二項與Poisson三種分配之間的關(guān)系課件

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 ObjectivesLO7-1 Describe the uniform probability distribution and use it to calculate probabil

2、ities.LO7-2 Describe the characteristics of a normal probability distribution.LO7-3 Describe the standard normal probability distribution and use it to calculate probabilities.LO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate prob

3、abilities.LO7-5 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(xù)隨機變數(shù)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：機率總和

4、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(shù)（可據(jù)此求機率），也能根據(jù)此函數(shù)求得平

5、均數(shù)、變異數(shù)、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(xué)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(xué)生等公車的時間從0

6、分至30分呈均等分配。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 distribut

7、ed from 0 to 30 minutes.1.請繪製此分配的圖形 2.請說明其總面積為13.學(xué)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(xué)生等公車超過25分鐘的機率？5.學(xué)生等公車介於10到20分鐘的機率？Draw a graph of this distribution.Show that the area of this uniform distribution is 1.00.How long will a student “typically” have to wait for a bus? In other words what is the me

8、an waiting time? What is the standard deviation of the waiting times?What is the probability a student will wait more than 25 minutes?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

9、between 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? Wha

10、t is the standard deviation 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 1

11、0 and 20 minutes?計算介於10到20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(xiàn)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=

12、14縱軸截距 (機率): 1/(b-a)=1/(14-8)=0.167(b) 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np7時，可用Poisson分配取代二項分配 超幾何、二項與Poisson三種分配之間的關(guān)係Normal Approximation to the Binomial (p.226)當二項分配中的觀察點數(shù)目很大時，計算其隨機變量對應(yīng)的機率，非常繁瑣！ e.g. 如果n =60, 求 P(x

13、=33) = 60C33 ()33 (1-)27 但二項分配的觀察點數(shù)很大時，其分配趨近常態(tài)分配 觀察點數(shù)目要多大？ n 5 and n(1-) 57-*Normal Approximation to the BinomialThe normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n. The normal probability distri

14、bution is generally a good approximation 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 BinomialUsi

15、ng the normal distribution (a continuous 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.22

16、8)由於二項分配是間斷機率分配，常態(tài)分配是連續(xù)機率分配，若以常態(tài)分配來代替二項分配求其機率的話，必須做連續(xù)性調(diào)整。也就是：將二項隨機變數(shù)之值加或減0.5這0.5稱為連續(xù)性調(diào)整因子(p.228)二項機率:P(X=60)常態(tài)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 probabi

17、lity distribution) is being approximated 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,

18、use the area above (X+0.5).3. For the probability 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(xù)性修正因子 1/3 圖 連續(xù)性修正因子1/2設(shè)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 ，=連續(xù)性修正因子 2/3設(shè)XB(n,)，當n5且n(1)5，則存在下列近似式P(Xc)P(

19、cX 5、且n(1-)=792 5P(x126) = P(z126+0.5-108/9.75) = P(z1.897)=0.9713常態(tài)分配的加法定理定理1：假設(shè)XN(,2), 若W=a+bx, 則W呈何種機率分配？WN(a+b, b22)若X是常態(tài)分配，其線性函數(shù)W=a+bx 也是常態(tài)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(shù)：Y=0.5+0.45X，請問：每包成本的平均數(shù)為何？變異數(shù)為何？y=a+bx=0.5+0.4512=5.9y=b2x2=(0.45) 2 0.52=0.05常態(tài)分配的加法定理定理2：假設(shè)XN(x,x2), YN(y,y2

20、)若W=aX+bY, 則W呈何種機率分配？ WN(ax+by, a2x2+b2y2)常態(tài)分配的加法定理範例假設(shè)ucc咖啡每包成本與售價為一常態(tài)分配。每包成本平均為5.9元，變異數(shù)為0.05。每包售價平均10元，變異數(shù)為1。批發(fā)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(shù)與變異數(shù)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)+

21、(0.75)2*(1)=0.05+0.5625=0.6125Standard deviation = 0.782 (元)7-*指數(shù)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(shù)定義，：平均發(fā)生頻率（次

22、數(shù)）Described by only one parameter, which we identify 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 “l(fā)ess skewed.”LO7-5 Describe the characteristics and compute probabilities using the exponential distribution.The exponenti

23、al distribution usually describes inter-arrival situations 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(shù)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

24、客上門的機率。E(X)V(X)=若用指數(shù)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/=1/6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(wǎng)路藥局的下單頻率服從指數(shù)分配，平均每20秒鐘接一個藥單。Orders for prescriptions arrive at a pharmacy management webs

25、ite according to an exponential probability distribution at a mean of one every twenty seconds.請問：下一個藥單1) 不到5秒內(nèi)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？先找=

26、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）因P(下一藥單抵達時間 x) = 1 - e-x故1) P(下一藥單抵達時間40) = e-(0.05)*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.232-233LO7-57-*Exponential Distribution Example p.234Compton電腦公司希望訂定該公司

27、新電源設(shè)備的保證最低使用年限，品保測試發(fā)現(xiàn)設(shè)備使用壽命服從指數(shù)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

28、hours is a mean and not a rate. Therefore, we must compute as 1/4000 or 0.00025 failures per hour. 該公司希望在保固期內(nèi)僅5電源設(shè)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 p.235因= 4000小時 （電源平均壽命）指數(shù)分配平均值1/40000.00025 (每小時電源壞掉