Transcript
Introduction to Probability and Statistics7th 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.
(Discrete) Poisson Distribution
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 :
(Discrete) Poisson Distribution
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
.
(Discrete) Poisson Distribution.
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
(Discrete) Poisson Distribution
E(X) increases with parameter or .The graph becomes broadened with increasing the parameter or
Characteristics of Poisson Distribution
Probability mass function Cumulative distribution function
(Discrete) Poisson Distribution
(Discrete) Poisson Distribution
(Discrete) Poisson Distribution
(Discrete) Poisson Distribution
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
f(x)
In a Period [a, b], f(x) is constant.
E(x):
(Continuous) Uniform Distributions
Var(X) :
F(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,
(2)
(3)
(4)
(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
(Continuous) Exponential Distribution
(3) μ=1/3, accordingly 10 days.
(1)
(2)
From a survey, the frequency of traffic accidents X is given by
f(x) = 3e-3x
(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?
• Survival function :
• Hazard rate, Failure rate:
λ=0.01 이므로 분포함수와 생존함수 :
F(x)=1-e-x/100
, S(x)=e-x/100
(1) 이 환자가 150 일 이내에 사망할 확률 :
(2) 이 환자가 200 일 이상 생존할 확률
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
P(X < 150) = F(150) = 1-e-1.5
= 1-0.2231 = 0.7769
P(X ≥ 200) = S(200) = e-2.0
= 0.1353
(Continuous) Exponential Distribution
(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.
⊙ Relation with Poisson Process
(Continuous) Gamma Distribution
(Continuous) Gamma Distribution
α : shape parameter, α > 0β : scale parameter, β > 0
α = 1 Γ (1, β) = E(1/β)
(Continuous) Gamma Distribution
(Continuous) Gamma Distribution
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).
⊙ Relation with Exponential Distribution
Exponential distribution is a special gamma distribution with = 1.
X1 : Time for the first accidentX2 : 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:
If the time to observe an traffic accident (X) in a region have the following probability distribution
f(x) = 3e-3x
, 0 < x < ∞
Estimate the probability to observe the first two accidents between the first and second months. Assume that the all accidents are independent.
(Continuous) Chi Square Distribution
A special gamma distribution α = r/2, β = 2
PD
E(X)
Var(X)
(Continuous) Chi Square Distribution
(Continuous) Chi Square Distribution
(Continuous) Chi Square Distribution
(Continuous) Chi Square Distribution
(Continuous) Chi Square Distribution
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
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
(Continuous) Chi Square Distribution
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?
(Continuous) Weibull Distribution
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?
(Continuous) Weibull Distribution
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.
(Continuous) Weibull Distribution
PDF :
CDF :
Survival
For positive α, β
(Continuous) Weibull Distribution
When a is 1, Weibull distribution reduces to the Exponential Model
(Continuous) Weibull Distribution
Failure function :
α = 3, β = 2CDF : F(x)SF : S(x)FF : h(x)
(Continuous) Weibull Distribution
Failure function :
(Continuous) Weibull Distribution
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) Weibull Distribution
(Continuous) Beta Distribution
Relationship between gamma and beta functions
Ex) Ratio of customers who satisfy the service
Ratio of time for watching TV in a day
Beta function :
PDF condition
α, β > 0
(Continuous) Beta Distribution
If α < 1, the graph goes to left. If β < 1, it goes to right.
α < 1 β < 1
(Continuous) Beta Distribution
If α < 1, the graph goes to left. If β < 1, it goes to right.
α < 1 β < 1
(Continuous) Beta Distribution
α, β > 0
If α = β, the graph is symmetric
With increasing α and β, the graph becomes narrow.
(Continuous) Beta Distribution
(Continuous) Beta Distribution
(1)
(2)
(3) P(X ≥ 0.7) :
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.
(Continuous) Normal Distribution
(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.
(Continuous) Normal Distribution
(Continuous) Normal Distribution
(Continuous) Normal DistributionStandardization
(Continuous) Normal Distribution
(Continuous) Normal Distribution
(Continuous) Normal Distribution ⊙ Use of the table
(Continuous) Normal Distribution ⊙ Use of the table
(Continuous) Normal Distribution
(Continuous) Normal Distribution
(Continuous) Normal Distribution
(Continuous) Normal Distribution
⊙ Approximation of Binomial distribution using Normal distribution
(Continuous) Normal Distribution
⊙ Approximation of Poisson distribution using Normal distribution
X P(∼ μ) X ≈ N(μ, μ) n → ∞
(Continuous) Normal Distribution
(1) X N(128.4, 19.6∼2)
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)
(Continuous) Central limit theorem
Central limit theorem
X1 , X2 , … , Xn : random variable with mean μ and variance σ2
Large n
(Continuous) Central limit theorem
(Continuous) Log-Normal Distribution
(Continuous) Student t Distribution
(Continuous) Student t Distribution
(Continuous) F Distribution
(Continuous) F 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) F Distribution
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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