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

2023
Views:

190

Pages:

4
Author:

Sabrina S.

CH 5: Probability Distributions 1 - The binomial, poisson, exponential, and normal distributions The Binomial Distribution May only be used when these conditions are satisfied: Random experiment which has a sample space with exactly two outcomes, one we label "success" and the other "failure." Random experiment is repeated n times (n ≥ 1). The outcome on any one repetition is not influenced by the outcome of any other repetition. We say we have n independent trials of a random experiment. Probability of success remains constant from trial to trial. Pr(success) = p; Pr(failure) = 1 - p or q = 1 - p. Our random variable X is the number of successes we observe in n trials. If the above conditions are satisfied, we have a binomial process, and random variable X has a binomial distribution. Binomial Distribution n independent trials Two outcomes Random variable X is the number of successes in n trials p(x) = * (n choose x) * p^x * (1 - p)^(n - x) if x = 0, 1, ..., n * 0 otherwise n and p are parameters of the distribution (n ≥ 1) and (0 < p < 1). X ~ B(n, p) Video 12: The Poisson and Exponential Distributions Excel spreadsheet containing pothole example: 1000 kms of road 850 potholes on this stretch of road On any repetition of a random experiment, we can't know what the outcome will be. We can use some theoretical model to give us a probability, so we have an idea of what to expect. Observed pattern gets closer to theoretical pattern with more repetitions. X = the number of potholes that occur in a given stretch of road (5 km) Discrete probability mass function The function will depend on only 1 parameter: the average rate of occurrence of potholes Average rate of occurrence = 850 / 1000 = 0.85 potholes per km ... = 150 * 0.85 = 4.25 potholes per 5 km of road X ~ P(λ = 4.25) p(x) = (e^(-λ) * λ^x) / x! for x = 0, 1, 2, 3, ... = 0 elsewhere This distribution depends on events occurring randomly but with an average rate of occurrence → POISSON PROCESS The random variable of interest is distance Y is a continuous random variable (distance) Y ~ E(λ) f(y) = λe^(-λy) for y > 0 f(y) = 0 elsewhere The average rate of occurrence of events that occur randomly: Ensure that λ uses the same units as the question asked E.g., λ for 5 km will be half that for 10 km Probability of exactly 2 potholes in a 5 km stretch of road (Poisson distribution): λ = 4.25 X ~ P(λ = 4.25) p(x) = (e^(-λ) * λ^x) / x! p(2) = (e^(-4.25) * 4.25²) / 2! = 0.129 Probability of 3 km stretch of (smooth) road between potholes (Exponential distribution): λ = 0.85 per km of road Pr(Y ≥ 3) = ∫₃^∞ f(y) dy = ∫₃^∞ 0.85e^(-0.85y) dy Pr(Y ≥ 3) = [-e^(-0.85y)]₃^∞ = 0.078 Probability of finding no potholes in 3 km stretch of road (Poisson): X ~ P(λ = 2.55 per 3 km of road) P(X = 0) = e^(-λ) * λ^0 / 0! = e^(-2.55) = 0.078 Poisson Distribution: Models the number of events occurring randomly With an average rate of occurrence per unit of time/space/distance Exponential Distribution: Models space or distance between events occurring randomly With an average rate of occurrence