Probability Distributions: Binomial, Poisson, Exponential, and Normal, Slides of Human-Computer Interaction Design

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1

Theoretical

Probability Models

2

Introduction

 Use Theoretical Probability Models When They Describe the

Physical Model “Adequately”

 The results of intelligent tests ~ Normal Distribution  The length of a telephone call ~ Exponential Distribution  The number of people arriving at a bank within an hour ~ Poisson Distribution  The number of defects in a bottle production line ~ Binomial Distribution

 Discrete Distributions

 Binomial, Poisson, etc.

 Continuous Distributions

 Exponential, Normal, Beta etc.

4

Binomial Distribution (Cont’d)

 Expected Value

 Variance

E ( X ) np

var( X )var( Y ) np ( 1  p )

Y = “# of failures in a sequence of n independent trials”, then Y = n – X

p y^ p n y

y

n

Y B n p Y y n p   

~ ( ,( 1  ))Pr(  | ,( 1  )) ( 1 ) ( )

 Cumulative Distribution Function

 

x

i

pi^ p n i

i

n

X B n p F x n p X x n p

0

~ ( , ) ( | , ) Pr( | , )

(See Appendix B)

E ( Y ) n ( 1  p )

5

Pr(Hit| X =5)=?

Let X = number of people out of 20 tasters who preferred the new pretzel

You are planning to sell a new pretzel, and you want to know whether it will be a success or not. If your pretzel is a “Hit”, you expect to gain 30% of the market. If it is a “Flop”, on the other hand, the market share is only 10%. Initially, you judged these outcomes to be equally likely. You decided to test the market first and found out that 5 out of 20 people preferred your pretzel to the competing product. Given the new data, what do you think of the chance of your pretzel being a Hit?

(Bayes Theorem)

Pr( X  5 )  Pr( X  5 Hit) Pr( X  5 Flop)

 Pr( X  5 | Hit)Pr(Hit) Pr( X  5 | Flop)Pr(Flop)

? (^) 0.5? 0.

Pretzel Example

Pr( 5 )

Pr( 5 |Hit)Pr(Hit) 

 X

X Pr(Hit | X = 5) =

7

Poisson Distribution

 Represent occurrences of events over a unit of measure (time or

space)

 e.g. number of customers arriving, number of breakdowns occurring

 Assumptions

 Events can happen at any point along a continuum  At any particular point, the probability of an event is small (i.e. events do not happen frequently)  Events happen independently of one another  The average number of events is constant over a unit of measure

 Probability Mass Function

X ~ Poisson( m )⇔Pr( X = x | m )=

X = “# of events in a unit of measure”

m is the average number of events in a unit of measure

(See Appendix C)

m

x

m

e

x 

8

Poisson Distribution (Cont’d)

 Expected Value

 Variance

E ( X ) m

var( X )  m

Y ~ Poisson( m • t )⇔Pr( Y = y | m • t )=

Y = “# of events in t units of measure” mt y

m t e

^ y^   !

( )

E ( Y )  m  t var( Y ) m  t

 Cumulative Distribution Function

m

x

i

i

X m F x m X x m m^ i^ e -

0

~ Poisson( )⇔ ( | ) Pr( | ) ∑! 

(See Appendix D)

10

In light of the new data, you feel that the chance of the current stand being a good location has slightly increased and thus you should stay.

| Good~ ( 0. 5 20 10 ) Pr( 7 |Good)^107!^100. 09

7

X Poisson m     X   e  

| Bad~ ( 0. 5 10 5 ) Pr( 7 |Bad)^57!^50. 104

7

X Poisson m     X   e  

| Dismal~ ( 0. 5 6 3 ) Pr( 7 |Dismal)^37!^30. 022

7

X Poisson m     X   e  

Pr( 7 |Dismal) Pr(Dismal) 0. 09 0. 7 0. 104 0. 2 0. 022 0. 1 0. 086

Pr( 7 ) Pr( 7 |Good) Pr(Good) Pr( 7 |Bad) Pr(Bad)         

        X

X X X

Pr(Good | X  7 ) Pr( = 7 ) = 0. 009. 086 •^0.^7 = 0. 733 Pr( = 7 |Good)Pr(Good) X

X

11

Exponential Distribution

 If the number of events occurring within a unit of measure follows a

Poisson distribution, then the time or space between the occurrence of

two events follows an exponential distribution

 Exponential distribution has the same assumptions as Poisson

distribution

 Probability Density Function

Let T =“Time (space) between two consecutive events”

F ( t | m ) Pr( T  t | m ) 1  e  mt

T ~ Exp( m )⇔ f ( t | m )= me -^ mt ( t ≥ 0 )

m is the same average rate used in Poisson distribution

 Cumulative Distribution Function

13

You wonder if you can provide fast service to your customers. It takes 3.5 minutes to cook a pretzel, so what is the probability that the next customer arrives before the pretzel is finished? As in the previous example, you assume customers arrive according to a Poisson process, and you consider your location being good, bad or dismal if you sell 20, 10, 6 pretzels per hour, respectively. Your prior belief is that Pr(Good)=0.7, Pr(Bad)=0.2, and Pr(Dismal)=0.1.

