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)

Use the formula from part (a) to compute the variance of the recorded score

h(X, Y) [ = max(X, Y)]

if the maximum of the two scores is recorded. (Round your answer to two decimal places.)
V[max(X, Y)] =

y

p(x, y)

        0      5      10      15    x 0 0.02 0.06 0.02 0.10 5 0.04 0.15 0.20 0.10 10 0.01 0.15 0.14 0.01

Explanation / Answer

The PMF for Max(X, Y) here is obtained as:

Let K = Max(X, Y) be the random variable defined. Then the PMF for K is obtained as:

P(K = 0) = 0.02,
P(K = 5) = 0.04 + 0.06 + 0.15 = 0.25,
P(K = 10) = 0.01 + 0.15 + 0.14 + 0.20 + 0.02 = 0.52,
P(K = 15) = 0.10 + 0.10 + 0.01 = 0.21

Now the expected value of K here is computed as:

E(K) = 0*0.02 + 5*0.25 + 10*0.52 + 15*0.21 = 9.6

Now the second moment of K here is computed as:

E(K2) = 0*0.02 + 52*0.25 + 102*0.52 + 152*0.21 = 105.5

Now the variance of K here is computed as:

Var(Max(X, Y) ) = E(K2) - [ E(K) ]2 = 105.5 - 9.62 = 13.34

Therefore 13.34 is the required variance here.

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