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Module 9 Modeling Uncertainty:
THEORETICAL PROBABILITY MODELS
Topics• Binomial Distribution• Poisson Distribution• Exponential Distribution• Normal Distribution• Beta Distribution
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Introduction
• Module 7:– basic probability– use in decision problems
• Module 8:– subjective probability– modeling for decision analysis
• Module 9:– theoretical distributions– application to decision analysis
• Module 9 software tutorial
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Theoretical Probability ModelsLearning Objectives
• Refresh knowledge:– Binomial distribution– Poisson distribution– Exponential distribution– Normal distribution
• Gain knowledge:– Beta distribution
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Binomial Distribution(a discrete distribution)
Model characteristics:
• Dichotomous outcomes– Two possible outcomes– One outcome can occur
• Constant probability of “success”
• Independence
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Binomial Distribution
Mathematical model:
• PB(R= r | n, p) = [n! / r! (n - r)!] pr(1-p)n-r
• B subscript = binomial probability
• R = binomial random variable
• r = number of successful outcomes
• n = number of events or trials
• p = probability of successful outcome
• E (R) = = np
Var (R) = = np(1- p)
2
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Binomial Distribution
Probability mass function notation:• f (x) = ( ) px qn-x
• X = binomial random variable• x = number of successful outcomes• n = number of events or trials• p = probability of a successful outcome• q = 1 – p = probability of an unsuccessful
outcome • E (X) = µ = np
V (X) = σ2 = µq
nx
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Binomial Distribution
Cumulative distribution function:
• Probability of k or fewer “successful” outcomes
• F (x ≤ k) = ∑ ( ) px qn-xk
X = 0
nx
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Binomial Distribution
• Frequent application situations:– Quality control– Reliability– Survey sampling
• Approximation by other models:– Poisson: p → 0 and n → ∞– Normal: p ≤ 0.5 and np > 5, OR
p > 0.5 and nq > 5
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Binomial Distribution
Example application:Items are manufactured in large lots, from each of which
twenty units are selected at random. The lot is accepted if the
sample contains three or fewer defectives. If the production
process yields, on the average, ten percent defectives, what is
the probability of lot acceptance?
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Binomial Distribution
Formulation:
Determine the probability of three or fewer
“successes” in 20 independent trials, each
having 0.1 probability of success.
F (x ≤ 3) = ∑ ( ) (0.1)x (0.9)20-x = 0.8673
X = 0
20 x
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Poisson Distribution(a discrete distribution)
Model characteristics:
• Time interval or spatial region
• Probability of an event is small
• Events are independent
• Events take place at a constant rate
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Poisson Distribution
Mathematical model:
• PP (X = k | m) = (e –m m k) / k !• P subscript = Poisson probability
• X = Poisson random variable
• k = number of events that occur
• e = natural logs base, 2.718…
• m = mean number of event occurrences
• E (X) = µ = m
Var (X) = σ2 = m
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Poisson Distribution
Probability mass function notation:• f (x) = (e – λ λ x ) / x !
• x = number of event occurrences• λ = mean number of event occurrences• e = natural logs base, 2.718…
• E (X) = µ = λV (X) = σ2 = λ
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Poisson Distribution
Cumulative distribution function:
• Probability of k or fewer occurrences within a temporal or spatial interval
• F (x ≤ k) = ( e – λ λ x ) / x !∑X = 0
k
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Poisson Distribution
• Frequent application situations:– Quality control– Reliability– Queuing– Physical properties
• Approximation by another model:– Normal: λ > 5
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Poisson Distribution
Example application:The probability that a person will have a negative
reaction to the injection of a certain serum is 0.001.
Determine the probability that two or more of 1,000
people will have a negative reaction to the injection.
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Poisson Distribution
Formulation:Determine the probability that zero or one
people will have a negative reaction
when λ = np = 1, and take the complement.
1 – F (x ≤ 1) = 1 - e-11x / x! = 0.264∑x= 0
1
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Exponential Distribution(a continuous distribution)
Model characteristics:
• Time or distance between two outcome occurrences
• Poisson model characteristics required
• Uses Poisson mean ( m or λ )
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Exponential Distribution
Mathematical model:
• f E ( T = t | m ) = me-mt
• E subscript = exponential probability
• T = exponential random variable
• t = interarrival period
• m = mean of Poisson distribution
• e = natural logs base, 2.718..
