DePuy 114

58


• Mean and Variance of Poisson Distribution

EM 561 GW DePuy 115

λ is average rate of occurrence of event

λ is average rate expressed in # / interval (i.e. #/hour, #/mm, #/kg, etc.)

3-9 Poisson DistributionEM 561 GW DePuy 116

Figure 3-14. Poisson distributions for selected values of parameter λ.

59

3-9 Poisson Distribution• Caution! Use Consistent Units

• Must use consistent units in calculation of probabilities, means and variances involving Poisson distribution

EM 561 GW DePuy 117

means, and variances involving Poisson distribution

• If you are interested in probability of X events occurring in interval, then λ must be expressed as rate for same interval• If X = # logons to computer site in 1 minute, then λ is avg. logon

rate in 1 minute

• If X = # logons to computer site in 8 minutes, then λ is avg. logon g p , g grate in 8 minutes

• If X = # logons to computer site in 2 hours, then λ is avg. logon rate in 2 hours

• Need to convert λ to proper units

3-9 Poisson Distribution• Suppose customers arrive at a bank according

to a Poisson distribution with mean rate 4 customers per 15 minutes

EM 561 GW DePuy 118

customers per 15 minutes.

1. Probability less than three customers arrive in a 15 minute interval?

2 P b bilit th 2 t i d i2. Probability more than 2 customers arrive during a 30 minute interval?

3. Probability 14 customers arrive between 10:00am and 11:00am?

60

How to do pmf calculations in Minitab• Minitab can calculate P(X=x) for Binomial,

Geometric, Negative Binomial, & Poisson distributions

EM 561 GW DePuy 119

distributions• Enter x values in Minitab worksheet

CalcProbability Distribution

Choose Probability Distribution (Binomial,

Minitab 15 English.lnk

Choose Probability Distribution (Binomial, Geometric, Poisson, etc)

• Choose ‘Probability’• Enter Parameter(s)• Enter/Choose Input Column

How to do pmf calculations in MinitabEM 561 GW DePuy 120

• Enter x values in Minitab worksheet

61

How to do pmf calculations in Minitab

Calc

EM 561 GW DePuy 121

Calc

Probability Distribution

Choose Probability Distribution (e.g. Poisson)

• Choose Probability

• Enter Parameter

• Enter/Choose Input Column

How to do pmf calculations in MinitabEM 561 GW DePuy 122

• P(X=x) values displayed in Sessions Window

62

How to do pmf calculations in Excel

• Excel can calculate P(X=x) for Binomial & Poisson distributions

EM 561 GW DePuy 123

• = poisson(x, mean, cumulative)• Find P(X=4) for Poisson with λ=7

• = poisson(4, 7, FALSE)

• = binomdist(x, trials, prob success, cumulative)( , , p _ , )• Find P(X=5) for Binomial with n=10 and p=0.3

• = binomdist(5, 10, .3, FALSE)

How to do pmf calculations in ExcelEM 561 GW DePuy 124

63



64

More Chapter 3 Problems• A test of weld strength involves loading welded

joints until a fracture occurs. For a certain type of weld, 80% of the fractures occur in the weld

EM 561 GW DePuy 127

itself, while the other 20% occur in the beam. Assume the location of fractures is independent from test to test.

1. What is the probability fewer than 5 weld fractures will be found in 9 tests?

2. What is the probability more than 3 beam fractures will be found in 7 tests?

3. What is the probability 5 weld fracture will be found in 9 or fewer tests?

More Chapter 3 Problems

• The number of messages received by a computer bulletin board is a Poisson r.v.

EM 561 GW DePuy 128

computer bulletin board is a Poisson r.v. with a mean of 7 messages per minute.

1. What is the probability 5 or fewer messages are received in a minute?

2 What is the probability more than 502. What is the probability more than 50 messages are received in 5 minutes?

65

More Chapter 3 Problems

• 1k, 3k, and 5k ohm resistors have been sampled from the production line and sent to QC for t ti A t b f 50 i t (25 1k h 15 3k

EM 561 GW DePuy 129

testing. A tub of 50 resistors (25 1k ohm, 15 3k ohm, 10 5k ohm) needs to be tested. The QC technician randomly, sequentially selects 2 resistors from the tub. Let X represent the number of 5k ohm resistors among the two selected resistors.

1. Find the probability distribution for X

2. What is the probability both of the selected resistors are 5k ohm?



READ THE BOOK!

EM 561 GW DePuy 130

• READ THE BOOK!

