Prob Distribution-Intro TVS

7/25/2019 Prob Distribution-Intro TVS

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Introduction to Probability

distributions

Dr.T.V.Subramanian


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Experiment RandomVariable

PossibleValues

Make 100 Sales Calls # Sales 0, 1, 2, ..., 100

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

Answer 33 uestions # Correct 0 1 2 ... 33

Count Cars at Toll # Cars 0, 1, 2, ...,

2 2


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Discrete

Probability Distribution

1. List of All Possible [ Xi, P(Xi) ] Pairs

Xi = Value of Random Variable (Outcome)

P(Xi) = Probabil ity Associated with Value

.

3. Collectively Exhaustive (Nothing Left Out)

4. 0 P(Xi) 1

5. P(Xi) = 1

3 3


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Discrete Probabil ity Distribution Example

Event: Toss 2 Coins. Count # Tails.

Values, Xi Probabilities, P(Xi)

0 1/4 = .25

=

2 1/4 = .25

4 4

1984-1994 T/Maker Co.


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Visualizing Discrete Probabil ity Distributions

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

st ng# Tails f(X i)

CountP(Xi)

.1 2 .50

2 1 .25

Graph Equation.50

P(X)

P x n

x n xp px n x( )

!

! ( )!( )=

1

.00

.25

X

5 5

0 1 2


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Discrete Random Variable Summary Measures

1. Expected ValueMean of Probabil ity Distribution

Weighted Average of All Possible Values

= E X =

X P X

2. Variance

e g e verage quare ev a on a ouMean

2

2

] =

2

6 6

i ] = i i


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Summary Measures Calculation Table

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

(Xi-)2

P(Xi)

Total

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

7 7


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Thinking Challenge

You toss 2 coins.

Youre interested in the

number of tails. What

standard deviation of this

random variable

number of tails? 1984-1994 T/Maker Co.

8 8


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Discrete Probability Distribution Function

1. Type of ModelRepresentation of Some P X x( | )= =

n er y ng enomenon

2. Mathematical Formula

xe-

3. Represents Discrete

Random Variable

4. Used to Get Exact

Probabilities

9 9


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Discrete Probability Distribution Models

DiscreteProbability

Distribution

Poisson HypergeometricBinomial

10 10


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Permutations

How many ways are there to select k objects from nobjects.

Example 2 out of 3: ABC, ACB, BCA,BAC,CAB,CBA = 6

Example 2 out of 4:ABCD BACD CABD DABC

ABDC BADC CADB DACB

ACBD BCAD CBAD DBAC =ACDB BCDA CBDA DBCA

ADBC BDAC CDBA DCAB

11 11


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Permutations

n!Permutations: The number of

n objects

6

123!3

=

=

12

212)!24( ===

12 12


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Combinations

How many ways are there to selectk

objects fromn

objects -- where order is insignificant!.

Example 2 out of 3: ABC, ACB, BCA,BAC,CAB,CBA = 3

Example 2 out of 4:ABCD BACD CABD DABC

ABDC BADC CADB DACB

ACBD BCAD CBAD DBAC =ACDB BCDA CBDA DBCA

ADBC BDAC CDBA DCAB

13 13


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Combinations

n!

Combinations: The number of ways ofselecting k objects out of n objects

n 123!3 3k 12)!23(!2 2

64

24

1212

1234

!24!2

!4==

=

=

4

14 14


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Binomial Distribution

1. Number of Successes in a Sampleof n Observations (Trials)

2. # Reds in 15 Spins of Roulette Wheel

. e ec ve ems n a a c o

Items

4. # Correct on a 33 Question Exam

15 15

.

100 Customers Who Enter Store


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Binomial Distribution Properties

1. Two Different Sampling MethodsInfinite Po ulation Without Re lacement

Finite Population With Replacement

. equence o n en ca r a s

3. Each Trial Has 2 Outcomes

Success (Desired Outcome) or Failure

16 16

.

