Question

3. Let (X, Y) be distributed according to the following joint probability mass function. p(1, 1) = 0.10, p(1, 2) = 0.12, p(1, 3) = 0.16 p(2, 1) = 0.08, p(2, 2) = 0.12, p(2, 3) = 0.10 p(3, 1) = 0.06, p(3, 2) = 0.06, p(3, 3) = 0.20 (a) [5 pts] Find E(X) and E(Y). (b) [5 pts] Find var(X) and var(Y). (c) [5 pts] Find the correlation between X and Y.

          3. Let (X, Y) be distributed according to the following joint probability mass function.

p(1, 1) = 0.10, p(1, 2) = 0.12, p(1, 3) = 0.16
p(2, 1) = 0.08, p(2, 2) = 0.12, p(2, 3) = 0.10
p(3, 1) = 0.06, p(3, 2) = 0.06, p(3, 3) = 0.20

(a) [5 pts] Find E(X) and E(Y).
(b) [5 pts] Find var(X) and var(Y).
(c) [5 pts] Find the correlation between X and Y.

3. Let (X, Y) be distributed according to the following joint probability mass function.

p(1, 1) = 0.10, p(1, 2) = 0.12, p(1, 3) = 0.16
p(2, 1) = 0.08, p(2, 2) = 0.12, p(2, 3) = 0.10
p(3, 1) = 0.06, p(3, 2) = 0.06, p(3, 3) = 0.20

(a) [5 pts] Find E(X) and E(Y).
(b) [5 pts] Find var(X) and var(Y).
(c) [5 pts] Find the correlation between X and Y.

Added by Kaitlin J.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adi S

06/16/2023

Step 1

The marginal probability mass function for X is: P_X(x) = P(x,1) + P(x,2) + P(x,3) P_X(1) = 0.10 + 0.12 + 0.16 = 0.38 P_X(2) = 0.08 + 0.12 + 0.10 = 0.30 P_X(3) = 0.06 + 0.06 + 0.20 = 0.32 The marginal probability mass function for Y is: P_Y(y) = P(1,y) + P(2,y) Show more…

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3. Let (X,Y) be distributed according to the following joint probability mass function. p(1, 1) = 0.10, p(1, 2) = 0.12, p(1, 3) = 0.16 p(2, 1) = 0.08, p(2, 2) = 0.12, p(2, 3) = 0.10 p(3, 1) = 0.06, p(3, 2) = 0.06, p(3, 3) = 0.20 (a) [5 pts] Find E(X) and E(Y). (b) [5 pts] Find var(X) and var(Y). (c) [5 pts] Find the correlation between X and Y.