Question

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).

Answer 1

Solution:

In throwing two dice,

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

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 - M_x)² = SS_x = 105

Y Values

∑ = 161

Mean = 4.472

∑(Y - M_y)² = SS_y = 70.972

X and Y Combined

N = 36

∑(X₁ - M_x)(Y - M_y) = 52.5

R Calculation

$r = ∑((X1 - My)(Y - Mx)) / \sqrt{(SSx)(SSy)} \\ r = 52.5 / \sqrt{(105)(70.972)} = 0.6082$

(ii):

Corr(Y, X₂) = 0.6082

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