Probability Distribution - Binomial, Exponential and Normal
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This document provides information about probability distributions and related concepts. It discusses random variables, probability mass functions of discrete random variables, probability density functions of continuous random variables, and the mean and variance of the binomial, exponential, and normal distributions. Examples are provided to illustrate key concepts such as defining a random variable, determining probabilities of outcomes, and calculating expected values.
Probability Distribution - Binomial, Exponential and Normal
- 1. BRAINWARE UNIVERSITY Name: SAMPAD KAR Roll No: BWU/BTA/22/225 Section: D Group: D2 Course Name: Probability & Statistics Course Code: BSCM202
- 2. Contents • Random Variables • Probability Distribution of Discrete random variables • Probability Distribution of Continuous random variables • Mean and Variance of Binomial Distribution • Mean and Variance of Exponential Distribution • Mean and Variance of Normal Distribution
- 3. Random Variables A random variable is a variable which represents the outcome of a trial, an experiment or an event. It is a number which is different each time the trial or event is repeated. For Example, A coin is tossed 3 times simultaneously. Find the Probability of getting 3 heads? S = {(H,H,,H),(H,H,T),(H,T,T),(T,H,T),(H,T,H),(T,H,H),(T,T,H),(T,T,T)} P(getting at least 3 heads) = 1 8
- 4. Probability Distribution of Discrete random variables A discrete variable is a variable that can "only" take-on certain numbers on the number line. The Probability Mass Function (PMF) is also called a probability function or frequency function which characterizes the distribution of a discrete random variable. Let X be a discrete random variable of a function, then the probability mass function of a random variable X is given by Px (x) = P( X=x ), For all x belongs to the range of X It is noted that the probability function should fall on the condition : •Px (x) ≥ 0 and •∑xϵRange(x) Px (x) = 1 Example When we roll a single dice, the possible outcomes are: 1, 2, 3, 4, 5, 6 The probability of each of these outcomes is 16. If we define the discrete variable X as: X: the number obtained when rolling a dice. Then this is a discrete random variable since the sum of the probabilities of each of these possible outcomes is equal to 1, indeed: 1 6 + 1 6 + 1 6 + 1 6 + 1 6 + 1 6 =1
- 5. A continuous random variable is a random variable that has only continuous values. Continuous values are uncountable and are related to real numbers. The probability density function (pdf) is used to describe the probabilities associated with a continuous random variable. Probability Distribution of Continuous random variables
- 6. Mean and Variance of Binomial Distribution For binomial distribution, pmf = P(X=x) = 𝑥 𝑛 𝐶𝑃𝑥 (1 − 𝑃)𝑛−𝑥 Mean = E(x) = 𝑥=1 𝑛 𝑥. 𝑥 𝑛 𝐶𝑃𝑥 (1 − 𝑝)𝑛−𝑥 = 𝑥=1 𝑛 𝑥. 𝑛! 𝑛−𝑥 ! 𝑥! 𝑃𝑥 (1 − 𝑃)𝑛−𝑥 = 𝑥=1 𝑛 𝑥. 𝑛! 𝑛−𝑥 ! 𝑥(𝑥−1)! 𝑃𝑥(1 − 𝑃)𝑛−𝑥 = 𝑥=1 𝑛 𝑛.(𝑛−1)! (𝑛−1)−(𝑥−1) ! (𝑥−1)! 𝑃. 𝑃𝑥−1 (1 − 𝑃)(𝑛−1)−(𝑥−1) = nP 𝑥=1 𝑛 (𝑛−1)! (𝑛−1)−(𝑥−1) ! (𝑥−1)! 𝑃𝑥−1 (1 − 𝑃)(𝑛−1)−(𝑥−1) = nP 𝑥=1 𝑛 𝑥−1 𝑛−1 𝐶 𝑃𝑥−1 (1 − 𝑃)(𝑛−1)−(𝑥−1) = 𝑛𝑃(P + (1 − 𝑃))(𝑛−1) = nP
- 7. E(x(x-1)) = 𝑥=1 𝑛 𝑥(𝑥 − 1). 𝑥 𝑛𝐶𝑃𝑥(1 − 𝑝)𝑛−𝑥 = 𝑥=1 𝑛 𝑥(𝑥 − 1). 𝑛! 𝑛−𝑥 ! 𝑥! 𝑃𝑥(1 − 𝑃)𝑛−𝑥 = 𝑥=1 𝑛 𝑥(𝑥 − 1). 𝑛! 𝑛−𝑥 ! 𝑥(𝑥−1)(𝑥−2)! 𝑃𝑥(1 − 𝑃)𝑛−𝑥 = 𝑥=1 𝑛 𝑛.(𝑛−1)(𝑛−2)! (𝑛−1)−(𝑥−1) ! (𝑥−2)! 𝑃2 . 𝑃𝑥−2 (1 − 𝑃)(𝑛−2)−(𝑥−2) = (𝑛2−𝑛)𝑃2 𝑥=1 𝑛 (𝑛−2)! (𝑛−2)−(𝑥−2) ! (𝑥−2)! 𝑃𝑥−2(1 − 𝑃)(𝑛−2)−(𝑥−2) = (𝑛2 −𝑛)𝑃2 𝑥=1 𝑛 𝑥−2 𝑛−2 𝐶 𝑃𝑥−2 (1 − 𝑃)(𝑛−2)−(𝑥−2) = (𝑛2−𝑛)𝑃2 (P + (1 − 𝑃))(𝑛−2) = (𝑛2−𝑛)𝑃2 E(𝑥2 ) = E(x(x-1)) + E(x) = (𝑛2 −𝑛)𝑃2 + nP Variance = E(𝑥2) – {E(x)}2 = (𝑛2 −𝑛)𝑃2 + nP – 𝑛2 𝑃2 = nP−𝑛𝑃2 = nP(1-P)
- 8. Mean and Variance of Exponential Distribution For exponential distribution, f(x) = Mean = E(x) = −∞ 0 0𝑑𝑥 + 0 ∞ 𝑥.λ 𝑒−λ𝑥 𝑑𝑥 = 0 + λ 0 ∞ 𝑥𝑒−λ𝑥𝑑𝑥 = λ. 𝛤(1) λ 2 = 1 λ Now, E(𝑥2 ) = λ 0 ∞ 𝑥2 𝑒−λ𝑥 𝑑𝑥 = λ. 𝛤(3) λ 3 = 2 λ 2 {λ 𝑒−λ𝑥 , x > 0 0 , elsewhere Variance = E(𝑥2) – {E(x)}2 = 2 λ 2 - 1 λ 2 = 1 λ 2
- 9. Mean and Variance of Normal Distribution For Normal Distribution, σ = Standard Deviation µ = Mean σ2= Variance Standard Normal Distribution, N(0,1): σ = 1 , μ = 0 f(x) = 1 2Π 𝑒− 𝑧2 2 z = 𝑥−𝜇 𝜎
- 10. THANK YOU