Stat400Lec13(Ch4.1)_ans - STAT 400 Chapter 4.1 Spring 2012...

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STAT 400 Chapter 4.1 Spring 2012 Chapter 2- 3: Univariate distirubitons (Review) Discrete r.v. o Uniform discrete, hypergeometric, bernoulli, binomial, poisson , negative binomial Continuous r.v. o Uniform, cauchy, exponential, gamma, chi-square, normal (gaussian) Important Statistics: o Mean, variance, standard deviation, moment generating function, median Distribution: o p.m.f. c.d.f (discrete) o p.d.f c.d.f (continuous) Chapter 4: Bivariate distributions Let X and Y be two discrete random variables. The joint probability mass function p ( x , y ) is defined for each pair of numbers ( x , y ) by p ( x , y ) = P ( X = x and Y = y ) . Let A be any set consisting of pairs of ( x , y ) values. Then P ( ( X, Y ) A ) =   y x A y x p , , . Let X and Y be two continuous random variables. Then f ( x , y ) is the joint probability density function for X and Y if for any two-dimensional set A P ( ( X, Y ) A ) =  A dy dx y x f , . Joint p.m.f. for two discrete r.v.s. joint p.d.f for two continuous r.v.s.
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1. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: x \ y 0 1 2 1 0.15 0.10 0 2 0.25 0.30 0.20 It is a valid probability model because  x y y x p . 1 , a) Find P ( X + Y = 2 ) . P ( X + Y = 2 ) = p ( 1, 1 ) + p ( 2, 0 ) = 0.10 + 0.25 = 0.35 . b) Find P ( X > Y ) . P ( X > Y ) = p ( 1, 0 ) + p ( 2, 0 ) + p ( 2, 1 ) = 0.15 + 0.25 + 0.30 = 0.70 . c) Find the (marginal) probability distributions p X ( x ) of X and p Y ( y ) of Y. y p Y ( y ) x p X ( x ) 0 0.40 1 0.25 1 0.40 2 0.75 2 0.20 d) Find E ( X ) , E ( Y ) , E ( X + Y ) , E ( X Y ) . E ( X ) = 1 0.25 + 2 0.75 = 1.75 . E ( Y ) = 0 0.40 + 1 0.40 + 2 0.20 = 0.8 . E ( X + Y ) = 1 0.15 + 2 0.25 + 2 0.10 + 3 0.30 + 3 0 + 4 0.20 = 2.55 . OR E ( X + Y ) = E ( X ) + E ( Y ) = 1.75 + 0.8 = 2.55 . E ( X Y ) = 0 0.15 + 0 0.25 + 1 0.10 + 2 0.30 + 2 0 + 4 0.20 = 1.5 .
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2. Alexis Nuts, Inc. markets cans of deluxe mixed nuts containing almonds, cashews, and peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight of the third type. Let X = the weight of almonds in a selected can and Y = the weight of cashews.
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