二项与Poisson三种分配之间的关系课件.ppt

1、Copyright 2015 McGraw-Hill Education.All rights reserved.No reproduction or distribution without the prior written consent of McGraw-Hill Education.Continuous Probability DistributionsChapter 77-*Learning ObjectivesnLO7-1 Describe the uniform probability distribution and use it to calculate probabil

2、ities.nLO7-2 Describe the characteristics of a normal probability distribution.nLO7-3 Describe the standard normal probability distribution and use it to calculate probabilities.nLO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate p

3、robabilities.nLO7-5 Describe the exponential probability distribution and use it to calculate probabilities.7-*均等分配 The Uniform Distribution均等分配是最簡單的連續隨機變數分配。The uniform probability distribution is perhaps the simplest distribution for a continuous random variable.此分配為方形分配，由最大值與最小值定義其範圍，此方形的面積1（亦即：機

4、率總和為1）This distribution is rectangular in shape and is defined by minimum(a)and maximum(b)values.LO7-1 Describe the uniform probability distribution and use it to calculate probabilities.7-*The Uniform Distribution Mean and Standard DeviationLO7-1在均等分配中，若知道最大值與最小值，我們就能定義均等分配的機率函數（可據此求機率），也能根據此函數求得平均

5、數、變異數、標準差。Knowing the minimum and maximum values of a uniform distribution,we can define the probability function,and calculate the mean,variance,and standard deviation of the distribution.7-*The Uniform Distribution Example(p.208)西南亞利桑納州立大學提供通勤公車，週一至週五從早上6點到晚上11點，每半小時一班公車（由北主街到校園），學生等公車的時間從0分至30分呈均

6、等分配。Southwest Arizona State University provides bus service to students.On weekdays,a bus arrives at the North Main Street and College Drive stop every 30 minutes between 6 a.m.and 11 p.m.Students arrive at the bus stop at random times.The time that a student waits is uniformly distributed from 0 to

7、 30 minutes.1.請繪製此分配的圖形 2.請說明其總面積為13.學生等公車一般要等多久？換句話說，平均等候時間為？等候時間的標準差為？4.學生等公車超過25分鐘的機率？5.學生等公車介於10到20分鐘的機率？1.Draw a graph of this distribution.2.Show that the area of this uniform distribution is 1.00.3.How long will a student“typically”have to wait for a bus?In other words what is the mean waitin

8、g time?What is the standard deviation of the waiting times?4.What is the probability a student will wait more than 25 minutes?5.What is the probability a student will wait between 10 and 20 minutes?LO7-17-*The Uniform Distribution Example(p.209)1.Graph of uniformly distributed waiting times between

9、0 and 30:P(X)=1/(30-0)=0.0333LO7-17-*The Uniform Distribution Example(p.209)2.Show that the area of this distribution is 1.00.LO7-17-*The Uniform Distribution Example(p.209)3.How long will a student“typically”have to wait for a bus?In other words what is the mean waiting time?What is the standard de

10、viation of the waiting times?LO7-17-*The Uniform Distribution Example(p.209-210)4.What is the probability a student will wait more than 25 minutes?計算介於25到30間的面積等候時間大於25分鐘的機率LO7-17-*The Uniform Distribution Example(p.210)5.What is the probability a student will wait between 10 and 20 minutes?計算介於10到2

11、0分鐘之間的面積其機率LO7-1Self-review 7-1(p.210)澳洲的牧羊犬壽命相對較短，他們的壽命介於8到14歲之間，且呈現均等分配。問：(a).請繪製此均等分配的圖形(b).說明此分配的面積為1(c).計算其平均值、標準差(d).某隻牧羊犬壽命介於10到14歲之機率？(e).牧羊犬壽命低於9歲的機率？(f).牧羊犬壽命正好等於9歲的機率？What is the probability a dog will live exactly 9 years?Self-review 7-1(p.210)(a)a=8,b=14縱軸截距(機率):1/(b-a)=1/(14-8)=0.167(b

12、)1/(14-8)*(14-8)=1(c)mean=(a+b)/2=(14+8)/2=11 s.d.=(14-8)2/12=1.732(d)P(10X14)=1/(14-8)*(14-10)=0.667(e)P(X20且且np7時，可用時，可用Poisson分配取代二項分配分配取代二項分配 超幾何、二項與超幾何、二項與Poisson三種分配之間的關係三種分配之間的關係Normal Approximation to the Binomial(p.226)l當二項分配中的觀察點數目很大時，計算其隨機變量對應的機率，非常繁瑣！e.g.如果n=60,求 P(x=33)=60C33()33(1-)27

