JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 Slide 1 Continuous Probability Distributions Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 Slide 2 Uniform Distribution Simplest – characterized by the interval endpoints, A and B. A ≤ x ≤ B = 0 elsewhere Mean and variance: and

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 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?

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 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 we’ll 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 μ and σ (sometimes μ and σ 2.)

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 Slide 5 Normal Distribution The density function of the normal random variable X, with mean μ and variance σ 2, is all x. (μ = 5, σ = 1.5)

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 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 μ = 0 and σ 2 = 1

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 Slide 7 Standard Normal Distribution Table A.3 Pages : “Areas under the Normal Curve”

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 Slide 8 Examples P(Z ≤ 1) = P(Z ≥ -1) = P(-0.45 ≤ Z ≤ 0.36) =

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 Slide 9 Name:________________________ 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) =

JMB Chapter 6 Part 1 v4 EGR 252 Spring 2012 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 μ = 40 and σ = 2 and x = 44 P(X<44) = P(Z< +2.0) = 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 1 v4 EGR 252 Spring 2012 Slide 11 The Normal Distribution “In Reverse” Example: Given a normal distribution with μ = 40 and σ = 6, find the value of X for which 45% of the area under the normal curve is to the left of X. Step 1 If P(Z < z) = 0.45, z = _______ (from Table A.3) Why is z negative? Step 2 X = _________ 45%