University:
McMaster University
Course:
STATS 1LL3 | Introduction to Probability and Statistics
Academic year:

2023
Views:

239

Pages:

4
Author:

Sabrina S.

Chapter 5: Pois distribution Poisson Distribution We are given a period of time during which events occur at random. The average rate at which events occur IS 2 events per time period. It is critical that the time period referred to in the rate must be the same as the time period during which the events are counted. Let the random variable during the time period. X be the number of events occuring Then X has the poisson distribution with parameter A, le X~P(2), and has the probability mass function p(x)=e x! X = 0,1,2,.. otherwise Example There is a large fleet of delivery trucks breakdowns per 5-day working week trucks on stand by On average there are 12 Each day there are two ↳ Let X be the random variable the number of trucks that break down in a given day Because we are interested in breakdowns per day, we need to convert the given weekly rate into a daily rate 12 Sdays 12 == 2.4 Thus we assume that X has Poisson distribution with 2 = 2.4 in Pr [x=x] = p(x) = -2.4 2.4x x! a.) What is the probability that on any day no standby is needed? Pr (no breakdowns) - Pr [x=o] -2.0 e 2.4 o! b.) What is the probability that number of standby trucks is inodequate? Pr (Inadequate standby) = Pr [x>2] - = - = 1- Pr [x2] (p(0) +p(1) + p(2)) [(0.0907) + (0.2177)+ (0.2613)] 0.4303 This means that 9% of days we will not use our standby trucks but on 43% of days we do not have enough. Deriving the Poisson Distribution.... lim n 2 lim n ()^= e^^ = 0 for any finite value a. events in time period J = acesses in 속 먹이도] Pr [x in J- Pr [x n trials =()()()^ see what happens when we let in tend to infinity Therefore the result we require p(x) = λ x! the probability mass function for the poisson distribution Exponential Distribution For the exponential distribution we consider the interval of time between events between events. Let the random variable X be the time Y is a continuous variable (it can take on any value in sample space s= {x1 x203) probability density functio If events are occuring at random with average rate & per unit of time then the probability density function for the random X, the length of time between events is given by variable X f(x) = Le otherwise Y is said to have the exponential distribution with parameter and we write XE() HAS TO BE ONE-DIMESIONAL SPACE Example A computer that operates continuously breaks down at random on average 1.6 times per week. This tells us A 1.5 per week and that the random variable X, the time between breakdowns, has density function +(x)= 1.5X ise 0 x>0 otherwise a.) What is the probability of no break downs for 2 weeks? -15x700 X > 2 exponential is continuous, evaluate probability 15€ dx -3 e+e - Pr[2]=S 00498 by integration b.) What is the probability of a breakdown within 3 days? make time compatible Wa e = 3 days -15x 3 -15x POCXC Se -0.6429 dx + e e 05258 +1 0.4742