Possion Distribution new slide.pptx

Poisson Distribution Definition The Poisson distribution is a discrete probability function that means the variable can only take specific values in a given list of numbers, probably infinite. A Poisson distribution measures how many times an event is likely to occur within “x” period of time.

1. Poisson distribution TVM E-Learning

2. Introduction The Binomial distribution describes a distribution of two possible outcomes designated as successes and failures from a given number of trials. The Poisson distribution focuses only on the number of discrete occurrence over some interval A Poisson experiment does not have a given number of trials (n) as binomial experiment does.

3. For example, whereas a binomial experiment might be used to determine how many black cars are in a random sample of 50 cars, a Poisson experiment might focus on the number of cars randomly arriving at a car wash during a 20-minute interval The Poisson distribution is the discrete probability distribution that applies to occurrences of some event over a specified interval . The random variable x is the number of occurrences of the event in an interval

4. The interval can be time Distance area volume Or Some similar unit The Poisson distribution is typically used as approximation to true underlying reality the number of car accidents in a day the number of dandelions in a square meter plot of land

5. Formula The only parameter involved in the Poisson distribution is the average ( lambda λ). X~P(λ) ; i.e. X is a RV that has a Poisson distribution. λ is the only parameter of the distribution, which is also the average rate at which that event occurs in the interval

6. Poisson probability mass function of lambda equal to 3 Poisson probability mass function

7. The average number of accidents on a public road is 7 times a week 1. What is the probability that no accident will happen in a particular week on the same road? example 2. The probability of a maximum of 3 accidents P(X, x≤4) P(X, x≤4) = P(0) + P(1) + P(2) + P(3) = 0.00091 + 0.00638 + 0.02234 + 0.05213 = 0.08176

8. The possibility of at least 4 accidents P(X, x≥4) P(X, x≥4) = 1 – [P(0) + P(1) + P(2) + P(3)] = 1 – [0.00091 + 0.00638 + 0.02234 + 0.05213] = 1 – 0.08176 = 0.91824 If the average number of accidents in a week is 3, then The possibility of more than two accidents in two weeks If 3 Events in a week Then μ = 3 × 2 = 6 in 2 weeks P( X,x >2) = 1 – [P(0) + P(1) + P(2)] 60 e–6 61 e–6 62 e–6 = 1 – ——— + ——— + ——— 0! 1! 2! = 1 – (0.0025 + 0.0149 + 0.0446) = 1 – 0.062 = 0.938

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