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Continuous Probability Distributions Part 2

Continuous Probability Distributions Part 2. Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull. Review: Standard Normal Random Variable.

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Continuous Probability Distributions Part 2

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  1. Continuous Probability Distributions Part 2 • Many continuous probability distributions, including: • Uniform • Normal • Gamma • Exponential • Chi-Squared • Lognormal • Weibull EGR 252 Spring 2012

  2. 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 EGR 252 Spring 2012

  3. Review:Standard Normal Distribution • Table A.3 Pages 735-736: “Areas under the Normal Curve” EGR 252 Spring 2012

  4. 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: Xis 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? EGR 252 Spring 2012

  5. 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 these equations: • Look at example 6.15, pg. 191 (9th edition) EGR 252 Spring 2012

  6. Look at example 6.15, pg. 191-192 p = 0.4 n = 100 μ= np = ___________ σ= ____________ For X < 30 z = ((30-0.5) – 40)/4.899 ) P (Z < ((30-0.5) – 40)/4.899 ) = P (Z < -2.14) = _________ (from table) EGR 252 Spring 2012

  7. Your Turn DRAW THE PICTURE!! • Refer to the previous example, • What is the probability that more than 50 survive? • What is the probability that exactly 45 survive? EGR 252 Spring 2012

  8. 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. EGR 252 Spring 2012

  9. 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 EGR 252 Spring 2012

  10. Exponential Distribution • Special case of the gamma distribution with α = 1. x > 0. • Describes the time until or time between Poisson events. μ = β σ2= β2 EGR 252 Spring 2012

  11. 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? EGR 252 Spring 2012

  12. 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. EGR 252 Spring 2012

  13. 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? 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 Using Excel: =GAMMADIST(1,2,1/2.7,TRUE) EGR 252 Spring 2012

  14. 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= αβα = 1(call) β = 1/2.7 μ = β = 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. EGR 252 Spring 2012

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