13、l但二項分配的觀察點數很大時，其分配趨近常態分配 觀察點數目要多大？n 5 and n(1-)57-*Normal Approximation to the Binomial The normal distribution(a continuous distribution)yields a good approximation of the binomial distribution(a discrete distribution)for large values of n.The normal probability distribution is generally a good app

14、roximation to the binomial probability distribution when n and n(1-)are both greater than 5.LO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate probabilities.7-*Normal Approximation to the BinomialUsing the normal distribution(a con

15、tinuous distribution)as a substitute for a binomial distribution(a discrete distribution)for large values of n seems reasonable because,as n increases,a binomial distribution gets closer and closer to a normal distribution.LO7-4Continuity Correction Factor(p.228)由於二項分配是間斷機率分配，常態分配是連續機率分配，若以常態分配來代替二項

16、分配求其機率的話，必須做連續性調整。也就是：將二項隨機變數之值加或減0.5這0.5稱為連續性調整因子連續性調整因子(p.228)二項機率:P(X=60)常態分配:P(59.5X60.5)7-*Continuity Correction FactorThe value.5 is subtracted or added,depending on the problem,to a selected value when a binomial probability distribution(a discrete probability distribution)is being approximat

17、ed by a continuous probability distribution(the normal distribution).LO7-47-*How to Apply the Correction FactorOnly one of four cases may arise:1.For the probability at least X occurs,use the area above(X-0.5).2.For the probability that more than X occurs,use the area above(X+0.5).3.For the probabil

18、ity that X or fewer occurs,use the area below(X-0.5).4.For the probability that fewer than X occurs,use the area below(X+0.5).LO7-4連續性修正因子 1/3 圖圖 連續性修正因子連續性修正因子 1/2n設XB(n,)，當n5且n(1)5，則存在下列近似式P(Xd)P(Xc)P(cXd)式中=n，=0.5dP Z0.5cP Z0.50.5cdPZ(1)n連續性修正因子 2/3n設XB(n,)，當n5且n(1)5，則存在下列近似式P(Xc)P(cX 5、且n(1-)=79

19、2 5P(x126)=P(z126+0.5-108/9.75)=P(z1.897)=0.9713常態分配的加法定理定理1：假設XN(,2),若W=a+bx,則W呈何種機率分配？WN(a+b,b22)若X是常態分配，其線性函數W=a+bx 也是常態分配例：ucc三合一咖啡每包重量XN(12,0.52),亦知每包成本Y是每包重量的函數：Y=0.5+0.45X，請問：每包成本的平均數為何？變異數為何？y=a+bx=0.5+0.4512=5.9y=b2x2=(0.45)2 0.52=0.05常態分配的加法定理定理2：假設XN(x,x2),YN(y,y2)若W=aX+bY,則W呈何種機率分配？WN(ax

20、+by,a2x2+b2y2)常態分配的加法定理範例假設ucc咖啡每包成本與售價為一常態分配。每包成本平均為5.9元，變異數為0.05。每包售價平均10元，變異數為1。批發給零售商的售價為75折。請問：ucc咖啡公司每包利潤的平均數與變異數為何？又每包利潤成何種分配？成本XN(5.9,0.05)售價YN(10,1)利潤Z=0.75Y-X=-X+0.75Y Z N(ax+by,a2x2+b2y2)Mean(z)=-1*5.9+0.75*10=7.5-5.9=1.6Variance(z)=(-1)2*(0.05)+(0.75)2*(1)=0.05+0.5625=0.6125Standard devi

21、ation=0.782(元)7-*指數分配 The Family of Exponential Distributions p.231Characteristics and uses:正偏，類似普瓦松分配Positively skewed,similar to the Poisson distribution(for discrete variables)Not symmetric like the uniform and normal distributions僅用一個參數定義，：平均發生頻率（次數）Described by only one parameter,which we ident

22、ify as,often referred to as the“rate”of occurrence parameter當遞減，分配則越來越不偏As decreases,the shape of the distribution becomes“less skewed.”LO7-5 Describe the characteristics and compute probabilities using the exponential distribution.The exponential distribution usually describes inter-arrival situati

23、ons such as:The service times in a system The time between“hits”on a web site The lifetime of an electrical component The time until the next phone call arrives in a customer service center指數分配 vs 普瓦松分配例如：餐廳每小時平均有6為顧客上門(=6)，普瓦松分配機率：求某特定時段，2位顧客上門的機率。E(X)V(X)=若用指數分配，平均每小時6位顧客上門，代表平均10分鐘有1位顧客上門，或=1/=1/

24、6小時P(x)=e-x、E(X)=1/、V(X)=2=1/2X:抵達時間P(抵達時間 x)=e-x 注意：P(抵達時間 x)=P(抵達時間x)因為 P(抵達時間=x)=07-*Exponential Distribution Example p.232-233某網路藥局的下單頻率服從指數分配，平均每20秒鐘接一個藥單。Orders for prescriptions arrive at a pharmacy management website according to an exponential probability distribution at a mean of one every

