COM2023 Mathematics Methods for Computing II

Transcription

1 COM223 Mathematics Methods for Computing II Lecture 11& 12 Gianne Derks Department of Mathematics (36AA4) Autumn 21 Use channel 4 on your EVS handset Overview Poisson distribution 3 Poisson distribution Eamples about the Poisson Distribution Graphs for Poisson Distributions (1) Graphs for Poisson Distributions (2) Binomial and Poisson Distribution (1) Binomial and Poisson distribution (3) Binomial and Poisson distribution (3) Epectation and variance Adding Poisson Distributions

2 Overview Distributions for discrete random variables: The binomial distribution (last hour) The Poisson distribution Poisson Distribution ( 4.6) Poisson distribution Poisson distributions are often used to describe the probabilities of the number of occurances of events which occur randomly (independently) at a constant average rate in space or time. For eample: the number of accidents per month on a given stretch of road; the number of faults in a piece of material; the number of telephone calls arriving at a switchboard in an hour; the number of fish caught in a given lake in 4 hours. The random variable X has a Poisson distribution with parameter λ if it has a probability mass function (pmf) given by p() = P(X = ) = λ! e λ, for =,1,2,3,... We denote this by X Po(λ). The number λ is the average rate with which the events occur. 2

3 Eamples about the Poisson Distribution The number of flaws in a fabric is on average 2 flaws per square meter. Assuming that the number of flaws X satisfies a Poisson distribution, find which one: X Po(2). For a square metric of fabric calculate the probability that it has no flaws: P(X = ) = 2 e 2!.1353 eactly 1 flaw: P(X = 1) = 21 e 2 1! or more flaws: P(X 2) = 1 p() p(1) = 1 3e A switchboard receives on average 5 calls per day. Assuming that the number of calls per day X satisfies a Poisson distribution, find which one: X Po(5). For an arbitrary day, calculate the probability that there are eactly 45 calls: P(X = 45) = 545 e 5 45! =.458; eactly 5 calls: P(X = 5) = 55 e 5 5! =.563. Graphs for Poisson Distributions (1) So the probability of getting the average is not very high for Poisson distributions with a large average (i.e., λ large). Using matlab, we can plot the distribution for Po(5) with between 2 and 8:.6 bar(2:8,poisspdf(2:8,5))

4 Graphs for Poisson Distributions (2) The Poisson distribution with a smaller average (i.e. λ small) looks quite different. Using matlab, we can plot the distribution for Po(2) with between and 1: bar(:1,poisspdf(:1,2)) Binomial and Poisson Distribution (1) The random variable X related to n independent trials, all with success probability θ, has a Binomial distribution (X B(n,θ)) if its pmf is p() = P(X = ) = ( n ) θ (1 θ) n, =,1,...,n The random variable X has a Poisson distribution with parameter λ (X Po(λ)) if it has a probability mass function (pmf) given by p() = P(X = ) = λ! e λ for =,1,2,3,... The Poisson distribution can be used to approimate to the Binomial distribution when the number of trials (n) is large and the probability of success (θ) is small. Just take λ = nθ. Below is a table with P(X = ). 1 Value of B(5, 1 ) B(1, 1 2 ) Po(1 2 )

5 Binomial and Poisson distribution (3) 1 Compare the binomial distribution B(1, 2 ) and Poisson distribution Po(5) graphically: bar(:2,binopdf(:2,1,.5)) bar(:2,poisspdf(:2,5)) Binomial and Poisson distribution (3) Eample The possibility that a person carries a rare gene is.2. A test is done with 1 people. Which binomial and approimate Poisson distributions should we use? The binomial distribution is X B(1,.2) and the approimate Poisson distribution is X Po(2). Using both the binomial and the approimate Poisson distributions, calculate the following probabilities in two ways: No person has the gene: The binomial probability is p B () = 1.2 (.998) 1 =.1351 and the Poisson probability is p P () = 2 e 2 /! =.1353; Less than 2 people have the gene: Binomial: P B (X < 2) = p B ()+p B (1) =.457; Poisson: P P (X < 2) = p P ()+p P (1) = (1+2)e 2 =

6 Epectation and variance The epectation and variance of a Poisson random variable are both equal to λ, i.e., E(X) = Var(X) = λ. Eample The number of flaws in concrete specimens is a Poisson random variable X with parameter λ = 3.5. (i) Find the standard deviation of X: σ = Var(X) = ; (ii) Find P(X = 4): P(X = 4) = (3.5)4 4! e ; (iii) Find P(X > 2): P(X > 2) = 1 p() p(1) p(2) = 1 (3.5)! e 3.5 (3.5)1 1! e 3.5 (3.5)2 2! e 3.5 = 1 ( /2)e Adding Poisson Distributions Property If X Po(λ) and Y Po(µ) and X and Y are independent, then X +Y Po(µ+λ). Eample The number of lorries per hour passing a point is modelled as a Poisson random variable X with parameter λ = 4. Calculate the probability that (i) less than 3 lorries pass in one hour: P(X < 3) = p()+p(1)+p(2) = (1+4+8)e 4 =.238; (ii) eactly 5 lorries pass in 2 hours: Define Z = X +X, then Z Po(4+4) = Po(8). Hence P(Z = 5) = 85 e 8 5! =.916; (iii) at most 1 lorry passes in 15 minutes: Define Ẑ = X/4, then Ẑ Po(4/4) = Po(1). Hence e 1 P(Ẑ 1) = p()+p(1) =! + e 1 1! =