Poisson Distribution tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án, bài tập lớn về tất cả các lĩnh vực kinh tế...
Đề tài :Mối quan hệ của nhị thức và phân phối poisson I. Phân phối nhị thức: a. Định nghĩa: ĐLNN rời rạc X được gọi là phân phối theo quy luật nhị thức, ký hiệu X B(n,p) nếu nó nhận các giá trị 0,1… n với xác xuất tương ứng: P(X=k)= ;k=0,1…,n q=1-p chú ý : trong trường hợp X B(n,p) với n=1 ta nói phân phối theo quy luật không- một, ký hiệu X A(p) khi đó bảng phân phối xác xuất của X là x 0 1 p Q p Ta dễ thấy rằng nếu X A(p), thì E(X)=p; Var(X)=pq. b. Các số đặc trưng của phân phối nhị thức Nếu X B(n,p) thì E(X)=np, Var(X)=npq. Còn Mod(X) là số nguyên thỏa mãn điều kiện :np-q Mod(X) np+q. II. Phân phối poisson: a. Định nghĩa: ĐLNN rời rạc X được gọi là phân phối theo quy luật poisson nếu nó có thể nhận các giá trị :0,1 n, với các xác xuất tương ứng P(X=k)= , k=0,1,2…. Trong đó 0 là tham số. Nếu X có phân phối poisson với tham số ta ký hiệu X P( ). b. Các số đặc trưng Nếu X P( ) thì E(X)= , Var(X)= . Còn Mod(X)là số nguyên thảo mãn điều kiện Mod(X) . III. Mối liên hệ giữa nhị thức và phân phối poisson. Giả sử X B(n,p). khi n lớn , p khá bé thì X có phân phối xấp xỉ phân phối Poisson với tham số =np. Khi đó P(X=k)= Chứng minh Giả sử X là đại lượng ngẫu nhiên có phân phối nhị thức với tham số(n,p)và =np. Trong đó n khá lớn và p khá bé Ta có P(X=K)= = = Do n khá lớn và p khá bé nên: ; ; Khi đó ta có thể thay công thức bernoulli bằng công thức poisson ( đpcm) IV. Ví dụ Bài ví dụ 1) Một trạm điện thoại trung bình một giờ có 240 lần gọi đến . tìm xác suất để trong 1 phút a) Không có lần gọi nào b) Có từ 2 đến 3 lần gọi. Giải Gọi X là số lần gọi điện thoại trong 1 phút Ta có X P( ) với =E(X)= =4. P(X=0)= = =0,01832 P(2 X 3)=P(X=2)+P(X=3)= + =0,34189 Bài ví dụ 2: Nước giải khát được chở từ sài gòn đi vũng tàu, mỗi xe chở 1000 chia bia sài gòn, 2000 chai coca và 800 chai nước trái cây. Xác suất để một chai mỗi loại bị bể trên đường đi tương ứng là 0,2%,0,11%,và 0,3%. Nếu không quá một chai bị bể thì lái xe được thưởng . a) Tính xác suất có ít nhất một chai bia sài gòn bị bể b) Tính xác suất để lái xe được thưởng c) Lái xe phải trở ít nhất mấy chuyến để xác suất có ít nhất một chuyến được thưởng không nhỏ hơn 0,9? Giải Gọi là ĐlNN chỉ số chai bai SG bị bể trong một chuyến. khi đó có phân phối nhị thức B( ) với =1000 và =0,002 Vì khá lớn , khá bé có phân phối poisson P( ) với = . =1000.0,002=2 X=P(2). bị bể trong một chuyến. Khi đó , có phân phối poisson P(2000;0,0011)=P(2,2) P(800;0,003)=P(2,4) a) Xác suất để có ít nhất một chai bia sài gòn bị bể là P( 1)= 1-P( =0) =1- =0,8647 b) Tính xác suất lái xe được thưởng Theo giả thiết, lái xe được thưởng khi có không quá một chai bị bể, nghĩa là + + 1 P(2); P(2,2) ; P(2,4) nên + + P(2+2,2+2,4)=P(6,6) lái xe được thưởng là P( + + 1 )=P( + + =0)+P( + + =1) = + =0,0103. c) Lái xe phải chở ít nhất mấy chuyến để xác suất có ít nhất 1 chuyến được thưởng không nhỏ hơn 0,9? Gọi n là số chuyến xe cần thực hiện A là biến cố có ít nhất 1 chuyến xe được thưởng Yêu Poisson Distribution Poisson Distribution By: OpenStaxCollege There are two main characteristics of a Poisson experiment The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event For example, a book editor might be interested in the number of words spelled incorrectly in a particular book It might be that, on the average, there are five words spelled incorrectly in 100 pages The interval is the 100 pages The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000) You will verify the relationship in the homework exercises n is the number of trials, and p is the probability of a "success." The random variable X = the number of occurrences in the interval of interest The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12 Of interest is the number of loaves of bread put on the shelf in five minutes The time interval of interest is five minutes What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three? Let X = the number of loaves of bread put on the shelf in five minutes