7주차
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7주차 7주차 Presentation Transcript

  • 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.
  • (Discrete) Poisson DistributionIt is often used as a model for the number of events (such as the number oftelephone calls at a business, number of customers in waiting lines, numberof defects in a given surface area, airplane arrivals, or the number ofaccidents 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 DistributionEx. 3. A small life insurance company has determined that on the average it receives 6death claims per day. Find the probability that the company receives at least sevendeath claims on a randomly selected day. P(x ≥ 7) = 1 - P(x ≤ 6) = 0.393697Ex. 4. The number of traffic accidents that occurs on a particular stretch of roadduring a month follows a Poisson distribution with a mean of 9.4. Find the probabilitythat less than two accidents will occur on this stretch of road during a randomlyselected month. P(x < 2) = P(x = 0) + P(x = 1) = 0.000860
  • (Discrete) Poisson Distribution
  • Characteristics of Poisson Distribution E(X) increases with parameter or . The graph becomes broadened with increasing the parameter or
  • (Discrete) Poisson DistributionProbability mass function Cumulative distribution function
  • (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 DistributionsWhat 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):
  •  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
  • 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 firstaccident?(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 theprobability 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
  • (Continuous) Exponential Distribution ⊙ 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
  • (Continuous) Gamma Distribution α : shape parameter, α > 0 β : scale parameter, β > 0α=1 Γ (1, β) = E(1/β)
  • (Continuous) Gamma Distribution
  • (Continuous) Gamma Distribution ⊙ Relation with Exponential DistributionExponential distribution is a special gamma distribution with = 1.IF X1 , X2 , … , Xn have independent exponential distribution with the sameexponent 1/β, the sum of these random variables S= X1 + X2 + … +Xn results ina gamma distribution, Γ(n, β).
  • If the time to observe an traffic accident (X) in a region have the followingprobability distribution -3x f(x) = 3e , 0<x<∞Estimate the probability to observe the first two accidents between the firstand 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 DistributionA 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 DistributionA random variable X follows a Chi Square Distribution with a degree offreedom 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
  • (Continuous) Chi Square DistributionWhy 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 lossesThe probability of failure can be modeled by the Weibull distributionEx) What is the failure ratio of the smart phone with time?
  • (Continuous) Weibull DistributionThe following are examples of engineering problems solved with Weibullanalysis:• A project engineer reports three failures of a component in serviceoperations during a three-month period. The Program Manager asks, "Howmany failures will we have in the next quarter, six months, and year?" Whatwill 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 willbe returned to depot for overhaul for each failure mode month-by-month nextyear?
  • (Continuous) Weibull DistributionProbability 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 For positive α, β PDF : CDF : Survival
  • (Continuous) Weibull DistributionWhen a is 1, Weibull distribution reduces to the Exponential Model
  • (Continuous) Weibull Distribution Failure function : α = 3, β = 2 CDF : F(x) SF : S(x) FF : h(x)
  • (Continuous) Weibull Distribution Failure function :
  • (Continuous) Weibull Distributionα : shape parameterβ : 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 Ex) Ratio of customers who satisfy the service Ratio of time for watching TV in a day α, β > 0Beta function : PDF condition  Relationship between gamma and beta functions
  • (Continuous) Beta DistributionIf α < 1, the graph goes to left. If β < 1, it goes to right. α<1 β<1
  • (Continuous) Beta DistributionIf α < 1, the graph goes to left. If β < 1, it goes to right. α<1 β<1
  • (Continuous) Beta DistributionIf α = β, the graph is symmetricWith increasing α and β, the graph becomes narrow. α, β > 0
  • (Continuous) Beta Distribution
  • (Continuous) Beta DistributionThe ratio of customers who ask about LTE service (X) to the total customer inthe 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
  • (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 Distribution Standardization
  • (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 n→∞ X ∼ P(μ) X ≈ N(μ, μ)
  • (Continuous) Normal Distribution
  • Chemical Oxygen Demand (COD) of the water samples are known to be described by anormal 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 theoremCentral limit theorem 2X1 , 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) 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