Question

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. y 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

Given chal у Piny) S 003 0:06 6.02 0.10 o'oy 0.20 0.16 01190 0.10 001 001 0:12 Mangance'l poblidity of x P(x=1) E P(X,Y) y (( x= 0) = 0.037 0.06+ 0.027 0.105 0.21 P(x =$ ) = 0 of + 0:/6 +020 +070 - 0:50 P(X=10 )= 0'01 +015+0:12+ oilo = 0 29 X 5 P(X=1) 0.21 0.29 marginal probability of tis P(n=9) E P(X=) 7=y) P(x = o) -0.03 + votto: 01- LPxz5)0:06 + 0:/6 + 0:15 - Plxalot 00270207 0.123 P(X215) 0:10 +010 to'o) 0.08 0:32 0:34 0:21 y 0 15 P(YY) : 0.08 0.37 0:34 021 Scanned with cs camscariner

Date 5.4 E(X)- EUP(X=1) ElY)= E y p (= y) = 8.4 y E(XY) E Eny P(x=4, Yey2 5x5x 0.16 +5X10x o zot 5x15x 0.10 & 16x 5x0.15 COX0X0.12 + TOX 15 X 00 42.5 ECXY) = 42.5 ECX)Ely) 5-4x8.4 2.86. 2.86 Cov CX Y = E( XY) 42.5 2 Llor(x,y) = E(X²) 41.5 พ Ex²P (X= 1) = Ex) HE(プ 2 Vorx) = 41.5- (5.4² = 12:39 EC42) Ey²P ( X = y) = 90.5 I v (4) = E(Y²) - (E(Y) ² = 90.5-18.432 19.gy - Cox (x1) Corcx ,) I var(x) barett 286 0182 J 12. 12.34 * 1994 5) V 2-013 cs Scanned with CamScanner