If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is 30 (12) = loaves of bread ( ) The probability question asks you to find P(x = 3) Try It The average number of fish caught in an hour is eight Of interest is the number of fish caught in 15 minutes The time interval of interest is 15 minutes What is the average number of fish caught in 15 minutes? ( 1560 )(8) = fish 1/14 Poisson Distribution A bank expects to receive six bad checks per day, on average What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day Let X = the number of bad checks the bank receives in one day If the bank expects to receive six bad checks per day then the average is six checks per day Write a mathematical statement for the probability question P(x < 5) Try It An electronics store expects to have ten returns per day on average The manager wants to know the probability of the store getting fewer than eight returns on any given day State the probability question mathematically P(x < 8) You notice that a news reporter says "uh," on average, two times per broadcast What is the probability that the news reporter says "uh" more than two times per broadcast This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast a What is the interval of interest? a one broadcast b What is the average number of times the news reporter says "uh" during one broadcast? b c Let X = What values does X take on? c Let X = the number of times the news reporter says "uh" during one broadcast x = 0, 1, 2, 3, d The probability question is P( ) d P(x > 2) Try It 2/14 Poisson Distribution An emergency room at a particular hospital gets an average of five patients per hour A doctor wants to know the probability that the ER gets more than five patients per hour Give the reason why this would be a Poisson distribution This problem wants to find the probability of events occurring in a fixed interval of time with a known average rate The events are independent Notation for the Poisson: P = Poisson Probability Distribution Function X ~ P(μ) Read this as "X is a random variable with a Poisson distribution." The parameter is μ (or λ); μ (or λ) = the mean for the interval of interest Leah's answering machine receives about six telephone calls between a.m and 10 a.m What is the probability that Leah receives more than one call in the next 15 minutes? Let X = the number of calls Leah receives in 15 minutes (The interval of interest is 15 minutes or hour.) x = 0, 1, 2, 3, If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives ( 18 )(6) = 0.75 calls in 15 minutes, on average So, μ = 0.75 for this problem X ~ P(0.75) Find P(x > 1) P(x > 1) = 0.1734 (calculator or computer) • • • • Press – and then press 2nd DISTR Arrow down to poissoncdf Press ENTER Enter (.75,1) The result is P(x > 1) = 0.1734 Note The TI calculators use λ (lambda) for the mean The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734: P(x > 1) = − poissoncdf(0.75, 1) 3/14 Poisson Distribution The graph of X ~ P(0.75) is: The y-axis contains the probability of x where X = the number of calls in 15 minutes Try It A customer service center receives about ten emails every half-hour What is the probability that the customer service center ... - 1 - CHƯƠNG I GIỚI THIỆU VỀ CÔNG FPT DISTRIBUTION 1.1.Qúa trình hình thành và phát triển của FPT DISTRIBUTION COMPANY (FDC): Công ty Phân phối FPT đã và đang dẫn đầu thị trường công nghệ thông tin (CNTT) và viễn thông tại Việt Nam, luôn sát cánh cùng các đối tác và hệ thống đại lý của mình mang đến cho người tiêu dùng Việt Nam những sản phẩm CNTT mới nhất, đa dạng về chủng loại và hoàn hảo về chất lượng. Là một công ty thành viên của Tập đoàn FPT, được chính thức thành lập từ ngày 13/4/2003 với trụ sở chính đặt tại Hà Nội và chi nhánh ở Tp. Hồ Chí Minh, Đà Nẵng, Cần Thơ, công ty luôn tự hào là đơn vị có thành tích kinh doanh nổi bật trong tập đoàn, với doanh thu năm 2007 vượt mức 516 triệu USD và tốc độ tăng trưởng hàng năm đạt hơn 59%. Hiệu quả hoạt động của Công ty Phân Phối FPT đã được khẳng định bởi chứng chỉ hệ thống quản lý chất