Question

please explain as much as possible, thanks

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let x = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table. p(x,y) 0 5 10 15 0 0.03 0.06 0.02 0.10 5 0.04 0.16 0.20 0.10 10 0.01 0.15 0.12 0.01 (a) Compute the covariance for X and Y. (Round your answer to two decimal places.) Cov(x, y) = (b) Compute p for X and Y. (Round your answer to two decimal places.) p =

Answer 1

P(x,y) o у 5 1o 15/Px(2) 0,03 0.06 0.02 0.10 0.21 0,20 0,16 X 5 0,10 0,5 10.04 10 0.01 0.15 0,12 1 0.010,29 Py@g) 10.08 0.37 0.34 0.21 Marginal distribution of a Х 5 х 10 0.5 (3) O. 21 0.29 Marginal distribution of y 15 5 to Y Pyly) 0.08 0.37 0.34 0.21 E[x]= 2 XPx600) = 0(0,20+ 50.5)+ 10(0.29) = 5.4 E[4] = By Py Cy) =0(0.08)+50.37)+10(0.34)+15(@24) = 8.4 E[x4] = say Pxyla, )

(0.03)+000.06)+0(0.02)+ 00.10)+0 (0.04)+25(0.16)+ 500. 20)+75(0.10)+06.01)+ 50 0.15)+100(0.12) +150 (0.01) = 42.5 a) Cov(x,y)= E(XY)-E[x] E[Y] - = 42.5– (5.4)(8,4) -2.86 Corcx,y) b) p. Corx,y) XTY E[x2]=SxPx (2) 00.21)+5.0.5)+10%90.29) 41.5 EfY= 2y py =0(0,08)+5%0.34) +10%0.34) +15%0.21) 90.5

V(x) = TX E[X3] –(E(X))? = 41.5-(5.4) ot 12.34 +0x = 3,5128 VCY)=07 = E[42]-CECY)) = 90.5 – (8.432 TL = 19.94 * ry= 4.4654 P= Cor(x,y) XY (3.5128) (4.4654) -2.86 P -0.1823 P= -0.18