Answer to Question #280308 in Statistics and Probability for Ravindran

Question #280308

Two ideal dice are thrown . Let X1 be the score on the first die and X2 be the score on the second die. Let Y = max(X1, X2) then

(i) Evaluate Corr(Y, X1).

(ii) Evaluate Corr(Y, X2).



1
Expert's answer
2021-12-20T04:01:39-0500

Solution:

In throwing two dice,

"S=\\{(1,1),(1,2)...(1,6)\n\\\\(2,1),...,(2,6)\n\\\\.\n\\\\.\n\\\\ (6,1),...(6,6)\n\n\\}"

X1=score on 1st die.

X2=score on 2nd die.

Y=max(X1,X2)

Thus, we have:

X1={1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6}

X2={1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6}

Y={1,2,3,4,5,6,2,2,3,4,5,6,3,3,3,4,5,6,4,4,4,4,5,6,5,5,5,5,5,6,6,6,6,6,6,6}

(i):

Corr(Y, X1)

X1 Values

∑ = 126

Mean = 3.5

∑(X1 - Mx)2 = SSx = 105


Y Values

∑ = 161

Mean = 4.472

∑(Y - My)2 = SSy = 70.972


X and Y Combined

N = 36

∑(X1 - Mx)(Y - My) = 52.5


R Calculation

"r = \u2211((X1 - My)(Y - Mx)) \/ \\sqrt{(SSx)(SSy)}\n\n\\\\\nr = 52.5 \/ \\sqrt{(105)(70.972)} = 0.6082"

(ii):

Corr(Y, X2) = 0.6082

It is same as Corr(Y, X1) because X1 and X2 contains same and equal number of observations.


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