Question

Use the following information to answer questions 9 – 12: The following joint probability distribution describes the number of bridesmaids and groomsmen at weddings. To avoid lengthy arithmetic, it assumes there are always 1, 2, or 3 bridesmaids and either 1, 2, or 3 groomsmen.

		# of Groomsmen (Y)
		1	2	3
# of Bridesmaids (X)	1	0.15	0.02	0.02	0.19
	2	0.08	0.25	0.05	0.38
	3	0.08	0.05	0.3	0.43
		0.31	0.32	0.37	1

9. What is the marginal probability that a wedding will have 2 bridesmaids?

10. Suppose you know that a wedding is going to have 2 groomsmen; what is the expected number of bridesmaids that will be at that wedding?

11. Is the number of bridesmaids independent of the number of groomsmen? (So, are these two random variables independent?) (Yes or No)

12. What is the correlation between the number of bridesmaids and the number of groomsmen? (Hint: this will take a while to do. You need to first calculate the expected value of each random variable, then the covariance between the variables, then each variables standard deviation, and then you can find the correlation) (Your answer will be between -1 and 1)

Answer 1

9)\

marginal probability that a wedding will have 2 bridesmaids =0.38

10)

E(X|Y=2) =ΣxP(x|y=2) =1*(0.02/0.32)+2*(0.25/0.32)+3*(0.05/0.32)=2.09375

11)

since P(X=1)=0.19 ; P(Y=1) =0.31 and P(x=1)*P(Y=1) =0.0589 is not equal to P(X=1,Y=1) =0.15

therefore X and Y are not independent

No

12)

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)
1	1	0.1500	0.1500	0.1500	0.15	0.15	0.15
1	2	0.0800	0.0800	0.1600	0.08	0.32	0.16
1	3	0.0800	0.0800	0.2400	0.08	0.72	0.24
2	1	0.0200	0.0400	0.0200	0.08	0.02	0.04
2	2	0.2500	0.5000	0.5000	1.00	1.00	1.00
2	3	0.0500	0.1000	0.1500	0.20	0.45	0.30
3	1	0.0200	0.0600	0.0200	0.18	0.02	0.06
3	2	0.0500	0.1500	0.1000	0.45	0.20	0.30
3	3	0.3000	0.9000	0.9000	2.70	2.70	2.70
	Total	1	2.0600	2.2400	4.92	5.58	4.95
	E(X)=ΣxP(x,y)=	2.060
	E(X2)=Σx2P(x,y)=	4.920
	E(Y)=ΣyP(x,y)=	2.240
	E(Y2)=Σy2P(x,y)=	5.580
	Var(X)=E(X2)-(E(X))2=	0.676
	Var(Y)=E(Y2)-(E(Y))2=	0.562
SD(X)=√Var(x)=		0.822
SD(X)=√Var(x)=		0.750
	E(XY)=ΣxyP(x,y)=	4.950
Cov(X,Y)=E(XY)-E(X)*E(Y)=			0.3356
Correlation ρxy=Cov(X,Y)/(SD(X)*SD(Y))=				0.5441