5. Trials Are Independent


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Binomial Probability Distribution Function

P X x n pn

p px n x( | , ) ! ( )= = 1

P(X= x | n,p) = probability that X = x givenn &p

n = sample size

p = probability of successx = number of successes in sample

17 17

(X = 0, 1, 2, ...,n)


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Binomial Probability Distribution Example

Event: Toss 1 Coin 4 times in a row. Note #.

P X x nn

x n x!

= = 1x n x

P X( | ,. )!( )!

. ( . )= = 3 4 5

3 4 35 1 53 4 3

18 18= .25


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Binomial Distribution Characteristics

n = 5 p = 0.1Mean .6P(X)

= =.0

.2

.

X

n = 5 p = 0.5

Standard Deviation

= P(X)

.2

.4

.

19 19

.

0 1 2 3 4 5


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Poisson Distribution

1. Number of Events that Occur in an Interval

Events Per Unit

, , ,

2. Examples# Customers Arriving in 20 minutes

# Strikes Per Year in the U.S.

# Defects Per Lot (Group) of VCRs

20 20


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Poisson Process

1. Constant Event Probability

60 1-Minute Intervals

.

Dont Arrive Together

. n epen ent vents

Arrival of 1 Person Does Not

21 21

1984-1994 T/Maker Co.


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Poisson Probability Distribution Function

x

= = e-

x !

P(X= x | ) = probability that X = x given

= expected (mean) number of successes

e = 2.71828 (base of natural logs)

x = number of successes per unit

22 22


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Poisson Distribution Characteristics

= 0.5Mean .6P(X)

N

=.0

.2

.

X

= 6i =1

P(X)

an ar ev a on

= .2.4

.

23 23

.

0 2 4 6 8 10


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Poisson Distribution Example

Customers arrive at

a rate of 72 per hour. x

= = e-

probability of 4

customers arriving in

x !

- 4m nu es

72 per hr. = 1.2 perP X( | . )

.

!= =

e .4 3 6

4.

= 3.6 per 3 = 0.1912

24 24

.

Interval


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Poisson Distribution Example

A typist enters 75 words per minute withsix errors per hour. What is theprobability of zero errors in a 255 word

paragraph?

25 25


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Expected Value & Variance Solution

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

(Xi-X)2

P(Xi)

0 .25 0 -1.00 1.00 .25

. .

2 .25 .50 1.00 1.00 .25

= 1.0 2 = .50

26 26


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Poisson Distribution Solution: Finding

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

= 4500 words/hr

6 errors/hr = 6 errors/4500 words

.

In a 255-word transaction (interval):

= (0.00133 errors/word )(255 words)

= 0.34 errors/255-word transaction

27 27


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Continuous Random Variable

1. An Event Expressed by a Numerical Value

Weight of a Student

Observe 115, 156.8, 190.1, 225

2. Continuous Random Variable

Obtained by Measuring

Infinite Number of Values in Interval

28 28


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Continuous Random Variable Examples

Event Random Possible

Weigh 100 People Weight 45.1, 78, ...

Measure Part Life Hours 900, 875.9, ...

Record Food Spending Money 54.12, 42, ...

Measure TimeBetween Arrivals

Inter-ArrivalTime

0, 1.3, 2.78, ...

29 29


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Continuous Probability Distribution Models

Continuous ProbabilityDistributions

Normal Exponential

30 30


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Normal Distribution

1. Bell-Shaped &Symmetrical f(X)

2. Mean, Median,

3. Middle Spread Is

Mean.

4. Random Variable HasMedianMode

31 31


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Importance of Normal Distribution

1. Describes Many Random Processes or

Continuous Phenomena

2. Basis for Statistical Inference

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Normal Distribution

211)(

= XeXf

= 3.14159; e = 2.71828

X = value of random variable (- < X < )

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Effect of Varying Parameters ( & )

f(X)

B

CA

X

34 34


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Normal DistributionProbability

ro a y s e

area under thecurve!

P c X d f X dx( ) ( ) ? =d

f(X)

c

X

35 35

c


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Infinite Number of

Normal distributions differ bymean & standard deviation.

