Question

I want all parts answers, and the answer needs to correct.

the answers need include from part a) to part n)

h.) E(X) (Give decimal answer to two places past the decimal.) Tries 0/5 1.) E(Y) (Give decimal answer to two places past the decimal.) ries 0/5 j.) E(XY) (Give decimal answer to two places past the decimal.) Tries 0/5 k.) Cov(X,Y) (Give decimal answer to THREE places past the decimal.) Tries 0/5 1.) Standard deviation of X (Give decimal answer to THREE places past the decimal.) Tries 0/5 m.) Standard deviation of Y (Give decimal answer to THREE places past the decimal.) Tries 0/5 n.) Corr(X,Y) (Give decimal answer to THREE places past the decimal.) Tries 0/5

Answer 1

The joint probability mass function (PMF) is given as follows: Y 1 0.13 p(x,y) 0 2 0 0.05 0.09 0.09 X 1 a 0.04 0.06 0.1 2 0.11 (a). Find a According to the properties of probability mass function, EXP(x, y) = 1 X y 0.05+0.13+0.04 +0.09+a+0.06 +0.09 +0.11 +0.1=1 a = 0.33 Hence, a=0.33

(b). From the above PMF table, P[X =1 and Y = 2]=0.06 (c) P[X s1]=P[x = 0, y = 0]+P [X = 0, y =1]+P [X = 0, Y = 2] +P[X = 1, Y =0]+P[X = 1, Y = 1] +P[ X =1, Y = 2] = 0.05+0.13+0.04 +0.09+0.33+0.06 = 0.70 Hence, the P[X Sl] is 0.70). (d).

P[X = 1 Y = 2) = P(X = 1, Y = 2) P(Y = 2) 0.06 0.06 +0.04 +0.1 =0.30 Hence, P[X = 1 Y = 2) is 0.30 (e) The E(Y| X = 0) can be calculated by using the table given as follows: 0 1 2 Y|X=0 P(Y|X=0) 0.217 0.228 0.2 E(Y| X = 0) = X(Y|X =0)P(Y|X = 0) = 0.228 +0.4 = 0.63

Hence, E (Y| X = 0) is 0.63 (f). The V (Y|X = 0) can be calculated as follows: V(Y|X = 0) = E[(Y|X = 0)*]-[E(Y|X =0)] e[(y\X =0)} ]=>(Y_X = 0).P[Y]X= 0) = 0.228 +0.8 =1.028 From part (e), E (Y| X =0)=0.63. V(Y|X =0) =1.028 – (0.63) = 0.63 Hence, V (Y|X = 0) is 0.63

The random variables are said to be independent if, P(x,y)=P(x).P(y) For example, P(X = 0, Y = 0) = 0.05 P(X = 0) =0.22 P(Y=0)=0.23 P(X = 0).P(Y = 0)=0.220.23 = 0.05 Similarly, P(X = 1, Y = 1) = 0.33 P(X = 1) = 0.48 P(Y = 1) = 0.57

P(X =1).P(Y =1) = 0.48x0.57 = 0.27 P(X = 1, Y =1)=P(X =1).P(Y=1) If, there is any one of the values of (x, y) for which P(x, y)=P(x).P(y) exist. Then two random variables are not independent. Hence, X and Y are not independent. (h). E (x)= XxP(x) x O X 1 2 p(x) 0.22 0.48 0.3 E(x)=0.48+0.6 =1.08

Hence, E(x)is 1.08 (i). E(y) = y.P(y) Y y o 1 2 ply) 0.23 0.57 0.2 E(y) = 0.57+0.4 = 0.97 Hence, E(y) is 0.97 G). E (xy)=2xy.P(xy)

=(0x0)(0.05)+(0x1)(0.13)+(0x2)(0.04)+(1x0)(0.09) +(1x1)(0.33)+(1x2)(0.06)+(1x1)(0.33)+(2x0)(0.09) +(2x1)(0.11)+(2x2)(0.1) =1.07 Hence, E (xy)is 1.07 (k). Cov(X,Y)= E(XY)- E(X) E (Y) = 1.07 – (1.08).(0.97) = 0.02 Hence, Cov(X,Y) is 0.02

(1). Standard deviation of X can be calculated as follows: VE (Y?)-[E(x)] E (x”)= Er?.P(x) The table of PMF of X is given as follows: X 0 1 2. p(x) 0.22 0.48 0.3 E (x2)=0.48+1.2 = 1.68 E(X)=1.08 1.68-41 1.68-(1.08)] = 0.72

Hence, standard deviation of X is 0.72 (m). Standard deviation of Y can be calculated as follows: 05 = VE(Y2)-[E(Y)] E(y?) = Xy?.P(y) The table of PMF of Y is given as follows: 0 1 > 2 0.2 ply) 0.23 0.57 E( y2)=0.57+0.8 =1.37

