Question

the number of points earned 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 on the second part. Suppose that the joint prut of X and Y is given in the accompanying table. * 0 5 10 0.01 0.04 0.01 0.06 0.13 0.15 0.02 0.20 0.17 (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

10 P(x,y) o o 0 0.01 0.06 x 510.04 0.13 1010.01 0.15 ply 10.06 0.34 10 0.02 0.2 0.17 0.39 15 P(x) 0.10 0.19 0.10 0.47 0.01 0.34 0.211 E(X)= Ex p () - 0+5(0.47) +10(0.34) -5.75 E() - Ey ply) - 0 + 5 (0.34) +10(0.39) +15(0.21) – 8.75 E(XY)- EExy play) - 0 + 25(0.13) +50(0.2)+75(0.10)+ 50(015)+ 100(0,17)+150 (0.01) = 46.75 E(X) - Exp(X) = 25 (0.47)+ 100(0.34) - 45.75 E(V9) - Eyºp (Y) = 25(0.34)+100(0.39)+225(0.21)= 94.75 v(x) = E(x3) - [E(X)]* = 45.75 - (5-75) - 12.6875 od 12.6875 » 9-112.6875 = 3.562 (Y)- E(Y) - [E()) = 94.75 - (8.75) - 18.1875 18.1875 → ,- 118.1875 - 4.265 a) Cov (x, y) = E(XY) - EQ)E(Y) = 46.75 - (5.75) (0.75) - - 3.5625 b) = Cov(X,Y) - 3.5625 - = - 0.23 (3.562)(4.265)