# JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 1 Continuous Probability Distributions Many continuous probability distributions, including: Uniform.

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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

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

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?

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 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 v2 EGR 252 Spring 2009 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 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 μ = 0 and σ 2 = 1

JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 7 Standard Normal Distribution Table A.3: “Areas Under the Normal Curve”

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) =

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) =

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 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 v2 EGR 252 Spring 2009 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. 1)If P(Z < k) = 0.45, k = ___________ 2)Z = _______ X = _________

JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 12 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 v2 EGR 252 Spring 2009 Slide 13 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 v2 EGR 252 Spring 2009 Slide 14 Your Turn Refer to the previous example, 1.What is the probability that more than 50 survive? 2.What is the probability that exactly 45 survive? DRAW THE PICTURE!!

JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 15 Gamma & Exponential Distributions Related to the Poisson Process (discrete) Number of occurrences in a given interval or region “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 v2 EGR 252 Spring 2009 Slide 16 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 σ 2 = αβ 2

JMB Chapter 6 Part 1 v2 EGR 252 Spring 2009 Slide 17 Exponential Distribution Special case of the gamma distribution with α = 1. x > 0. Describes the time until or time between Poisson events. μ = β σ 2 = β 2

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?

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? β = 1 / λ = 1 / 2.7 = 0.3704 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 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? β = 1/ 2.7 = 0.3704 α = 2 P(X < 1) = (1/ β 2 ) x e -x/ β 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

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 = 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.

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