Day4_Feb_25_M360 - The p.d.f for a random variable X is...

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The p.d.f. for a random variable, X , is given by the function: Is this function a valid p.d.f.? , 0 ( ) 0 ,all other values of x x x e x f x - =
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2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 -0.2 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 f x ( 29 = x e -x 3 1 x x e dx - =
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Continuous joint density function: Suppose that X and Y are continuous random variables with a p.d.f. given by : (2 3 ) ,0 2, 0 2 ( , ) 20 0, all other values of , x y x y f x y x y + = What does the graph of this function look like?
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Is this a valid pdf ?
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2 2 0 0 2 3 20 x y dy dx + = ∫ ∫
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Determine the probability that X < ½ and Y < 1
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A D B C 1/2 1 0 0 (2 3 ) 20 x y dy dx + Determine the probability that X < ½ and Y < 1 X Y Z
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1/2 1 0 0 (2 3 ) 20 x y dy dx + Determine the probability that X < ½ and Y < 1
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Determine the probability that X > 1
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Determine the probability that X > 1 A D B C
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Determine the probability that X > 1 3 5 =
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Evaluate this integral and interpret the result: g(x) is the marginal distribution of X and 2 0 2 3 2 3 ( ) 20 10 x y x g x dy + + = = ( ) ( ) b a P a X b g x dx < < = 2 1 2 3 3 ( 1) (1 2) 10 5 x P X P X dx + = < < = = ( ) g x
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Evaluate this integral and interpret the result: h(x) is the marginal distribution of Y and 2 0 2 3 ( ) 20 x y h y dx + = 2 0 2 3 3 2 ( ) 20 10 x y y h y dx + + = = ( ) ( ) b a P a Y b h y dy < < =
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A snack food company markets cans of deluxe mixed nuts containing almonds, cashews, and peanuts. The net weight of each can is exactly 1 pound, but the weight contribution of each type of nut is random. Because the three weights sum to one, a joint pdf for any two of the three gives all necessary information about the third type. Let X = the weight of the almonds in a random can and Y = the weight of the cashews.
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