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JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 1 Continuous Probability Distributions Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull Slide 2 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 2 Uniform Distribution Simplest characterized by the interval endpoints, A and B. A x B = 0 elsewhere Mean and variance: and Slide 3 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 3 Example: Uniform Distribution A circuit board failure causes a shutdown of a computing system until a new board is delivered. The delivery time X is uniformly distributed between 1 and 5 days. What is the probability that it will take 2 or more days for the circuit board to be delivered? Slide 4 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 4 Normal Distribution The bell-shaped curve Also called the Gaussian distribution The most widely used distribution in statistical analysis forms the basis for most of the parametric tests well perform later in this course. describes or approximates most phenomena in nature, industry, or research Random variables (X) following this distribution are called normal random variables. the parameters of the normal distribution are and (sometimes and 2.) Slide 5 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 5 Normal Distribution The density function of the normal random variable X, with mean and variance 2, is all x. ( = 5, = 1.5) Slide 6 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 6 Standard Normal RV Note: the probability of X taking on any value between x 1 and x 2 is given by: To ease calculations, we define a normal random variable where Z is normally distributed with = 0 and 2 = 1 Slide 7 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 7 Standard Normal Distribution Table A.3: Areas Under the Normal Curve Slide 8 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 8 Examples P(Z 1) = P(Z -1) = P(-0.45 Z 0.36) = Slide 9 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 9 Your turn Use Table A.3 to determine (draw the picture!) 1. P(Z 0.8) = 2. P(Z 1.96) = 3. P(-0.25 Z 0.15) = 4. P(Z -2.0 or Z 2.0) = Slide 10 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 10 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 = 40 and = 2 and x = 44 P(X JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 16 Gamma Distribution The density function of the random variable X with gamma distribution having parameters (number of occurrences) and (time or region). x > 0. = 2 = 2 Slide 17 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 17 Exponential Distribution Special case of the gamma distribution with = 1. x > 0. Describes the time until or time between Poisson events. = 2 = 2 Slide 18 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 18 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 3.5 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process How long before the next call? Slide 19 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 19 Poisson 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? = 1 / = 1 / 2.7 = 0.3704 = 2 What is the value of P(X 1) Can we use a table? No We must use integration. Slide 20 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 20 Poisson 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? = 1/ 2.7 = 0.3704 = 2 P(X < 1) = (1/ 2 ) x e -x/ dx = 2.7 2 x e -2.7x dx = [-2.7xe -2.7x e -2.7x ] 0 1 = 1 e -2.7 (1 + 2.7) = 0.7513 Slide 21 JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 21 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 = = = 1 = 1/2.7 = = 0.3407 min. We expect the next call to arrive in 0.3407 minutes. When = 1 the gamma distribution is known as the exponential distribution.