• We covered all sections of Chapter 3

66

IE 360 GW DePuy 131

4-1 Continuous Random Variables• Remember: Random Variable is a function that

assigns a number to each outcome in the sample space of a random experiment

EM 561 GW DePuy 132

space of a random experiment.

67

Probability Distributions• The probability distribution of a random


EM 561 GW DePuy 133



General



Binomial

Geometric

Negative Binomial

Poisson

Uniform

Normal

Exponential

Uniform

4-2 Probability Distributions and Probability Density Functions

• The probability distribution of a random variable X is a description of the probabilities associated

EM 561 GW DePuy 134

with the possible values of X

• Distribution can be specified by a formula

Figure 4-2 Probability determined from the area under f(x).

68


• Definition

EM 561 GW DePuy 135

• Same general idea as discrete distribution, now integrate instead of sum


• Example: The probability density function of the net weight (in lbs) of a can of chemical

EM 561 GW DePuy 136

the net weight (in lbs) of a can of chemical herbicide is f(x) = 2.0 for 49.75<x<50.25

• What is the probability a can weighs less than 50.1 lbs?

69


• For continuous r.v. does not make sense to talk about probability of a particular outcome occurring

EM 561 GW DePuy 137

about probability of a particular outcome occurring, but rather a range of outcomes

• Probability an exact outcome occurs = 0

• Example: Let X= weight of herbicide can (in lbs)• P(X<50.1) = P(X ≤ 50.1) = 0.70

4-3 Cumulative Distribution Function

• Definition

IE 360 GW DePuy 138

• Example: Cumulative distribution function for• Example: Cumulative distribution function for herbicide

⎪⎩

⎪⎨

⎧

≤≤≤

<=

xxx

xxF

25.50125.5075.490.2

75.490)(

70

4-4 Mean and Variance of a Continuous Random Variable

• Definition

EM 561 GW DePuy 139

4-4 Mean and Variance of a Continuous Random Variable

• Example: The probability density function of

EM 561 GW DePuy 140

the net weight (in lbs) of a can of chemical herbicide is f(x) = 2.0 for 49.75<x<50.25

• Find Mean and Variance of net weight of cans of herbicide

71

4-5 Continuous Uniform Distribution

• Definition

EM 561 GW DePuy 141

• Mean and Variance

4-5 Continuous Uniform DistributionEM 561 GW DePuy 142

Fi 4 8 C ti if b bilit d it f tiFigure 4-8 Continuous uniform probability density function.

Area under the curve = ?

Area of rectangle = length * height

72

4-5 Continuous Uniform Distribution

• A new battery supposedly with a charge of 1.5 volts actually has a voltage with a uniform di t ib ti b t 1 43 d 1 60 lt

EM 561 GW DePuy 143

distribution between 1.43 and 1.60 volts.

1. Probability the battery has a voltage less than 1.53 volts?

2. Probability the battery has a voltage higher than 1.47 volts?

3. Expected or mean voltage?

4. Suppose the specification for batteries is 1.5 ±0.04 volts. What proportion of batteries are out of spec?

4-6 Normal DistributionHistory• Most widely used distribution

EM 561 GW DePuy 144

• The sum or average of random variables becomes normal as n→large (Central Limit Theorem)

• proved by De Moivre in 1733 (lost for 100 years)• Re-discovered by Gauss (Gaussian distribution)• Many items in nature are result of the sum of

RV’s• We’ll discuss Central Limit Theorem later. Now

let’s learn how to calculate probabilities for a normal r.v.

73

4-6 Normal Distribution• Definition

EM 561 GW DePuy 145

Normal Distribution has 2 parameters: μ and σ2

4-6 Normal DistributionEM 561 GW DePuy 146

Fi 4 10 N l b bilit d it f ti fFigure 4-10 Normal probability density functions for selected values of the parameters μ and σ2.

Notation used: N(μ,σ2)

above e.g. N(5,1); N(5,4); N(15,1)

74

4-6 Normal Distribution• Normal Distribution is a symmetric distribution

EM 561 GW DePuy 147

4-6 Normal Distribution• How to find probabilities for normal r.v.?• Integrate f(x)?

EM 561 GW DePuy 148

g ( )• Use Tables• How many possible normal distributions?

– How many tables?

• One special normal distribution, standard normal distribution N(0 1)distribution N(0,1)– One table ☺

• Convert any normal X~N(μ,σ2) r.v. to standard normal Z~N(0,1) r.v. and use standard normal table

75

4-6 Normal DistributionEM 561 GW DePuy 149

4-6 Normal Distribution

• First let’s practice using standard normal table.

