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) 0 5 10 15 0 0.01 0.06 0.02 0.10 X 5 0.04 0.17 0.20 0.10 10 0.01 0.15 0.13 0.01 (a) Compute the covariance for X and Y. (Round your answer to two decimal places.) Cov(X, Y) = -2.80 x (b) Compute p for X and Y. (Round your answer to two decimal places.) p= 0x

Answer 1

	y
x	0	5	10	15	Total
0	0.01	0.06	0.02	0.1	0.19
5	0.04	0.17	0.2	0.1	0.51
10	0.01	0.15	0.13	0.01	0.30
Total	0.06	0.38	0.35	0.21	1.00

marginal distribution of x:
x	P(x)	xP(x)	x^2P(x)
0	0.190	0.000	0.000
5	0.510	2.550	12.750
10	0.300	3.000	30.000
total	1.000	5.550	42.750
E(x)	=		5.5500
E(x^2)	=		42.7500
Var(x)=	E(x^2)-(E(x))^2=		11.9475

marginal distribution of y:
y	P(y)	yP(y)	y^2P(y)
0	0.060	0.000	0.000
5	0.380	1.900	9.500
10	0.350	3.500	35.000
15	0.210	3.150	47.250
total	1.000	8.550	91.750
E(y)	=		8.5500
E(y^2)	=		91.7500
Var(y)=σy=	E(y^2)-(E(y))^2=		18.6475

E(XY) =∑xyP(x,y)= 43.7500

a)

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

-3.70

b)

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

-0.25