# Continuous Probability Distributions Many continuous probability distributions including

- Slides: 21

Continuous Probability Distributions • Many continuous probability distributions, including: ü Uniform ü Normal ü Gamma ü Exponential ü Chi-Squared ü Lognormal ü Weibull JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 1

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 v 2 EGR 252 Fall 2011 Slide 2

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 v 2 EGR 252 Fall 2011 Slide 3

Normal Distribution • The “bell-shaped curve” • Also called the Gaussian distribution • The most widely used distribution in statistical analysis ü forms the basis for most of the parametric tests we’ll perform later in this course. ü describes or approximates most phenomena in nature, industry, or research • Random variables (X) following this distribution are called normal random variables. ü the parameters of the normal distribution are μ and σ (sometimes μ and σ2. ) JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 4

Normal Distribution • The density function of the normal random variable X, with mean μ and variance σ2, is all x. (μ = 5, σ = 1. 5) JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 5

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 σ2 = 1 JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 6

Standard Normal Distribution • Table A. 3 Pages 735 -736: “Areas under the Normal Curve” JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 7

Examples • P(Z ≤ 1) = • P(Z ≥ -1) = • P(-0. 45 ≤ Z ≤ 0. 36) = JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 8

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 v 2 EGR 252 Fall 2011 Slide 9

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 1 v 2 EGR 252 Fall 2011 Slide 10

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 45% X = _____ JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 11

Normal Approximation to the Binomial • If n is large and p is not close to 0 or 1, or if n is smaller but p is close to 0. 5, then the binomial distribution can be approximated by the normal distribution using the transformation: • NOTE: add or subtract 0. 5 from X to be sure the value of interest is included (draw a picture to know which) • Look at example 6. 15, pg. 191 (8 th edition) JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 12

Look at example 6. 15, pg. 191 -192 p = 0. 4 n = 100 μ = ______ σ = _______ if x = 30, then z = ___________ and, P(X < 30) = P (Z < _____) = _____ JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 13

Your Turn • Refer to the previous example, DRAW THE PICTURE!! 1. What is the probability that more than 50 survive? 2. What is the probability that exactly 45 survive? JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 14

Gamma & Exponential Distributions • Related to the Poisson Process (discrete) ü Number of occurrences in a given interval or region ü “Memoryless” process • 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. • Typical problem statement: The length of time in days between student complaints about too much homework follows a gamma distribution with α = 3 and β = 7. What is the probability that up to two weeks will elapse before 6 complaints are heard? JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 15

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 = αβ 2 JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 16

Exponential Distribution • Special case of the gamma distribution with α = 1. x > 0. ü Describes the time until or time between Poisson events. μ=β σ2 = β 2 JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 17

Is It a Poisson Process? • For homework and exams in the introductory statistics course, you will be told that the process is Poisson. ü 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? ü 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? JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 18

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? ü β = 1 / λ = 1 / 2. 7 = 0. 3704 ü α=2 What is the value of P(X ≤ 1) Can we use a table? No We must use integration. JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 19

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? β = 1/ 2. 7 = 0. 3704 α = 2 P(X < 1) = (1/ β 2) x e-x/ β dx = 2. 72 x e -2. 7 x dx = [-2. 7 xe-2. 7 x – e-2. 7 x] 01 = 1 – e-2. 7 (1 + 2. 7) = 0. 7513 JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 20

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 β = 1/2. 7 μ = β = 0. 3407 min. We expect the next call to arrive in 0. 3407 minutes. When α = 1 the gamma distribution is known as the exponential distribution. JMB Chapter 6 Part 1 v 2 EGR 252 Fall 2011 Slide 21

- Continuous Probability Distributions Many continuous probability distributions including
- Continuous Probability Distributions Many continuous probability distributions including
- Continuous Probability Distributions Many continuous probability distributions including
- Continuous Probability Distributions Many continuous probability distributions including
- Continuous Probability Distributions Many continuous probability distributions including