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The table on the right shows the joint probability distribution between two random variables – X and Y. (In a joint probability distribution table, numbers in the cells of the table represent the probability that particular values of X and Y occur together.
What is the mean of the sum of X and Y?
(A) 1.2
(B) 3.5
(C) 4.5
(D) 4.7
(E) None of the above.
(A) 1.2
(B) 3.5
(C) 4.5
(D) 4.7
(E) None of the above.
Solution: The correct answer is D. The solution requires three computations: (1) find the mean (expected value) of X, (2) find the mean (expected value) of Y, and (3) find the sum of the means. Those computations are shown below, beginning with the mean of X.
E(X) = Σ [ xi * P(xi) ]
E(X) = 0 * (0.1 + 0.1) + 1 * (0.2 + 0.2) + 2 * (0.2 + 0.2) = 0 + 0.4 + 0.8 = 1.2
Next, we find the mean of Y.
E(Y) = Σ [ yi * P(yi) ]
E(Y) = 3 * (0.1 + 0.2 + 0.2) + 4 * (0.1 + 0.2 + 0.2) = (3 * 0.5) + (4 * 0.5) = 1.5 + 2 = 3.5
And finally, the mean of the sum of X and Y is equal to the sum of the means. Therefore,
E(X + Y) = E(X) + E(Y) = 1.2 + 3.5 = 4.7
Note: A similar approach is used to find differences between means. The difference between X and Y is E(X – Y) = E(X) – E(Y) = 1.2 – 3.5 = -2.3; and the difference between Y and X is E(Y – X) = E(Y) – E(X) = 3.5 – 1.2 = 2.3
Problem 2
The table on the left shows the joint probability distribution between two random variables – X and Y; and the table on the right shows the joint probability distribution between two random variables – A and B.
E(X) = 0 * (0.1 + 0.1) + 1 * (0.2 + 0.2) + 2 * (0.2 + 0.2) = 0 + 0.4 + 0.8 = 1.2
Next, we find the mean of Y.
E(Y) = Σ [ yi * P(yi) ]
E(Y) = 3 * (0.1 + 0.2 + 0.2) + 4 * (0.1 + 0.2 + 0.2) = (3 * 0.5) + (4 * 0.5) = 1.5 + 2 = 3.5
And finally, the mean of the sum of X and Y is equal to the sum of the means. Therefore,
E(X + Y) = E(X) + E(Y) = 1.2 + 3.5 = 4.7
Note: A similar approach is used to find differences between means. The difference between X and Y is E(X – Y) = E(X) – E(Y) = 1.2 – 3.5 = -2.3; and the difference between Y and X is E(Y – X) = E(Y) – E(X) = 3.5 – 1.2 = 2.3
Problem 2
The table on the left shows the joint probability distribution between two random variables – X and Y; and the table on the right shows the joint probability distribution between two random variables – A and B.
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