普瓦松分配汇编

1、7-*Learning ObjectivesnLO7-1 Describe the uniform probability distribution and use it to calculate probabilities.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 Approxim

2、ate the binomial probability distribution using the standard normal probability distribution to calculate probabilities.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 distributio

3、n 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

4、-*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 distribu

5、tion. 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. an

6、d 11 p.m. Students arrive at the bus stop at random times. The time that a student waits is uniformly distributed 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？nDraw a graph of this distribution.nShow that the area of this unifo

7、rm distribution is 1.00.nHow long will a student “typically” have to wait for a bus? In other words what is the mean waiting time? What is the standard deviation of the waiting times?nWhat is the probability a student will wait more than 25 minutes?nWhat is the probability a student will wait betwee

8、n 10 and 20 minutes?LO7-17-*The Uniform Distribution Example (p.209)1. Graph of uniformly distributed waiting times 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.

9、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 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

10、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

11、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) 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

12、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 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

13、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 approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.LO7-4 Approximate the binom

14、ial probability distribution using the standard normal probability distribution to calculate probabilities.7-*Normal Approximation to the BinomialUsing the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seem

15、s 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 Corr

16、ection 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 approximated by a continuous probability distribution (the normal distribution).LO7-47-*How to Apply the Correction

17、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 probability that X or fewer occurs, use the area below (X -0.5).4. For the probability that fewer than X o

18、ccurs, 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-)=792 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

19、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+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，

20、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 deviation = 0.782 (元)7-*指數分配 The Family of Exponential Distributio

21、ns 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 identify as , often referred to as the “rate” of occurrence p

22、arameter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 situations such as: The service times in a system The tim

23、e 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/=1/6小時P(x) = e-x 、E(X) = =1/ 、 V(X) =2= 1/ 2X:抵達時

24、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 twenty seconds.請問：下一個藥單1)

25、 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-(0.05)*4

26、0 = 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

27、發現設備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

28、 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 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 = 0.324