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# Lec10 - IOE/Stat IOE/Stat 265 Fall 2009 Lecture#10(Uniform...

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IOE/Stat 265, Fall 2009 Lecture #10: (Uniform), Normal, and Exponential Distributions Ch 4-3 – 4-4 1 Continuous Uniform Distribution A random variable X has a continuous uniform A random variable X has a continuous uniform distribution if f(x) = 1/(b - a) for a x b 1 a x b f(x) (b a) 0 otherwise = Distribution Plot Uniform, Lower=5, Upper=11 a b E(x) + μ = = 0.18 0.16 0.14 0.12 0 10 y ( ) 2 2 b 0.10 0.08 0.06 0.04 0 02 Density 2 2 2 a E(x ) 12 σ = − μ = 11 10 9 8 7 6 5 0.02 0.00 X CDF for Continuous Uniform Distribution 0 ( ) ( ) x a x a F(X x) a x b b a < = = x b 1 > CDF: F(x) vs x 1.0 0.8 0 6 0.6 0.4 0.2 F(x) 3 16 14 12 10 8 6 4 2 0 0.0 x 4 3 Normal Distribution Properties 4-3 Normal Distribution Properties The normal distribution is undoubtedly the most The normal distribution is undoubtedly the most important and useful one of them all! Mean, Median, Mode all coincide. Symmetrical. N t ti Distribution Plot Normal, Mean=10, StDev=3 Notation: X ~ N ( μ , σ 2 ) 0.14 0.12 0.10 0.08 0.06 0.04 Density σ 4 μ 20 15 10 5 0 0.02 0.00 X μ

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Applications Normal Distributions are approximately EVERYWHERE! Examples: IQ Scores Measurement Error Economic Measures Most outputs of a random process that only have inherent variation present (SPC Charts). Combinations of other distributions (Central Limit Theorem). In fact, problem solving in many arenas often tries to figure out why data are NOT normally distributed. 5 Probability Distributions Probability Density Function for Normal 2 2 ( ) /(2 ) 1 ( ; ) - x f x e x μ σ μ σ = ∞ < < ∞ ( ; , ) 2 πσ and 0 −∞ < μ < ∞ σ > Example: Suppose X~N(100,1 2 ), find f(X=100) Æ S X N(100 5 2 ) fi d f(X 100) Æ Suppose X~N(100,5 ), find f(X=100) Suppose X~N(100,10 2 ), find f(X=100) Æ h l ff b d f df 6 Notice the scaling effect ~ remember condition 2 of pdf 0.4 1 StDev Distribution Plot Normal, Mean=100 0.3 ty 5 10 0.2 0.1 Densit 120 110 100 90 80 0.0 X 8 Standardizing a Variable To simplify the integration of f(x ) we will map To simplify the integration of f(x, μ , σ ), we will map our data into Standardized Form, aka “Z-score”. Z t d di i bl X th t Z-score standardizes a variable, X, so that Z has a mean = 0, and variance = 1. We may standardize our variable, X, by performing the following Z transform: x z − μ = σ 9
Normal Distribution Mapping Map any X into Z distribution. X ~ N ( μ , σ 2 ) −3σ −2σ −1σ +1σ +2σ +3σ μ N X − μ x -3 -2 -1 +1 +2 +3 0 Z Z ~ N (0,1 2 ) μ = σ z 10 PDF and CDF of Z (Standard Normal Random Variable) 2 z /2 1 f(z;0,1) e - z 2 = ∞ < < ∞ π PDF: CDF: (z) P(Z z) ϕ = Use Tables or Software to evaluate the CDF(z): Appendix A3 (page 668) Excel =normsdist(z) or =normdist(x, μ , σ ,true) 11 Minitab Calc >>Probability Distributions >>Normal Normal Table Symmetry Useful Equations: P(Z > z) = 1 – P(Z z) P(Z > z) = P(Z -z) Do we need positive and negative Z-tables? ) ( z Z P 12 Z Normal Table Positive Z Normal Table – Positive Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.50000 0.50399 0.50798 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.53586 0.1 0.53983 0.54380 0.54776 0.55172 0.55567 0.55962 0.56356 0.56749 0.57142 0.57535 0.2 0.57926 0.58317 0.58706 0.59095 0.59483 0.59871 0.60257 0.60642 0.61026 0.61409 0.3 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.65173 0.4 0.65542 0.65910 0.66276 0.66640 0.67003 0.67364 0.67724 0.68082 0.68439 0.68793 0.5 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.72240 0.6 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.75490 0.7 0.75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.78524 0.8 0.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.81327 0.9 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.83891 1.0 0.84134 0.84375 0.84614 0.84849 0.85083 0.85314 0.85543 0.85769 0.85993 0.86214 1.1 0.86433 0.86650 0.86864 0.87076 0.87286 0.87493 0.87698 0.87900 0.88100 0.88298

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