Continuous Probability Distributions Part 2 Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull Uniform Normal – Gamma Exponential Chi-Squared Lognormal Weibull - JMB Chapter 6 Part 2 EGR 252 2019

Review: Standard Normal Random Variable Normal Distribution Review: the probability of X taking on any value between x1 and x2 is given by: To ease calculations, we define a normal random variable where Z is normally distributed with μ = 0 and σ2 = 1 JMB Chapter 6 Part 2 EGR 252 2019

Review: Standard Normal Distribution Table A.3 Pages 735-736: “Areas under the Normal Curve” JMB Chapter 6 Part 2 EGR 252 2019

Applications of the Normal Distribution A certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard deviation of 2 ohms. What percentage of the resistors will have a resistance less than 44 ohms? Solution: X is normally distributed with μ = 40 and σ = 2 and x = 44 P(X<44) = P(Z< +2.0) = 0.9772 Therefore, we conclude that 97.72% will have a resistance less than 44 ohms. What percentage will have a resistance greater than 44 ohms? JMB Chapter 6 Part 2 EGR 252 2019

Gamma & Exponential Distributions Related to the Poisson Process: Number of occurrences (discrete Ch.5) in a given interval or region Sometimes we’re interested in the number of events that occur in an area (eg flaws in a square yard of cotton). Sometimes we’re interested in the time until a certain number of events occur. Area and time are variables that are measured (continuous). Gamma distribution may apply. JMB Chapter 6 Part 2 EGR 252 2019

Gamma Distribution The density function of the random variable X with gamma distribution having parameters α (number of occurrences) and β (time or region). x > 0. μ = αβ σ2 = αβ2 Describes the time until a specified # of Poisson events occurs. JMB Chapter 6 Part 2 EGR 252 2019

Exponential Distribution Special case of the gamma distribution with α = 1. x > 0. Describes the time until Poisson event occurs Describes the time between Poisson events μ = β σ2 = β2 JMB Chapter 6 Part 2 EGR 252 2019

Is It a Poisson Process? For homework and exams in the introductory statistics course, you will be told that the process is Poisson. An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. What is the probability that up to a minute will elapse before 2 calls arrive? An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. How long before the next call? JMB Chapter 6 Part 2 EGR 252 2019

Poisson/Gamma Example Problem An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. What is the probability that up to 1 minute will elapse before 2 calls arrive? β = 1 / λ = 1 / 2.7 = 0.3704 α = 2 calls x = 1 minute What is the value of P(X ≤ 1)? Can we use a table? No We must use integration. β = 1/λ = 1/(2.7) = 0.3704 α = 2 JMB Chapter 6 Part 2 EGR 252 2019

Poisson/Gamma Example Solution An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. What is the probability that up to 1 minute will elapse before 2 calls arrive? The time until a number of Poisson events occurs follows the gamma distribution. β = (1/ 2.7) = 0.3704 α = 2 (calls) P(X < 1) = (1/ β2) x e-x/ β dx = 2.72 x e -2.7x dx = [-2.7xe-2.7x – e-2.7x] 01 = 1 – e-2.7 (1 + 2.7) = 0.7513 P = 75.13% Using Excel: =GAMMADIST( 1, 2, 1/2.7, TRUE ) β = 1/λ = 1/(2.7) = 0.3704 α = 2 P(X < 1) = 0∫1 (1/ β2) x e-x/ β dx = 2.72 0∫1 x e -2.7x dx = [-2.7xe-2.7x – e-2.7x]01 = 1 – e-2.7 (1 + 2.7) = 0.7513 JMB Chapter 6 Part 2 EGR 252 2019

Another Type of Question An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. What is the expected time before the next call arrives? Expected value = μ = α β α = 1(call) β = 1/2.7 μ = α β = (1) (0.3704) min. We expect the next call to arrive in 0.3704 minutes. When α = 1 the gamma distribution is known as the exponential distribution. Exponential distribution JMB Chapter 6 Part 2 EGR 252 2019