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Unformatted text preview: 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 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 . Nonstandard normal. n ( x ; μ,σ 2 ) = f ( x ; μ,σ ) = 1 √ 2 πσ e 1 2 ( x μ σ ) 2 , x ∈ R . Standardizing a nonstandard normal. Suppose X , 1 ∼ N ( x ;0 , 1) and X μ,σ ∼ N ( x ; μ,σ 2 ) . Then in distribution X μ,σ := σX , 1 + μ ∼ N ( x ; μ,σ 2 ) = F ( x ; μ,σ ) . 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 nonstandard 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 , 1 ≤ 1 . 62] = . 9474 . Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1 x P( X <= x ) 1.62 0.947384 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 , 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 , 1  ≤ 1 . 64] = P [  X , 1  ≤ η ( . 94)] = . 90 by symmetry. 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 , 1  ≤ c ] = P [ c < X , 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 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....
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This note was uploaded on 03/05/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell.
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