E(Y) = 0.97 0,= (1.37 (0.97)] = 0.65 Hence, standard deviation of Y is 0.65 (n). Corr(X,Y)= Cov(X,Y) Or.o 0.02 0.72 x 0.65 = 0.04 Hence, Corr (X,Y) is 0.04

Answer 2

sunce E Question & pler,y) = 1 nie. 0.05+0.09 +0:09+0.13+ a toil+0:04+0:06 +0.1 = 1 le 0.67 + a=1 le a= 0.33 Answer 6- P(x=1 and y=2) = 0.06 Auswer Y 1 sunce 2 Total =+B) O 0:05 0:13 0.04 0.22 X ㅗ 0.09 0.33 0.06 0.48 2 0109 0:11 oil 0.3 Total =ply) 10.23 0:57 0.2 ㅗ 022 x=0 Thus, 0.23 0.57 p(a) and ply)= ; 2=1 0.40 0.30 ; y=0 ; y=1 y=2 0.20 ; X = 2 ;

P( x1) = P(x=0)+ P(x-1) = 0.227 0.48 = 0.70 Auswer 0.06 = 0.30 0.20 Answer 0.05 5 = 0.22 22 @ PCX=1/4=2) = P(x = 1, Y= 2) Ply=2) bly/x-o) = ply=0/x=0) = Þ(X=0, y=0) Þ(x=0) P/s = \/x=0) = Þrx=0, Y =1) Þ(y=2/x=0) = P(x=0, y =2) p(x=0) Þ(x=0) Thus, 2 - 2 ;y=0 ply/x-o ) 13 s y = 1 0.13 13 = 0.22 22 0.04 0122 22 22 A ;y=2 Ely/x=o) – Žy. Þ/a/x=) 2016 (420/x+o) +). Þfu=1&20) +2.°/4=2/x=0) = 0+1(103)+2) = 21 0.95 22 Auswer

yo (3) EC4%8=0)= 를 y? pry/x=o) = (0)? ply=0/x=0) +(13. pſy=1/x-o)+(2). Þ[7=240 = 0+1).(.) + (4) (4) - 29 Var (Y/x=o) = E(4%x=0) - [E(1/800)) = 29- (21) = 638 – 44/ 484 197 484 0.41 Auswer g)- P(x=0, Y=0)= 0.05 P(x=0). Pſy=0) = (0.22) (0.23) = 0.0506 Suice, P(x=0, Y=o) + P(x=0). P(y=0) Thus x aud y are not widependent Correct Answer is No Auswer Answer

(h) E(X)= 3 x.p(e) = p.pro) +1.501)+2.p(2) =0)(022) + (1) (0.48) 4(2)(0-20) x=0 = 1:08 - Answer (1) - Ely) = y.Ply) = 0.680)+1.661) +2.5/2) = (0)6-23) +(1)(0.57) +(2)(0-20) y=0 = 0.97 Auswer x y ) E(XY) = Es ny.pla, y) (0) (0) (0.05)+(1) 6) (0.09)+(2) (0) (0-09)+(a)(1)(113)+C)(1)(0-33) H230)6-11 +0)(2)(0.04)+(11(2)(0.06)+(2)/2) (0:1) 1.07 = 0+0+ 0 +0 +0.33 +0.22+ 0 +0.12 +0.4 - Answer (k)- Cor(x,y) - Elxy) – E(X) E(Y) = 1.07-(1.00) (0.97) = 1.07–10476 = 0.0224 2 2 - E(X²) = {x = 0.022 = Answer 7120 prae) (0)2plo) + (0)266) +(2)?/2) - 0 (0.22) + (48)+ 4 (0:30) = 1.68

Var (x) = E(X") - [E6x))?= 1.68 -(1.03)= 1968 – 1.1664 = 0.5136 Standard deviation of x= /var (X) = 10.5/36 odaberes 2 0.7166 58915 = 0.717 - Answer y=0 (m) EC42)= {y? ply) =60]?.pro) +017.50) +(2)? Þ(2) 0(0.23) + 110.57) + 460.20) = 1.37 Var (y) – Ely?)=E(4))! 1.37—(0.97)>= 1:37 – Boe 0.9409 = 0.4291 Standard deviation of y= V Var (y) = 10.4291 = = 0.6550 57249 = 0.655 Answer (6) Cou(x,y) = (x,y) = Cor(x,y) Vad x) Vally) 0.0224 (8.7/6658915) (0-655057249) 0.0224 0.469452617 = 0.0477151457 = 0.048 Auswer