25、 twenty seconds.請問：下一個藥單1)不到5秒內可接到的機率是？2)超過40秒才接到的機率是？3)介於5到40秒接到的機率是？Find the probability the next order arrives in:1)less than 5 seconds,2)more than 40 seconds,3)or between 5 and 40 seconds.LO7-5P.232 例題如何解？n先找=1/20=0.05（因為平均每20秒接一個藥單,故=20,=1/，每秒接0.05個藥單）n因P(下一藥單抵達時間 x)=1-e-xn故1)P(下一藥單抵達時間40)=e-(0

26、.05)*40 =0.13533)P(5下一藥單抵達時間 40)=P(下一藥單抵達時間 40)-P(下一藥單抵達時間5)=(1-0.1353)-0.2212=0.64357-*Exponential Distribution Example p.232-233221207788011155201.)()()(eArrivalPLO7-513530864701114014040201.)()()()(eArrivalPArrivalP7-*Exponential Distribution Example p.234Compton電腦公司希望訂定該公司新電源設備的保證最低使用年限，品保測試發現設備

27、使用壽命服從指數分配，平均可用4000小時，注意：4000小時為平均值，而非頻率，因此，1/40000.00025(每小時的耗損率)Compton Computers wishes to set a minimum lifetime guarantee on its new power supply unit.Quality testing shows the time to failure follows an exponential distribution with a mean of 4000 hours.Note that 4000 hours is a mean and not a

28、 rate.Therefore,we must compute as 1/4000 or 0.00025 failures per hour.該公司希望在保固期內僅5電源設備耗損，他們該訂定多長的保固期？Compton wants a warranty period such that only five percent of the power supply units fail during that period.What value should they set for the warranty period?LO7-5Exponential Distribution Example

29、 p.235因=4000小時（電源平均壽命）指數分配平均值1/40000.00025(每小時電源壞掉的的頻率）普瓦松分配平均值且希望訂出x，使得電源壽命低於x的機率不大於0.05 0.05=P(使用壽命、時間4500)=1-P(時間4500)=1-(1-e-0.00025*4500)=e-0.00025*4500=0.3247例題某公司平均10天接3筆訂單，問：3天內接到下一筆訂單的機率為？至少需要5天才能接到下一筆訂單的機率為？解答：此為指數分配=3/10=0.3（平均每天接0.3筆訂單）普瓦松分配的平均值P(時間3)=1-e-0.3*3=0.5934P(時間5)=1-P(時間5)=e-0.

30、3*5=0.2231注意：此指數分配的平均值為10/3=3.33天，亦即：平均每3.33天接一筆訂單例題由A點開到B點的公車每隔15分鐘開一班車，某人沒看時刻表，就到站牌等車，請問他至少要等5分鐘的機率為？請用普瓦松分配與指數分配分別計算之。解答：普瓦松分配普瓦松分配：（至少等5分鐘，所以5分鐘內來0班車）而5分鐘區間平均來(5/15)班車 E(X)=1/3 普瓦松平均值普瓦松平均值(5(5分鐘區間分鐘區間)P(x=0)=(1/3)0 e-(1/3)/0!=e-(1/3)=0.7165指數分配指數分配：E(X)=15 分鐘（平均15分鐘來一班車）=1/15=0.067（平均每分鐘來0.067班

31、車）普瓦松平均值普瓦松平均值(1(1分鐘區間分鐘區間)P(時間5)=e-0.067*5=e-(1/3)=0.7165指數分配 1/3Poisson隨機變數指數隨機變數(a)20分鐘內，平均5部車子開進停車場(=5輛20分鐘)(a)平均每隔4分鐘有一部車子開進停車場(=4分鐘輛)(b)高速公路一每10公里平均有5個窪洞(=5個10公里)。(b)高速公路上，平均每2公里有1個洞(=2公里個)。(c)某一機器30分鐘內平故障3次(=3次30分鐘)。(c)某一機器平均每10分鐘故障1次(=10分鐘次)表表 Poisson隨機變數與指數隨機變數之比較隨機變數與指數隨機變數之比較n設X為指數隨機變數，代表單位時間之平均次數，代表平均時間(每次)-/(),0,01(),0,0方方法法一一：方方法法二二：xxf xexf xex(6-21)(6-22)指數分配 2/3圖圖 指數分配機率之計算指數分配機率之計算()1/,0 xf xex指數分配 3/3各種分配的觀念與使用情形架構圖各種分配的觀念與各種分配的觀念與使用情形架構圖使用情形架構圖