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Introduction to Probability and Statistics
7th Week (4/19)

Special Probability Distributions (2)

(Discrete) Poisson Distribution
- Describe an event that rarely happens.
- All events in a specific period are mutually independent.
- The probability to occur is proportional to the length of the period.
- The probability to occur twice is zero if the period is short.

It is often used as a model for the number of events (such as the number of
telephone calls at a business, number of customers in waiting lines, number
of defects in a given surface area, airplane arrivals, or the number of
accidents at an intersection) in a specific time period.

If z > 0

Satisfy the PF
condition

 Probability function :

.
Ex.1. On an average Friday, a waitress gets no tip from 5 customers. Find the
probability that she will get no tip from 7 customers this Friday.

The waitress averages 5 customers that leave no tip on Fridays: λ = 5.
Random Variable : The number of customers that leave her no tip this Friday.
We are interested in P(X = 7).

Ex. 2 During a typical football game, a coach can expect 3.2 injuries. Find the
probability that the team will have at most 1 injury in this game.

A coach can expect 3.2 injuries : λ = 3.2.
Random Variable : The number of injuries the team has in this game.
We are interested in

.

Ex. 3. A small life insurance company has determined that on the average it receives 6
death claims per day. Find the probability that the company receives at least seven
death claims on a randomly selected day.
P(x ≥ 7) = 1 - P(x ≤ 6) = 0.393697

Ex. 4. The number of traffic accidents that occurs on a particular stretch of road
during a month follows a Poisson distribution with a mean of 9.4. Find the probability
that less than two accidents will occur on this stretch of road during a randomly
selected month.

P(x < 2) = P(x = 0) + P(x = 1) = 0.000860

Characteristics of Poisson Distribution
E(X) increases with parameter or .
The graph becomes broadened with increasing the parameter or


Probability mass function Cumulative distribution function


Comparison of the Poisson
distribution (black dots) and the
binomial distribution with n=10 (red
line), n=20 (blue line), n=1000 (green
line). All distributions have a mean of
5. The horizontal axis shows the
number of events k. Notice that as n
gets larger, the Poisson distribution
becomes an increasingly better
approximation for the binomial
distribution with the same mean

Discrete Probability Distributions: Summary

• Uniform Distribution
• Binomial Distributions
• Multinomial Distributions
• Geometric Distributions
• Negative Binomial Distributions
• Hypergeometric Distributions
• Poisson Distribution

Continuous Probability Distributions

What kinds of PD do we have to know to
solve real-world problems?

(Continuous) Uniform Distribution

(Continuous) Uniform Distributions
In a Period [a, b], f(x) is constant.

 f(x)

 E(x):

If X ∼ U(0, 1) and Y = a + (b - a) X,
(1)Distribution function for Y
(2)Probability function for Y
(3)Expectation and Variance for Y
(4) Centered value for Y

(1)

Since y = a + (b - a) x so 0 ≤ y ≤ b,

(Continuous) Uniform Distributions

(Continuous) Exponential Distribution

▶ Analysis of survival rate
▶ Period between first and second earthquakes
▶ Waiting time for events of Poisson distribution

For any positive

From a survey, the frequency of traffic accidents X is given by
-3x
f(x) = 3e (0 ≤ x)

(1)Probability to observe the second accident after one month of the first
accident?
(2)Probability to observe the second accident within 2 months
(3)Suppose that a month is 30 days, what is the average day of the accident?

(1)

(2)

(3) μ=1/3, accordingly 10 days.

• Survival function :

• Hazard rate, Failure rate:

A patient was told that he can survive average of 100 days. Suppose that the
probability function is given by

(1) What is the probability that he dies within 150 days.
(2) What is the probability that he survives 200 days

λ=0.01이므로 분포함수와 생존함수 :
-x/100 -x/100
F(x)=1-e , S(x)=e

(1) 이 환자가 150일 이내에 사망할 확률 :
-1.5
P(X < 150) = F(150) = 1-e = 1-0.2231 = 0.7769

(2) 이 환자가 200일 이상 생존할 확률
-2.0
P(X ≥ 200) = S(200) = e = 0.1353

⊙ Relation with Poisson Process

(1) If an event occurs according to Poisson process with the ratio λ , the waiting
time between neighboring events (T) follows exponential distribution with the
exponent of λ.

(Continuous) Gamma Distribution


α : shape parameter, α > 0
β : scale parameter, β > 0

α=1 Γ (1, β) = E(1/β)

⊙ Relation with Exponential Distribution

Exponential distribution is a special gamma distribution with = 1.

IF X1 , X2 , … , Xn have independent exponential distribution with the same
exponent 1/β, the sum of these random variables S= X1 + X2 + … +Xn results in
a gamma distribution, Γ(n, β).