lượng đạt tiêu chuẩn ISO9001:2000. Công ty Phân phối FPT đã chứng minh được vị thế số 1 trong lĩnh vực phân phối các sản phẩm CNTT và Viễn thông trên thị trường Việt Nam. Công ty Phân phối FPT có cơ cấu tổ chức chặt chẽ và thống nhất trên toàn quốc với đội ngũ nhân viên đông đảo, nhiệt tình, năng động, sáng tạo, có trình độ chuyên môn và năng suất lao động cao, trong đó trên 92% số nhân viên có kinh nghiệm hoạt động trong lĩnh vực CNTT, viễn thông và phân phối. Hệ thống thông tin đóng vai trò hết sức quan trọng trong thành công của Công ty, trong đó phải kể đến hệ thống thông tin tài chính và thông tin quản lý: phần mềm kế toán Oracle, FIFA (FPT Information Finance Architecture), MIS (Management Information System), SCM (Supply Chain Management), CRM (Customer Relationship Management), HRM (Human Resource Management), FDC Inside . Với những thế mạnh sẵn có cùng tôn chỉ hướng tới khách hàng, Công ty Phân phối FPT cam kết tiếp tục mang đến cho khách hàng của mình những giá trị gia tăng, giữ vững niềm tin và uy tín với các đối tác, tiếp tục đứng vững trên thị trường trong nước và vươn ra thị trường nước ngoài -Tên giao dịch : FPT Distribution -Ngày thành lập : 13/04/2003 (kết hợp từ 3 trung tâm phân phối của FPT) -Trụ sở : 298G Kim Mã, Quận Ba Đình, Hà Nội -Vốn đầu tư : 516 triệu USD - 67 - CHARACTERISTICS AND DISTRIBUTIONS OF NITROUS OXIDE-PRODUCING DENITRI- FYING FUNGI IN NATURAL ENVIRONMENTS K. Oishi and T. Kusuda Department of Civil Engineering, Faculty of Engineering, Kyushu University, Hakozaki 6-10-1, Higashi-ku, Fukuoka 812-8581, Japan Abstract Tea field soils, and sediments of an irrigation pond and a tidal river, in which a variety of organic matter was supplied as energy sources, were collected. The activities of bacterial and fungal denitrifications in these samples were determined. Denitrifying fungi in all of these samples produced N 2 O from nitrate and nitrite as a final product, whereas denitrifying bacteria produced N 2 . Nitrous oxide produced by fungi was reduced to N 2 by bacteria. The fungal denitrification potentials were the highest in the submerged litter on the pond sediment, followed by the farmyard manure-amended soil, the inorganic fertilizer-amended soil, the litter-free pond sediment, and the tidal river sediment. The enrichments of denitrifying fungi in natural envi- ronments were related with the distributions of the organic material such as straws and litter. The contributions of fungal denitrification to total denitrification were large in soil environments, especially in the farmyard manure-amended soil, and were small in aquatic environments such as the sediments of pond and river. The pH in situ was not related with the fungal denitrification potentials. Keywords fungal denitrification; bacterial denitrification; nitrous oxide; organic matter; sediments; soils Introduction Fungi generally are found in lakes, ponds, rivers, estuaries, marine, wastewater, and soils. Despite their wide occurrence, little attention has been given to the presence and ecological significance of fungi. Especially, characteristics and contributions of fungal denitrification in natural environments are poorly understood. Deni- trification is a process in which nitrite and/or nitrate are reduced to N 2 gas through N 2 O. The process has been considered to be mainly caused by bacteria. However, pure culture experiments have shown that fungi such as Fusarium sp., Trichoderma hamatum, Chaetomium sp., Gibberella fujikuroi etc., can reduce nitrite and several strains can reduce nitrate as well, but the final product is mainly N 2 O rather than N 2 ( Bleakley and Tiedje, 1982; Burth and Ottow, 1983; Shoun and Tanimoto, 1991; Shoun et al., 1992). The distributions of denitrifying fungi, which produce N 2 O as a final product, would be ecologically important to understand the contribution of natural systems to the atmospheric concentration of N 2 O. Previous studies have been con- ducted with pure cultures of fungi. However, the characteristics and distributions of denitrifying fungi