• Then we’ll talk about how to convert any normal

EM 561 GW DePuy 150

• Then we ll talk about how to convert any normal X~N(μ,σ2) r.v. to standard normal Z~N(0,1) r.v. (called standardizing)

• Table III in back of book– Shows cumulative probabilities for standard normal

distribution e.g. P(Z<z)

• Remember normal distribution is symmetric

76

EM 561 GW DePuy 151

EM 561 GW DePuy 152

Be careful reading table!

First page reads 0.09 to 0.00, second page

d 0 00 t 0 09reads 0.00 to 0.09

77

4-6 Normal Distribution1. P(Z < 1.56)2. P(Z < -0.86) Helpful hint & reminders:

EM 561 GW DePuy 153

( )3. P(Z > -1.37)4. P(Z > 2.12)5. P(-1.25 < Z < 0.37)6. P(0.22 < Z < 3.02)

• Draw a picture

• Norm Dist is symmetric

• Total area under curve = 1

• Table gives P(Z<z)7. P(Z < -4.3)8. find the value of z such that P(Z > z) = 0.019. find the value of z such that P(-z < Z < z) = 0.95

Table gives P(Z<z)


• We know how to use standard normal table.

N h t t l X N( 2) t

EM 561 GW DePuy 154

• Now, how to convert any normal X~N(μ,σ2) r.v. to standard normal Z~N(0,1) r.v. (called standardizing)

78


μ−=

XZ

IE 360 GW DePuy 155

• Standardizing finds the # of standard deviations(σ) X is from its mean (μ)

• Remember: Z~N(0 1) i e μ=0 and σ=1

σ

• Remember: Z~N(0,1) i.e. μ=0 and σ=1 • Z = 1.3 is 1.3 standard deviations above μ=0

• Z = -0.4 is 0.4 standard deviations below μ=0

4-6 Normal Distribution• X~N(10,4). Find P(X < 13)

• Z = (13 – 10)/2 = 1 5

EM 561 GW DePuy 156

Z (13 10)/2 1.5

• P(X < 13) = P(Z < 1.5) = 0.933193

79


• Example: Suppose X ~ N(5.2, 3.9)

1 Fi d P(X 4 8)

EM 561 GW DePuy 157

1. Find P(X < 4.8)

2. Find P(X > 8.0)

3. Find P(2.0 < X < 5.0)

4-6 Normal Distribution• The thickness of glass sheets produced by a

certain process are normally distributed with a f 3 00 f

EM 561 GW DePuy 158

mean of 3.00mm and a standard deviation of 0.12mm

1. Probability a glass sheet is thinner than 3.15mm?

2. Probability a glass sheet is thicker than 2.78mm?

3. Probability a glass sheet’s thickness is between 2.95 and 3.09mm?

80

4-8 Exponential Distribution• Definition

EM 561 GW DePuy 159

M d V i• Mean and Variance

4-8 Exponential DistributionEM 561 GW DePuy 160

Exponential distribution has 1 parameter: λ

81

4-8 Exponential Distribution• Exponential distribution is related to Poisson distribution

• If the number of events is distributed Poisson, then the ti b t t i E ti ll di t ib t d

EM 561 GW DePuy 161

time between events is Exponentially distributed

Number of events ~ P(λ)

ixi exf λλ −=)(

x1 x2 x3 x4 x5x1 x2 x3 x4 x5

4-8 Exponential Distribution

• Often easier to work with cumulative distribution function for exponential distribution rather than

EM 561 GW DePuy 162

integrating for each problem

Cumulative Distribution Function for Exponential Distribution

F(x) = P(X ≤ x) = 1 – e-λx for x≥0

• Be careful to use consistent units for X and λ• If λ in minutes, then X in minutes

• If λ in cm, then X in cm

82

4-8 Exponential Distribution

• Suppose customers arrive at a bank according to a Poisson distribution with mean rate 16

t h

EM 561 GW DePuy 163

customers per hour.

1. What is the probability the time until the next customer arrives is less than 10 minutes?

2. What is the probability the time until the next customer arrives is more than 15 minutes?

3. What is the probability the next customer will arrive in 2 to 3 minutes?

4. What is the average time between customer arrivals?

4-8 Exponential Distribution• An interesting property of an exponential random variable

is the lack of memory property.

