stat1801_2

# stat1801_2 - RandomVariables Example:gendersof3children 3 3...

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X = No. of Boys 3 1 0 2 1 2 1 2 X(BBB) = 3 X(GBG) = 1 X(GGG) = 0 …… Y = Difference in no.        of each gender 3 1 1 3 1 1 1 1 Y(BBB) = 3 Y(GBG) = 1 Y(GGG) = 3 Random Variables Example: genders of 3 children BBB BBG BGB BGG GBB GBG GGB GGG { } { } GBB BGB BBG X , , 2 = = ( 29 { } ( 29 375 . 0 , , 2 = = = GBB BGB BBG P X P { } { } φ = = = 2 Y ( 29 ( 29 0 2 = = = P Y P : X Discrete random variable

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Probability Mass Function (X = 0) =  ({GGG}) = 0.125 (X = 1) =  ({BGG, GBG, GGB}) = 0.375 (X = 2) =  P  ({BBG, BGB, GBB}) = 0.375 (X = 3) =  ({BBB}) = 0.125 (0) = (1) = (2) = (3) = Space of  X  : ( 29 { } 3 , 2 , 1 , 0 = X Properties of pmf ( 29 0 . 1 x p ( 29 ( 29 1 . 2 = X x x p ( 29 ( 29 = A x x p A X P . 3 pmf of X :  ( 29 ( 29 ( 29 ( 29 125 . 0 3 375 . 0 2 375 . 0 1 125 . 0 0 = = = = p p p p ( 29 ( 29 3 , 2 , 1 , 0 , 5 . 0 3 3 = = x x x p pmf of X :
Cumulative Distribution Function ( 29 ( 29 ( 29 < < - = = x t p x X P x F x t , Example:  (0)=0.125,    (1)=0.375,    (2)=0.375,    (3)=0.125 ( 29 ( 29 ( 29 125 . 0 0 0 0 = = = p X P F ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 = = = = = = = 99999 . 0 96 . 0 78 . 0 125 . 0 0 5 . 0 5 . 0 F F F p X P F ( 29 ( 29 ( 29 ( 29 5 . 0 1 0 1 1 = + = = p p X P F ( 29 ( 29 ( 29 ( 29 1 3 0 3 3 = + = = p p X P F 0 1 2 3 0.125 0.5 0.875 1 x x x x { (2)=0.375 Step function

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Cumulative Distribution Function ( 29 ( 29 ( 29 < < - = = x t p x X P x F x t , Properties  Non-decreasing ( 29 ( 29 b F a F b a  Right continuous ( 29 ( 29 b F x F b x = + lim  Bounded ( 29 ( 29 1 , 0 = = - F F  Characterize  X ( 29 ( 29 ( 29 a F b F b X a P - = <
Continuous Random Variable Recall the relative frequency histogram width class frequency relative density =

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Continuous Random Variable
width class

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## This note was uploaded on 02/01/2012 for the course STAT 1801 taught by Professor Mrchung during the Fall '10 term at HKU.

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stat1801_2 - RandomVariables Example:gendersof3children 3 3...

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