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. p(x, y) I o 5 10 15 0 0.02 0.06 0.02 0.10 X 5 0 .04 0.16 0.20 0.10 100 .01 0.15 0.13 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.)

Answer 1

	y
x	0	5	10	15	Total
0	0.02	0.06	0.02	0.1	0.20
5	0.04	0.16	0.2	0.1	0.50
10	0.01	0.15	0.13	0.01	0.30
Total	0.07	0.37	0.35	0.21	1.00

marginal distribution of x:
x	P(x)	xP(x)	x^2P(x)
0	0.200	0.000	0.000
5	0.500	2.500	12.500
10	0.300	3.000	30.000
total	1.000	5.500	42.500
E(x)	=		5.5000
E(x^2)	=		42.5000
Var(x)=	E(x^2)-(E(x))^2=		12.2500

marginal distribution of y:
y	P(y)	yP(y)	y^2P(y)
0	0.070	0.000	0.000
5	0.370	1.850	9.250
10	0.350	3.500	35.000
15	0.210	3.150	47.250
total	1.000	8.500	91.500
E(y)	=		8.5000
E(y^2)	=		91.5000
Var(y)=σy=	E(y^2)-(E(y))^2=		19.2500

E(XY) =∑xyP(x,y)= 43.500

a)

Covar(y,x)=E(XY)-E(X)*E(Y)=

-3.25

b)

Correlation coefficient ρ=Cov(X,Y)/√(σx*σy)=

-0.21