in natural environments are unknown. In this study, tea soils, and the sediments of an irrigation pond and a tidal river, to which different types of organic matter were supplied, were collected. The final products of fungal denitrification in these soils and sediments were determined, and the distribution of denitrifying fungi in natural system was estimated. Materials Tea field soils Soils were collected at the surface (0-10cm) in two tea fields. One was mainly amended with a farmyard manure (organic soil), and the other with inorganic fertilizer Journal of Water and Environment Technology, Vol.3, No.2, 2005 - 169 - INFLUENCE DISTRIBUTIONS OF ACID DEPOSITION IN MOUNTAINOUS STREAMS ON A TALL CONE-SHAPED ISLAND, YAKUSHIMA S. Ebise* and O. Nagafuchi** *Dept. of Civil and Environmental Engineering System, Setsunan University, 17-8 Ikeda-Nakamachi, Neyagawa, Osaka, 572-8508, Japan **Dept. of Environmental System, Chiba Institute of Science, 3 Shiomi, Choshi, Chiba 288-0025 ABSTRACT Yakushima, facing at 800 km east of Shanghai in the East China Sea, is a tall cone-shaped island with seven exceeding 1800 m peaks. The prevailing winds of westrelie on the island blow mostly fromnorthwest and west. It has been exposed to acid rain of pH 4.7 and precipitation 8000 mm in the central highland. More than sixty mountainous streams were observed at downstream points seasonally for past twelve years. The alkalinity of streamwaters in the southwestern part was lower than others. The concentrations of SO 4 2- in the northwestern part were higher than others. The high concentrations of SO 4 2- , dissolved SiO 2 and other ions in the southwestern part with high canopy density of evergreen broadleaved forest were caused by higher air temperature, less rainfall and higher evapotranspiration than other parts. The alkalinity, pH and EC in the catchment of north stream in the upstream branch of the R. Anboh became lower than those in the catchment of south stream. The height of catchment boundaries, the direction of the main axis of a catchment and the prevailing winds govern the influence of acid deposition on water quality of branch streams. KEYWORDS Acid deposition; mountainous stream; influence distribution; cone-shaped; prevailing wind INTRODUCTION Yakushima, an island lying 800 km east of Shanghai on the boundary between the Pacific Ocean and East China Sea (Fig. 1), is a World Heritages Area and a National Park of Japan. Yakushima is famous for its yaku-sugi (Crytomeria japonica), one of which, called Jomon-sugi, is the oldest living organism in Japan. The prevailing winds on the island were northwestern (strongly so in winter) and western except during the short typhoon season, when the wind comes from the southeast. Consequently, Yakushima is exposed to acid deposition with an annual mean pH of 4.7 (MOE, 2004). The annual precipitation ranges 4300 mm at the coast to above 8000 mm in the central highland where seven peaks exceeds 1800 m. The tallest peak is Mt. Miyanouradake at 1935 m. The island has a steep mountainous landform and is covered with a thin soil layers overlying granite. Therefore, the soil’s ability to neutralize acid deposition is very weak (Ebise, 1996). Journal of Water and Environment Technology, Vol.3, No.2, 2005 - 170 - Figure 1 Yakushima Island and main rivers All 14,000 residents of the island live on the coast. There are no sources of artificial pollutants in the mountains, where the only visitors are mountain climber and backpackers. Back-trajectory and Pb-isotope trace methods show that most of the acid deposition coming from the Asia Continent is transported overseas by the prevailing westerlies in the upper air region, until it strikes the high mountains on the island. The major species in the acid deposition is sulfuric acid (Nagafuchi et al. 1995, 2001a, 2001b). The pH of wet deposition tends to be lower in Probability in Computing © 2010, Quoc Le & Van Nguyen Probability for Computing 1 LECTURE 7: CONTINUOUS DISTRIBUTIONS AND POISSON PROCESS Agenda Continuous random