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Lecture8standard - Statistics 511 Statistical Methods Dr...

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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Lecture 10: Other Continuous Distributions and Probability Plots Devore: Section 4.4-4.6 October, 2011 Page 1
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Gamma Distribution Gamma function is a natural extension of the factorial For any α > 0 , Γ( α ) = Z 0 x α - 1 e - x dx Properties: 1. If α > 1 , Γ( α ) = ( α - 1)Γ( α - 1) 2. Γ( n ) = ( n - 1)! for any n Z + 3. Γ ( 1 2 ) = π October, 2011 Page 2
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 The natural definition of a density based on the gamma function is f ( x ; α ) = x α - 1 e - x Γ( α ) if x 0 0 otherwise A gamma density with parameters α > 0 , β > 0 is f ( x ; α, β ) = 1 β α Γ( α ) x α - 1 e - x/β if x 0 0 otherwise α is a shape parameter , β is a scale parameter The case β = 1 is called the standard gamma distribution October, 2011 Page 3
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Gamma pdf: graphical illustration October, 2011 Page 4
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Gamma Distribution Parameters The mean and variance of a random variable X having the gamma distribution f ( x ; α, β ) are E ( X ) = αβ and V ( X ) = αβ 2 Let X have a gamma distribution with parameters α and β Then P ( X x ) = F ( x ; α, β ) = F ( x/β ; α ) In the above, F ( x ; α ) = Z x 0 y α - 1 e - y Γ( α ) dy is an incomplete gamma function . It is defined for any x > 0 . October, 2011 Page 5
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Exponential Distribution as a Special Case of Gamma Distribution Assume that α = 1 and β = 1 λ . Then, f ( x ; λ ) = λe - λx if x 0 0 otherwise Its mean and variance are E ( X ) = 1 λ and V ( X ) = 1 λ 2 Note that μ = σ in this case October, 2011 Page 6
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Exponential pdf: graphical illustration October, 2011 Page 7
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Exponential cdf Exponential cdf can be easily obtained by integrating pdf, unlike the cdf of the general Gamma distribution The result is F ( x ; λ ) = 1 - e - λx if x 0 0 otherwise October, 2011 Page 8
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Example I Suppose the response time X at an on-line computer terminal has an exponential distribution with expected response time 5 sec E ( X ) = 1 λ = 5 and λ = 0 . 2 The probability that the response time is between 5 sec and 10 sec is P (5 X 10) = F (10; 0 . 2) - F (5; 0 . 2) = 0 . 233 In R: pexp(10,0.2)-pexp(5,0.2) October, 2011 Page 9
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Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2011 Example II On average, 3
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