a. Construct the joint distribution for X and Y by filling in the table below. R

Question

a. Construct the joint distribution for X and Y by filling in the table below. Reduce all fractions!

b. Compare the marginal distributions for X and Y.

c.Why should the results in part b have been anticipated?

d. Find the expected value and variance for X and Y.

e. Find the Covariance for X and Y.

f. Find the correlation between X and Y, and interpret it.

g. Why does the sign of the correlation make sense?

h. Find the expected value and variance for Z.

i. Returning to the original problem of drawing two cards without replacement and counting the number of diamonds, identify the name of the discrete distribution involved.

j. Turn to the section of chapter 4 that discusses the distribution identified in part i and use the formulas provided by the text to find the mean and variance for the number of diamonds in two draws without replacement from a poker deck. Compare your answers to those for part h.

Two cards are drawn without replacement from a poker deck. Let if the first card is a diamond i if the second card is a diamond and Y 0 if the first card is something else 0 if the second card is something else Z-X Y,i.e., Z equals the number of diamonds selected and let

Explanation / Answer

cards are drawn without replacement from a poker deck of 52 cards .

X=1 if first card selected is diamond

=0 otherwise

now there are 13 diamonds

so P[X=1]=13/52=1/4=0.25 so P[X=0]=1-P[X=1]=1-1/4=3/4=0.75

Y=1 if second card is diamond

=0 otherwise

cards are drawn without replacement.

so value of Y depends on value of X

so P[Y=1|X=1]=P[second card is diamond when first card is already diamond]=P[drawing a diamond from 12 diamonds from 51 total cards]=12/51

P[Y=1|X=0]=P[second card is diamond when first card is not diamond]=P[drawing a diamond from 13 diamonds from 51 total cards]=13/51

so P[Y=1]=P[Y=1,X=1]+P[Y=1,X=0]=P[Y=1|X=1]*P[X=1]+P[Y=1|X=0]*P[X=0]=(12/51)*(1/4)+(13/51)*(3/4)=0.25

so P[Y=0]=1-P[Y=1]=1-0.25=0.75

a)

so P[Y=1,X=1]=P[Y=1|X=1]*P[X=1]=12/(51*4)=0.0588

P[Y=1,X=0]=P[Y=1|X=0]*P[X=0]=(13/51)*(3/4)=0.1912

P[Y=0|X=0]=P[not drawing a diamond when at first draw no diamond were drawn]=P[drawing from 38 cards apart from diamonds from 51 cards]=38/51

P[Y=0|X=1]=P[not drawing a diamond when at first draw diamond was drawn]=P[drawing from 39 cards apart from diamonds from 51 cards]=39/51

so P[Y=0,X=0]=P[Y=0|X=0]*P[X=0]=(38/51)*(3/4)=0.5588

P[Y=0,X=1]=P[Y=0|X=1]*P[X=1]=(39/51)*(1/4)=0.1912

so the joint distribution of X and Y is

Y 0 1

0 0.5588 0.1912

X

1 0.1912 0.0588

b) so the marginal distributions of X is

P[X=1]=0.25 P[X=0]=0.75

X: 1 0

P[X=x]: 0.25 0.75

so the marginal distribution of Y is

P[Y=1]=P[Y=1,X=1]+P[Y=1,X=0]=P[Y=1|X=1]*P[X=1]+P[Y=1|X=0]*P[X=0]=(12/51)*(1/4)+(13/51)*(3/4)=0.25

P[Y=0]=1-P[Y=1]=0.75

Y: 1 0

P[Y=y]: 0.25 0.75

so the marginal distributions of X and Y are same.

c) X=1 when first card is diamond Y=1 when second card is diamond

=0 otherwise =0 otherwise

cards are drawn without replacement.

now under simple random sampling without replacement drawing a particular at any given draw is basically same.

so X and Y are similar. hence both of their distributions are same.

d) expectation of X is E[X]=1*0.25+0*0.75=0.25

expectation of Y is E[Y]=1*0.25+0*0.75=0.25

variance of X is V[X]=E[X2]-E2[X]=12*0.25+02*0.75-0.252=0.25-0.252=0.1875

variance of Y is V[Y]=E[Y2]-E2[Y]=12*0.25+02*0.75-0.252=0.25-0.252=0.1875

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