Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Normal Distribution 1 f (x) = e σ 2π 1 % x− µ ( − ' * 2& σ ) For Standard Normal Distribution µ = 0, σ = 1 € 1 f (x) = e 2π 1 2 − x 2 2 Standard Normal Distribution Area Under the Curve = 1 2π ∫ b a e −x 2 2 dx Numerical Integration = 1 for a = -∞, b = ∞ € = 0.67 for a = -1, b = 1 = 0.95 for a = -2, b = 2 = 0.025 for a = -∞, b = -2 (left tail probability) = 0.025 for a = 2, b = ∞ (right tail probability) Standard Normal Distribution Maximum Frequency at Mean The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. f (x) = 1 e 2π 1 − x2 2 Take first derivative, set equal to 0, solve for x: Derivative of e u = due u € f "(x) = € 1 2π 1 1 % − x2 ( − x2 −x e 2 '−xe 2 * = 2π & ) = 0 only if x = 0 = mean of distribution € Standard Normal Distribution Inflection Points at Mean ± Standard Deviation The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. Take second derivative, set equal to 0, solve for x: −x f "(x) = e 2π −1 f ""(x) = e 2π € −1 2 x 2 1 − x2 2 from above 1 2( −1 2 % − x x −x −1 2 2 2 + e (1− x ) '−xe * = 2π & ) 2π = 0 only if x = −1 or 1