Let T=the time between two consecutive customers

Pr( 3. 5 |Good)Pr(Good) Pr( 3. 5 |Bad)Pr(Bad) Pr( 3. 5 |Dismal)Pr(Dismal)

Pr( 3. 5 ) Pr( 3. 5 Good) Pr( 3. 5 Bad) Pr( 3. 5 Dismal)     

           T T T

T T T T

Pr(T<3.5) =?

0.7 0.2^ 0.

??^?

Pretzel Example (Cont’d)

14

|Dismal~ ( = /min = /min)⇔Pr( < 3. 5 |Dismal)= 1 - = 0. 295

|Bad~ ( = /min = /min)⇔Pr( < 3. 5 |Bad)= 1 - = 0. 442

|Good~ ( = /min = /min)⇔Pr( < 3. 5 |Good)= 1 - = 0. 689

• • 3. 5 10

1 60

6

• • 3. 5 6

1 60

10

• • 3. 5 3

1 60

20

101

61

(^13)

T Exp m T e

T Exp m T e

T Exp m T e

Pr( 3. 5 |Dismal)Pr(Dismal)

Pr( 3. 5 ) Pr( 3. 5 |Good)Pr(Good) Pr( 3. 5 |Bad)Pr(Bad)

T

T T T

In other words, about 60% of your customers will have to wait until the pretzel is ready. Therefore, the fast service does not seem very appealing.

16

Normal Distribution (Cont’d)

 Standard Normal Distribution

 Convert to Standard Normal Distribution

Pr( X  a |,)Pr( Z  a  |  0 ,   1 )

,

X ~ N ( , ) Z ~ N ( 0 , 1 ) X 

     

(See Appendix E for Cumulative Probability)

z P ( Z≤z )

X ~ N (μ=10, σ^2 =400), then the probability X is less than or equal to 35 isPr( (^) X 35 ) Pr(Z ) Pr(Z 1. 25 ) 0. 8944 (Appendix E) 400   ^35 ^10   

17

Normal Distribution (Cont’d)

 Other Important Probabilities

Pr( ) Pr( )

Pr( ) Pr( ) Pr( )



 





 

 

 

 



 

    

   

        b a

a X b a b

Z Z

a X b Z

Because standard normal distribution is symmetric around zero,

Pr( 1. 25 ) Pr( 1. 25 ) 0. 8944 0. 1056 0. 7888

Pr( 15 35 ) Pr(^15400103540010 ) Pr( 1. 25 1. 25 )       

              Z Z

X Z Z

Pr( Z - z )Pr( Z  z ) 1 - Pr( Z  z )

= 1 - 0. 8944 = 0. 1056

Pr( X ≤- 15 )=Pr( Z ≤^15400 10 )=Pr( Z ≤- 1. 25 )=Pr( Z ≥ 1. 25 )= 1 - Pr( Z ≤ 1. 25 )

X ~ N (μ=10, σ^2 =400), then

19

Your plant manufactures disk drivers for personal computers. One of your machines produces a part that is used in the final assembly. The width of the part is important to the proper functioning of the disk driver. If the width falls below 3.995 or above 4.005 mm, the disk driver will not work properly and must be repaired at a cost of $10.40. The machine can be set to produce parts with width of 4mm, but it is not perfectly accurate. In fact, the width is normally distributed with mean 4 and the variance depends on the speed of the machine. The standard deviation of the width is 0. at the lower speed and 0.0026 at the higher speed. Higher speed means lower overall cost of the disk driver. The cost of the driver is $20.45 at the higher speed and $20.75 at the lower speed. Should you run the machine at the higher or lower speed?

Quality Control Example

20

Let X = width of a disk driver

(P 1 =?) $20.75+$10.40=$31. $20.

$20.45+$10.40=$30. $20.

Cost/Driver Low Speed

High Speed

X ≤3.995 or X ≥4.005 (Defective)

3.995 ≤ X ≤4.005 (Not Defective)

X ≤3.995 or X ≥4.005 (Defective)

3.995 ≤ X ≤4.005 (Not Defective)

(P 2 =?)

P 1 =Pr(Defective | Low Speed) P 2 =Pr(Defective | High Speed)

1 0. 9914 0. 0086

Pr(Defective|Low Speed) 1 Pr(Not Defective|Low Speed)   

 

Pr(Not Defective|Low Speed) Pr( 3. 995 4. 005 | 4 , 0. 0019 )

|Low Speed~ ( 4 , 0. 0019 )      

   

  X

X N

Pr( 30.^995. 00194 40.^005. 00194 ) Pr( 2. 63 2. 63 ) Pr( 2. 63 ) Pr( 2. 63 )

 ^  Z      Z   Z   Z 