Question

The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table:

y
x	0	1	2	3	4
0	0.03	0.01	0.02	0.01	0.01
1	0.09	0.05	0.08	0.03	0.01
2	0.08	0.05	0.07	0.04	0.02
3	0.06	0.04	0.08	0.04	0.02
4	0.04	0.03	0.04	0.03	0.02

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Find σ_{X + Y}.

σ_{X + Y} =

Answer 1

from given data:

x	y	f(x,y)	x*f(x,y)	y*f(x,y)	x^2f(x,y)	y^2f(x,y)	xy*f(x,y)
0	0	0.03	0	0	0	0	0
1	0	0.09	0.09	0	0.09	0	0
2	0	0.08	0.16	0	0.32	0	0
3	0	0.06	0.18	0	0.54	0	0
4	0	0.04	0.16	0	0.64	0	0
0	1	0.01	0	0.01	0	0.01	0
1	1	0.05	0.05	0.05	0.05	0.05	0.05
2	1	0.05	0.1	0.05	0.2	0.05	0.1
3	1	0.04	0.12	0.04	0.36	0.04	0.12
4	1	0.03	0.12	0.03	0.48	0.03	0.12
0	2	0.02	0	0.04	0	0.08	0
1	2	0.08	0.08	0.16	0.08	0.32	0.16
2	2	0.07	0.14	0.14	0.28	0.28	0.28
3	2	0.08	0.24	0.16	0.72	0.32	0.48
4	2	0.04	0.16	0.08	0.64	0.16	0.32
0	3	0.01	0	0.03	0	0.09	0
1	3	0.03	0.03	0.09	0.03	0.27	0.09
2	3	0.04	0.08	0.12	0.16	0.36	0.24
3	3	0.04	0.12	0.12	0.36	0.36	0.36
4	3	0.03	0.12	0.09	0.48	0.27	0.36
0	4	0.01	0	0.04	0	0.16	0
1	4	0.01	0.01	0.04	0.01	0.16	0.04
2	4	0.02	0.04	0.08	0.08	0.32	0.16
3	4	0.02	0.06	0.08	0.18	0.32	0.24
4	4	0.02	0.08	0.08	0.32	0.32	0.32
	Total	1	2.14	1.53	6.02	3.97	3.44
	E(X)=ΣxP(x,y)=					2.14
	E(X2)=Σx2P(x,y)=					6.02
	E(Y)=ΣyP(x,y)=					1.53
	E(Y2)=Σy2P(x,y)=					3.97
	Var(X)=E(X2)-(E(X))2=					1.44
	Var(Y)=E(Y2)-(E(Y))2=					1.63
	E(XY)=ΣxyP(x,y)=					3.44
	Cov(X,Y)=E(XY)-E(X)*E(Y)=					0.1658

σ_{X + Y} = sqrt(Var(x)+Var(Y)+2*Cov(x,y)) =sqrt(1.44+1.63+2*0.1658)

=1.8442