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) 05 10 15 0 0.03 0.06 0.02 0.10 x 5 0.04 0.17 0.20 0.10 10 0.01 0.15 0.11 0.01 (a) If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score E(X + Y)? (Enter your answer to one decimal place.) (b) If the maximum of the two scores is recorded, what is the expected recorded score? (Enter your answer to two decimal places.)

Explanation / Answer

a) Here for X we have the marginal distribution as:

P(X = 0) = 0.03 + 0.06 + 0.02 + 0.1 = 0.21
P(X = 5) = 0.04 + 0.17 + 0.20 + 0.1 = 0.51
P(X =10) = 0.01 + 0.15 + 0.11 + 0.01 = 0.28

Therefore, E(X) = 0.21*0 + 0.51*5 + 0.28*10 = 5.35

Now for Y, the distribution is obtained as:

P(Y = 0) = 0.03 + 0.04 + 0.01 = 0.08
P(Y = 5) = 0.06 + 0.17 + 0.15 = 0.38
P(Y = 10) = 0.02 + 0.20 + 0.11 = 0.33
P(Y = 15) = 0.10 + 0.10 + 0.01 = 0.21

Therefore, E(Y) = 0*0.08 + 5*0.38 + 10*0.33 + 15*0.21 = 8.35

Therefore we get here:

E(X + Y) = E(X) + E(Y) = 5.35 + 8.35 = 13.7

Therefore 13.7 is the expected value here.

b) The max value for each combination is recorded as:

The expected value of the maximum value is now computed by adding the products of the above maximum values with their corresponding probabilities as:

E( Max) = 0*0.03 + 5*( 0.06 + 0.04 + 0.17 ) + 10*( 0.02 + 0.2 + 0.11 + 0.15 + 0.01 ) + 15*(0.1 + 0.1 + 0.01 )

E( Max ) = 0 + 1.35 + 4.9 + 3.15 = 9.4

Therefore 9.4 is the expected value here.

0 5 10 15 0 0 5 10 15 5 5 5 10 15 10 10 10 10 15

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