• E (T) = μ = 1 / m
Var (T) = σ2 = 1 / m2
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Exponential Distribution
Probability density function notation:
• f (x) = λ e -λx
• x = length of interval between
occurrences
• λ = mean of Poisson distribution
• e = natural logs base, 2.718…
• E (X) = μ = 1 / λ
V (X) = σ2 = 1 / λ2
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Exponential Distribution
Cumulative distribution function:
• Probability that the interval between two occurrences is of length k or less
• F ( x ≤ k ) = ∫ λ e – λ x d x = 1 – e – λ k0
k
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Exponential Distribution
• Frequent application situations:– Quality control– Reliability– Queuing– Physical phenomena
• Applicable when underlying process is Poisson
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Exponential Distribution
Example Application:The life of a certain brand of light bulb can be approximated
by the exponential probability density function. If the mean
life of the light bulb is 1,000 hours, determine the probability
that the bulb will last more than 1,000 hours
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Exponential Distribution
Formulation:
• Determine the probability that a light bulb will have a life of 1000 or fewer hours, when µ = 1000, and take the complement
• µ = 1/λ = 1000 so λ = 1/1000 = 0.001
• 1 – F(x ≤ 1000) = 1 - ∫ 0.001 e- 0.001x dx
= 1 - {1- e-0.001(1000) } = 1 - {1 – e-1 } = 0.3680
1000
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Normal (Gaussian) Distribution(a continuous distribution)
Model characteristics:
• “Bell–shaped” curve
• Effective for measured phenomena
• Effective for multiple sources of uncertainty
• Two parameters, µ and σ
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Normal Distribution
Mathematical model:• fn ( y | µ,σ2 ) = (2πσ2)-1/2 exp [ -(y-µ)2 / 2σ2]
• N subscript = normal probability• y = value(s) of random variable Y• π = 3.14159…• e = natural logs base 2.718…• µ = the distribution mean• σ = the distribution standard deviation
• E (Y) = µ Var (Y) = σ2
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Normal Distribution
Probability density function:• f (x) = (2πσ2)-1/2 exp [ -(x-µ)2 / 2σ2]
- ∞ < x < ∞
- ∞ < µ < ∞
- σ > 0
- π, e are constants
- µ, σ are parameters
• E (X) = µ
Var (X) = σ2
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Normal Distribution
Cumulative distribution function:• Probability that a value of x or less will occur• F (x) = (2πσ2)-1/2 ∫ exp [ -(v-µ)2 / 2σ2] dv• P (a < X ≤ b) = F (b) – F (a)
= (2πσ2)-1/2∫ exp [ -(v-µ)2 / 2σ2] dv
x
-∞
a
b
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Normal Distribution
• Frequent application situations:– measured phenomena
– measured results of multiple additive phenomena
– approximation of other distributions
– numerous inferential statistical methods
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Normal Distribution
Example application:The ages of employees at a company are
normally distributed with a mean of 50 years
and a standard deviation of 5 years. Determine
the percentage of employees whose ages are
likely to be between 50 and 52.5 years.
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Normal Distribution
Formulation:• Determine the probability of age ≤ 50 years, and determine the
probability of age ≤ 52.5 years. Take the difference, and multiply the
result by 100 to obtain the percentage.
• Using the standard normal variate, Z = (x –μ) / σ, and cumulative distribution function probabilities from either tabled values or appropriate software, we obtainF(x ≤ 50) = F(Z ≤ 0) = 0.5000 and
F(x ≤ 52.5) = F(Z ≤ 0.5) = 0.6915
• ThenP(50 < x ≤ 52.5) = 0.6915 – 0.5000 = 0.1915 or 19.15 %
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Beta Distribution
Model characteristics:
• Variable bounded at both ends
• Numerous distribution shapes
• Events are independent
• Models proportions
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Beta Distribution
Generalized beta probability density function:
• f (x | λ1, λ2, ω1, ω2)
=
where - ω1 ≤ x ≤ ω2
- λ1, λ2, ≥ 0 - Γ(λ) = (λ – 1)! when λ Є I+, OR
____ 1
ω2-ω1
_________(λ1+λ2)Γ
Γ (λ1) Γ (λ2) (______x – ω1ω2 – ω1)
λ1-1(1 - _____x – ω1ω2 – ω1)
λ2-1
= ∫ xλ-1 e-x dx otherwise∞
0
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Beta Distribution
More frequently,
set ω1 = 0 and ω2 =1 • So f (x | λ1, λ2) = xλ
1-1 (1-x)λ
2-1
Where 0 ≤ x ≤ 1 λ1, λ2 ≥ 0 Γ (λ) = (λ-1)! Where λ Є I+, OR
=
__________Γ (λ1+ λ2)
Γ (λ1) Γ (λ2)
∫ xλ-1 e-x dx otherwise∞
0
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Beta Distribution
• When λ1 and λ2 are positive integers:
• Probability density function
f (x | λ1, λ2) = x λ1-1 (1 –
x)λ2-1
where the random variable X = x
and 0 ≤ x ≤ 1
_____________(λ1+ λ2 – 1)!
(λ1 – 1)! (λ2 – 1)!
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Beta Distribution
• Let n = λ1 + λ2 and r = λ1
• f β ( q | r,n ) = q r-1 ( 1 – q )n-r-1
• β subscript = beta probability
• Q = Beta random variable
• q = proportion value between 0 and 1
• r = number of “successes”
• n = number of trials
• E (Q) = µ = r / n
Var (Q) = σ2 = r ( n – r) / [ n2 ( n + 1) ]
( r – 1) ! ( n – r – 1) !________________(n – 1)!
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Beta Distribution
Cumulative distribution function:• F (Q ≤ q | r,n) = ∫ υr-1 (1 – υ) dυ
• Called Incomplete Beta Function– Equivalence to binomial cumulative distribution function
• Values from tables or software
______________(n – 1)!(r – 1)! (n – r – 1) 0
q
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Beta Distribution
• Frequent application situations:– Proportions or percentages
– Physical variables in restricted intervals
– Tolerances in quality control and reliability
– PERT networks (generalized form)
– Bayesian analysis (informative a priori)
• Approximated by difference between two normal distributions when λ1+ λ2 ≥ ~ 30
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Beta Distribution
Example Application:• Suppose that the percentage of employees who submit
medical benefits claims each year is a beta random variable. As the benefits coordinator for a small firm with 40 employees, you currently expect that about 30 percent will most likely submit claims this year. Determine the probability that more than 16 employees will submit claims.
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Beta Distribution
Formulation:• Determine the probability that 16 or fewer of 40
employees will submit claims, and take the complement.
• μ = r/n = 0.30 = r = (0.30)(40) = 12
q = 16/40 = 0.40
• From table values or software,
Fβ (Q ≤ 0.40 | 12, 40) = 0.91
• Thus Pβ (Q > 0.40) = 1 – 0.91 = 0.09
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Summary
• Theoretical distributions to model uncertainty• Five distributions:
– Binomial– Poisson– Exponential– Normal– Beta
• Each distribution:– Mathematical formulation– Mean and variance– Graphical representations– Application situations– Example problem