Lect10_2700_s09 - ENGRD 2700 Basic Engineering Probability...

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Normal CLT Gamma Density Other Probability Plots; QQ- . . . Title Page JJ II J I Page 1 of 25 Go Back Full Screen Close Quit ENGRD 2700 Basic Engineering Probability and Statistics Lecture 10: More Densities David S. Matteson School of Operations Research and Information Engineering Rhodes Hall, Cornell University Ithaca NY 14853 USA [email protected] February 23, 2009
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Normal CLT Gamma Density Other Probability Plots; QQ- . . . Title Page JJ II J I Page 2 of 25 Go Back Full Screen Close Quit 1. The normal density (cont). Reminder: Standard normal. Form of the standard normal pdf: n ( x ; 0 , 1) = f ( x ; 0 , 1) = 1 2 π e - 1 2 x 2 , x R . Non-standard normal. n ( x ; μ, σ 2 ) = f ( x ; μ, σ ) = 1 2 πσ e - 1 2 ( x - μ σ ) 2 , x R . Standardizing a non-standard normal. Suppose X 0 , 1 N ( x ; 0 , 1) and X μ,σ N ( x ; μ, σ 2 ) . Then in distribution X μ,σ := σX 0 , 1 + μ N ( x ; μ, σ 2 ) = F ( x ; μ, σ ) .
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Normal CLT Gamma Density Other Probability Plots; QQ- . . . Title Page JJ II J I Page 3 of 25 Go Back Full Screen Close Quit The non-standard variable X μ,σ can be standardized by subtracting the mean and dividing by the sd to get the standardized variable : X μ,σ - μ σ which has mean 0 and variance 1. Examples. 1. P [ X 0 , 1 1 . 62] = . 9474 . Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1 x P( X <= x ) 1.62 0.947384
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Normal CLT Gamma Density Other Probability Plots; QQ- . . . Title Page JJ II J I Page 4 of 25 Go Back Full Screen Close Quit 2. Find the 95th percentile η ( . 95) for N (0 , 1); that is the solution to P [ X 0 , 1 η ( . 95)] = . 95 . From Mtab: Inverse Cumulative Distribution Function Solution: η ( . 95) = 1 . 64485 . Minitab output: Normal with mean = 0 and standard deviation = 1 P( X <= x ) x 0.95 1.64485 3. Note P [ | X 0 , 1 | ≤ 1 . 64] = P [ | X 0 , 1 | ≤ η ( . 94)] = . 90 by symmetry.
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Normal CLT Gamma Density Other Probability Plots; QQ- . . . Title Page JJ II J I Page 5 of 25 Go Back Full Screen Close Quit 4. Find c such that P [ | X 0 , 1 | ≤ c ] = P [ - c < X 0 , 1 < c ] = 0 . 95 . Think: Means probability outside [ - c, c ] is .05. So probability above c is 0.025 by symmetry. So c = η (0 . 975), the 97.5th percentile. From Mtab: η (0 . 975) = 1 . 96 . (It is sometimes handy to remember this is 2 for rough back of the envelope calculations.) Mtab output: Inverse Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1 P( X <= x ) x 0.975 1.95996
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Normal CLT Gamma Density Other Probability Plots; QQ- . . . Title Page JJ II J I Page 6 of 25 Go Back Full Screen Close Quit z α notation. Frequently a hypothesis test is set up to reject the null hypothesis if a normally distributed test statistic is unusually large. To control error probabilities we need to know probabilities a normal variate is large. Definition. z α is the solution to P [ X 0 , 1 z α ] = α, 0 < α < 1 . So P [ X 0 , 1 z α ] = 1 - α, and thus z α = η (1 - α ) the 100(1 - α )th percentile. Commonly used values z . 025 = η ( . 975) = z . 05 = η ( . 95) =
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Normal CLT Gamma Density Other Probability Plots; QQ- . . .
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