An instructor has given a short quiz consisting of two parts. For a randomly sel

Question

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table.

p(x, y)

(a) Compute the covariance for X and Y. (Round your answer to two decimal places.)
Cov(X, Y) =  

(b) Compute for X and Y. (Round your answer to two decimal places.)
=

y

p(x, y)

   0 5 10 15 x 0 0.01 0.06 0.02 0.10 5 0.04 0.15 0.20 0.10 10 0.01 0.15 0.15 0.01

Explanation / Answer

The question is asking for the expected value of E (X+Y), lets first define the expected value of a jointly distributed random variable

E [h (x, y)] = å å h(x, y) § p(x, y) X, Y discrete x y «« h(x, y) § f(x, y) dx dy X , Y continuous

For this problem the h (X, Y) = X + Y by substituting this into the above definition for X and Y being discrete variables and getting corresponding values from Table1, E(X + Y) = å å (x + y) § p(x, y) = (0 + 0)(.02) + (0 + 5)(0.06) x y + (0 + 10)(0.02) + (0 + 15)(0.10) + (5 + 0)(0.04) + (5 + 5)(0.15) + (5 + 10)(0.20) + (5 +15)(0.10) + (10 + 0)(0.01) + (10 + 5)(0.15) + (10 + 10)(0.14) + (10 + 15)(.01) = 14.10 points ALSO: E(X + Y) = E(X) + E(Y) = å å (x) § p(x, y) + å å (y) § p(x, y) = 14.10 points

For part b the question is asking for the expected value of the maximum of the two scores. Let Z = max (x,y) shown in Table2a by using the corresponding values from Table2b. The max value for (0,0) is 0, the max value for (0,5) is 5, the max value for (5,5) is 5, and the max value of (5,0) is 5 and so on. Since the distribution is independent we can add up the probabilities for each max of h(x,y) = z.

T(0)(0.02) + (5)(0.25) + (10)(0..52) + (15)(.21) = 9.60 dollars

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