prob distribution-intro tvs
TRANSCRIPT
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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
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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
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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
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(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
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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
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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
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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
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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
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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, ...
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Continuous Probability Distribution Models
Continuous ProbabilityDistributions
Normal Exponential
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Normal Distribution
1. Bell-Shaped &Symmetrical f(X)
2. Mean, Median,
3. Middle Spread Is
Mean.
4. Random Variable HasMedianMode
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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
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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
37 37
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= .
Shaded Area Exaggerated
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
Shaded Area Exaggerated
= 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?
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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
49 49
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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