variables. Uniform distribution Exponential distribution © 2010, Quoc Le & Van Nguyen Probability for Computing 2 Poisson process Queuing theory Continuous Random Variables Consider a roulette wheel which has circumference 1. We spin the wheel, and when it stops, the outcome is the clockwise distance X from the “0” mark to the arrow. Sample space Ω consists of all real numbers in [0, 1). Assume that any point on the circumference is equally likely to © 2010, Quoc Le & Van Nguyen Probability for Computing 3 Assume that any point on the circumference is equally likely to face the arrow when the wheel stops. What’s the probability of a given outcome x? Note: In an infinite sample space there maybe possible events that have probability = 0. Recall that the distribution function F(x) = Pr(X ≤ x). and f(x) = F’(x) then f(x) is called the density function of F(x). Continuous Random Variables f(x)dx = probability of the infinitesimal interval [x, x + dx). Pr(a ≤X<b) = b dx x f ) ( © 2010, Quoc Le & Van Nguyen Probability for Computing 4 Pr(a ≤X<b) = E[X] = E[g(X)] = a dx x f ) ( dxxfxg )()( dxxxf )( Joint Distributions Def: The joint distribution function of X and Y is F(x,y) = Pr(X ≤ x, Y ≤ y). = where f is the joint density function. f(x, y) = y x dudvvuf ),( ),( 2 yxF yx © 2010, Quoc Le & Van Nguyen Probability for Computing 5 Marginal distribution functions F X (x)=Pr(X ≤x) and F Y (y)=Pr(Y ≤y). Example: F(x,y) = 1- e -ax – e -by + e -(ax+by) , x, y >= 0. F X (x)=F(x,∞) = 1-e -ax . F Y (y)=1-e -by . Since F X (x)F Y (y) = F(x, y) X and Y are independent. Conditional Probability What is Pr(X≤3|Y=4)? – Both numerator and denominator = 0. Rewriting Pr(X≤3|Y=4)? = ) 4 4 | 3 Pr( lim Y X © 2010, Quoc Le & Van Nguyen Probability for Computing 6 Rewriting Pr(X≤3|Y=4)? = Pr(X≤x|Y=y) = ) 4 4 | 3 Pr( lim 0 Y X x u Y du yf yuf )( ),( Uniform Distribution Used to model random variables that tend to occur “evenly” over a range of values Probability of any interval of values proportional to its width Used to generate (simulate) random variables from virtually any distribution © 2010, Quoc Le & Van Nguyen Probability for Computing 7 Used to generate (simulate) random variables from virtually any distribution Used as “non-informative prior” in many Bayesian analyses elsewhere 0 1 )( bya ab yf by bya ab ay ay yF 1 0 )( Uniform Distribution - expectation )(3 ))(( )(33 11 2)(2 ))(( )(22 11 )( 2 2 22333 22 222 ab abbaab ab aby ab dy ab yYE ab ab abab ab aby ab dy ab yYE b a b a b a b a © 2010, Quoc Le & Van Nguyen Probability for Computing 8 )(2887.0 12 12 )( 12 )( 12 2 12 )2(3)(4 23 )( )()( 3 )( 2 2222222 2 22 2 2 2 2 ab abab ababbaabababba ababba YEYEYV abba Additional Properties Lemma 1: Let X be a uniform random variable on [a, b]. Then, for c ≤ d, Pr(X ≤c|X ≤d)= (c-a)/(d- a). That is, conditioned on the fact that X ≤d, X is uniform on [a, d]. © 2010, Quoc Le & Van Nguyen Probability for Computing 9 uniform on [a, d]. Lemma 2: Let X 1 , X 2 , …, X n be independent uniform random variables over [0, 1], Let Y 1 , Y 2 , …, Y n be the same values as X 1 , X 2 , …, X n in increasing sorted order. Then E[Y k ] = k/(n+1). Exponential Distribution Right-Skewed distribution with maximum at y =0 Random variable can only take on positive values Used to model inter-arrival times/distances for a Poisson process © 2010, Quoc Le & Van Nguyen Probability for Computing 10 Poisson process [...]... Van Nguyen Probability for Computing 14 ... P(x = 100) = poissonpdf(104.1667, 100) ≈ 0.0366 P(x ≤ 100) = poissoncdf(104.1667, 100) ≈ 0.3651 The Poisson distribution can be used to approximate probabilities for a binomial distribution This... both the binomial and Poisson distributions Let X = the number of defective bulbs in a string Using the Poisson distribution: • μ = np = 100(0.03) = • X ~ P(3) • P(x ≤ 4) = poissoncdf(3, 4) ≈ 0.8153... are independent Notation for the Poisson: P = Poisson Probability Distribution Function X ~ P(μ) Read this as "X is a random variable with a Poisson distribution. " The parameter is μ (or λ); μ