If the time to observe an traffic accident (X) in a region have the following
probability distribution
-3x
f(x) = 3e , 0<x<∞

Estimate the probability to observe the first two accidents between the first
and second months. Assume that the all accidents are independent.
X1 : Time for the first accident
X2 : Time between the first and second accidents

Xi ∼ Exp(1/3) , I = 1, 2

S = X1 + X2 : Time for two accidents

S ∼ Γ(2, 1/3)
Probability function for S :

Answer:

(Continuous) Chi Square Distribution
A special gamma distribution α = r/2, β = 2

 PD

 E(X)

 Var(X)


A random variable X follows a Chi Square Distribution with a degree of
freedom of 5, Calculate the critical value to satisfy P(X < x0 )=0.95

Since P(X < x0 )=0.95, P(X > x0 )=0.05.

From the table, find the point with d.f.=5 and α=0.05

Why do we have to be bothered?

(Continuous) Weibull Distribution

The Weibull distribution is used:

•In survival analysis
•In reliability engineering and failure analysis
•In industrial engineering to represent
manufacturing and delivery times
•In extreme value theory
•In weather forecasting
•In communications systems engineering
•In General insurance to model the size of
Reinsurance claims, and the cumulative
development of Asbestosis losses

The probability of failure can be modeled by the Weibull distribution

Ex) What is the failure ratio of the smart phone with time?


The following are examples of engineering problems solved with Weibull
analysis:

• A project engineer reports three failures of a component in service
operations during a three-month period. The Program Manager asks, "How
many failures will we have in the next quarter, six months, and year?" What
will it cost? What is the best corrective action to reduce the risk and losses?

• To order spare parts and schedule maintenance labor, how many units will
be returned to depot for overhaul for each failure mode month-by-month next
year?


Probability distributions for reliability engineering

• Exponential
– “Random failure”
• Gamma
• Log-Normal
• Weibull
- The Weibull is a very flexible life distribution model
with two parameters.
- It has the ability to provide reasonably accurate failure
analysis and failure forecasts with extremely small
samples.

For positive α, β

 PDF :

 CDF :

 Survival


When a is 1, Weibull distribution reduces to the Exponential Model


 Failure function :

α = 3, β = 2
CDF : F(x)
SF : S(x)
FF : h(x)


 Failure function :

α : shape parameter
β : scale parameter

• β < 1.0 indicates infant mortality
• β = 1.0 means random failures (independent of age)
• β > 1.0 indicates wear out failures

(Continuous) Beta Distribution
Ex) Ratio of customers who satisfy the service
Ratio of time for watching TV in a day

α, β > 0
Beta function :

PDF condition

 Relationship between gamma and beta functions


If α < 1, the graph goes to left. If β < 1, it goes to right.

α<1 β<1


If α = β, the graph is symmetric
With increasing α and β, the graph becomes narrow.

α, β > 0

The ratio of customers who ask about LTE service (X) to the total customer in
the S* Telecom was represented by a beta distribution with α=3 and β=4.
(1) Probability distribution function for X
(2) Mean and variance
(3) Probability that 70% of the total customer asked.

(1)

(2)

(3) P(X ≥ 0.7) :

(Continuous) Normal Distribution


[Ref]
μ is the center of the graph and σ is the degree of variance.

μ is different and σ is same μ is same and σ is different.

Standardization

⊙ Use of the table


⊙ Approximation of Binomial distribution using Normal distribution


⊙ Approximation of Poisson distribution using Normal distribution

n→∞
X ∼ P(μ) X ≈ N(μ, μ)

Chemical Oxygen Demand (COD) of the water samples are known to be described by a
normal distribution with mean of 128.4 mg/L and standard deviation of 19.6 mg/L.

(1)Probability that the COD for a random water sample is less than 100 mg/L.
(2) Probability that the COD for a random water sample is between 110 and 130 mg/L.
2
(1) X ∼ N(128.4, 19.6 )

(2)

(Continuous) Central limit theorem

Central limit theorem
2
X1 , X2 , … , Xn : random variable with mean μ and variance σ

Large n

(Continuous) Central limit theorem

(Continuous) Log-Normal Distribution

(Continuous) Student t Distribution

(Continuous) F Distribution
⊙ Effect of the degree of freedom

For α, 100(1- α)% : fα(m, n)

(1) P(X ≥ fα(m, n) ) = α

(2) P(f1-α/2(m, n) ≤ X ≤ fα/2(m, n)) = α

(3) F ∼ F (m, n) ⇒ 1/F ∼ F (n, m)

Continuous Probability Distributions: Summary

• Uniform Distribution
• Exponential Distributions
• Gamma Distributions
• Chi-square Distributions
• Weibull Distributions
• Beta Distributions
• Normal Distribution
• Student t Distribution
• F Distribution

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