55010Solution4 - STAT 550 HW #4 Solution 2.1-5 (a) p(0) F...

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1 STAT 550 HW #4 Solution 2.1-5 (a) (0) (0) .2 (1) (0) .5 .2 .3 pF p F F  Therefore, we have the probability mass function of X is: 0 1 2 3 4 ( ) .2 .3 .3 .1 .1 x px (b) ( 3) (3) (4) .2 P X p p 2.1-6 (a) 1 .1 .3 .05 .25 .21 .09 k     (b) ( 2) ( 2) .1 ( 1) ( 2) .4 (0) ( 2) (0) .49 Fp F p p F p p p Therefore, we have the CDF of Z is: 2 1 0 1 2 4 ( ) .1 .4 .49 .54 .79 1 z Fz  (c) ( 1 1) ( 2) .54 .1 .44 ( 0) 1 (0) .51 ( 2) (2) .79 ( 2 2) .54 P Z F F P Z F P Z F P Z F       2.1-7 1 1 1 0 ( ) ( ) (.9) 0.1 0.1 (.9) x x x ii X i i i F x p i 2.2-1
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2 (a) ( 2, 1) (3,2) (4,2) .05 .05 .1 ( 1 ( 2) 1 (.05 .1 .05 .05 .05) .7 P X Y p p P Y P Y     (b) 0 1 2 ( ) .3 .4 .3 y py (c) (0,0) .05 1 ( 0| 0) ( 0) .15 3 p P Y X PX (d) 0 1 2 3 4 ( ) .15 .40 .25 .12 .08 x px 0 1 2 ( ) .3 .4 .3 y (0,0) .05 (0) (0) .045 .
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55010Solution4 - STAT 550 HW #4 Solution 2.1-5 (a) p(0) F...

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