EM 561 GW DePuy 164

• suppose no bank customers arrive from 12:00 to 12:15 • the probability that no customer arrive from 12:15 to 12:30 p y

(i.e. 15 minutes) is still 0.018316. • Because we have already been waiting for 15 minutes, we

feel that we are “due.” That is, the probability of an arrival in the next 15 minutes should be greater than 0.018316. However, for an exponential distribution this is not true.

83

4-8 Exponential DistributionLack of Memory Property

• Suppose an electronic component fails due to voltage surges. The time in hours between voltage surges follows

EM 561 GW DePuy 165

surges. The time in hours between voltage surges follows an exponential distribution with λ = 0.2 hr = 1 failure/5 hours. i.e. the expected time between failures is 1/λ = 5 hours

• Suppose an electronic component has been operational for 3 hours, how much longer would you expect the electronic component to be operational?

• Suppose an electronic component has been operational for 40 hours, how much longer would you expect the electronic component to be operational?

4-8 Exponential DistributionLack of Memory Property

• The memoryless property of the exponential distrib tion makes it attracti e for modeling man

EM 561 GW DePuy 166

distribution makes it attractive for modeling many processes, however it is unsuitable for modeling other processes.

• If an electronic component fails due to random voltage surges then it may be appropriate to model the failure time with an exponentialmodel the failure time with an exponential distribution.

• However, if the failure of an electronic component is due to wearout, then the exponential distribution would not be appropriate to model failure times.

84

How to do cdf calculations in Minitab• Minitab can calculate P(X<x) for Normal, Uniform,

and Exponential distributions

EM 561 GW DePuy 167


Calc


Choose Probability Distribution (Normal,

Minitab 15 English.lnk

Uniform, Exponential)• Choose ‘Cumulative Probability’

• Enter Parameter(s)


How to do cdf calculations in MinitabEM 561 GW DePuy 168


85

How to do cdf calculations in Minitab

Calc

EM 561 GW DePuy 169

Calc


Choose Probability Distribution (e.g. Normal)

• Choose Cumulative ProbabilityProbability

• Enter Parameters


Note: For Normal Distr., enter st. dev. not variance

How to do cdf calculations in MinitabEM 561 GW DePuy 170

• P(X<x) values displayed in Sessions Window

86

How to do cdf calculations in Excel

• Excel can calculate P(X<x) for Normal & Exponential distributions

EM 561 GW DePuy 171

• = NORMDIST(x,mean,standard_dev,cumulative)

• Find P(X<4.8) for N(5.2, 3.9)

• = normdist(4.8, 5.2, 1.97484, true)

= EXPONDIST(x lambda cumulative)• = EXPONDIST(x,lambda,cumulative)

• Find P(X<10) for Expo with λ=0.26667

• = expondist(10, 0.26667, true)

How to do cdf calculations in ExcelEM 561 GW DePuy 172

87


• The compressive strength of samples of cement can be modeled by a normal distribution with a

f 5800 k / 2 d t d d d i ti f

EM 561 GW DePuy 173

mean of 5800 kg/cm2 and a standard deviation of 150 kg/cm2.

1. What is the probability that a sample’s strength is less than 6200 kg/cm2?

2 What is the probability that a sample’s strength is2. What is the probability that a sample s strength is between 5900 and 6000 kg/cm2?

3. What strength is exceeded by 95% of the samples?

More Chapter 4 Examples• The number of taxi arrivals at an

intersection is Poisson distributed with mean of 6 taxi arrivals per hour

EM 561 GW DePuy 174

mean of 6 taxi arrivals per hour.

1. What is the probability you wait longer than 1 hour for a taxi?

2. Suppose you have already been waiting 1 hour pp y y gfor a taxi, what is the probability a taxi arrives in the next 20 minutes?

3. What is the probability 5 taxis arrive at the intersection during the next 30 minutes?

88


• Suppose the time it takes a hematology cell counter to complete a test on a blood

EM 561 GW DePuy 175

psample is uniformly distributed between 50 and 75 seconds.

1. What is the probability a test requires more than 70 seconds to complete?

2. What percentage of the tests require less than 60 seconds to complete?



READ THE BOOK!

EM 561 GW DePuy 176

• READ THE BOOK!

• We covered all sections 4-1 thru 4-8 except 4-7

89

Test #1 here

• Open book

O t

EM 561 GW DePuy 177

• Open notes

• Bring a calculator

• No talking/communicating

• Covers Chapters 1, 2, 3, 4