Each distribution wouldrequire its own table.

f(X)

X

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Thats an infinite number!


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Standardize theNormal Distribution

Normal Standardized

=Z

= 1

Distribution Normal Distribution

Z = 0 ZX

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One table!


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Standardizing Example

12.0.

=

== ZormaDistribution

an ar ze

Normal Distribution

Z==

38 38

ZZ= 0 .12X= .


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Obtaining the Probability

tandardized Normal

Probability Table (Portion)

Z= 1Z .00 .01

0.0 .0000 .0040 .0080

.02

.0398 .0438

.

0.1 .0478

ZZ= 0 0.120.2 .0793 .0832 .08710.3 .1179 .1217 .1255

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Probabilities

Exaggerated

E l


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ExampleP(3.8 X 5)

12.0.

=

== ZormaDistribution

an ar ze

Normal Distribution

Z=

0.0478

=

40 40

ZZ= 0-0.12Shaded Area Exaggerated

X=3.8

E l


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ExampleP(2.9 X 7.1)

Z= = =

10

21.

Normal

DistributionStandardized

Normal DistributionZ= = =

1021

..

Z

.1664

.0832.0832

41 41

0-.21 .21

Shaded Area Exaggerated

52.9 7.1 X

E l


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ExampleP(X 8)

Z=

=

=

10 30.orma

Distribution

an ar ze

Normal Distribution

Z

.5000

.1179

.

42 42

ZZ= .


X= 5 8

Example


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ExampleP(7.1 X 8)

Z= = =

10

21.

Normal

DistributionStandardized

Normal DistributionZ= = =

1030.

Z==

.0832

.

.0347

43 43

z= 0 .30 Z.21


= 5 87.1 X


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Normal Distribution Thinking Challenge

You work in Quality Control

for GE. Light bulb life has a

= 2000 hours & = 200

hours. Whats the probabil itythat a bulb will last

A. between 2000 & 2400

hours?

B. less than 1470 hours?

44 44

Solution


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SolutionP(2000 X 2400)

Z

X

=

=

=

2400 2000

200 2 0.

orma

Distribution

tan ar ze

Normal Distribution

Z =

.4772

=

45 45

ZZ= 0 2.0X= 2000 2400

Solution


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SolutionP(X 1470)

Z

X

=

=

=

1470 2000

200 2 65.

orma

Distribution

an ar ze

Normal Distribution

Z =

.5000

=

.4960.0040

46 46

ZZ= 0-2.65X= 20001470

Finding Z Values


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Finding Z Valuesfor Known Probabilit ies

an ar ze orma

Probability Table (Portion)

at s ven

P(Z) = 0.1217?

Z .00 0.2

0.0 .0000 .0040 .0080

Z= 1

.1217.01

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871ZZ= 0 .31

47 47

. .. .Exaggerated

Finding X Values


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Finding X Valuesfor Known Probabilit ies

= 1= 10

Normal Distribution Standardized Normal Distribution

.1217 .1217

== .

48 48Shaded Area Exaggerated

.. ===

A i N lit


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Assessing Normality

1. Compare Data

Characteristics toProperties of Normal

Normal Probability Plotfor Normal Distribution

Distribution

2. Evaluate Normal

90

Probabil ity Plot

Create on30

60

Z

X

Plot of Data Values& Standardized

-2 -1 0 1 2

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Quantile Values Look for Straight Line!

N l P b bilit Pl t


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Normal Probability Plots

- -

90 90

30

-2 -1 0 1 2

Z 30

-2 -1 0 1 2

Z

Rectangular U-Shaped

90 90

30

60

Z

X

30

60

Z

X

50 50

- - - -

E ti l Di t ib ti


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Exponential Distribution

P arrival time< X X

( )=1 - e

e = the mathematical constant.

= the population mean of arrivals

X = any value of the continuous random

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Examples of the Exponential Distribution


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Examples of the Exponential Distribution

1. Customers arriving at an ATM machine

.

3. Drivers arrivin at a toll booth

4. Computer failures

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Prob Distribution-Intro TVS

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