second part. rot and Yis given in the accompanying table of points ea rned on th
ID: 3315260 • Letter: S
Question
second part. rot and Yis given in the accompanying table of points ea rned on the second part. Suppose that the joint p earned on two parts. has glven a An instructor has given a short quiz consisting or wo parts. For a randomly selected student, let -t e number or points earned on the r st part and -t e rumber pix, y) 0 5 1015 0 0.03 0.06 0.C2 0.10 x 5 0.04 0.14 0.20 0.10 1 ID·010.150.14 D.01 pernd recarded scare x? (Enter your answer to one decimal place.) two parts, what is the ex (a) the score roorded in the grace baok is the total number at points eamed on the t (b I the maximum of the two scores is recorded, what is the expected recorded score? (Enter your answer to two decimal places.)Explanation / Answer
a) The PDF for X here is obtained as:
P(X =0) = 0.21,
P(X = 5) = 0.48,
P(X = 10) = 0.31
Therefore, E(X) = 0*0.21 + 5*0.48 + 10*0.31 = 5.5
Now for Y the distribution here is obtained as:
P(Y = 0) = 0.08,
P(Y = 5) = 0.35,
P(Y = 10) = 0.36,
P(Y = 15) = 0.21
Therefore, the expected value here is computed as: E(Y) = 0*0.08 + 5*0.35 + 10*0.36 + 15*0.21 = 8.5
ThereforeE(X + Y) = E(X) + E(Y) = 5.5 + 8.5 = 14
Therefore 14 is the expected value here.
b) Here the maximum of the X and Y are obtained for each of the combination as:
Therefore,
P(M = 0) = 0.03,
P(M = 5) = 0.24,
P(M = 10) = 0.52,
P(M = 15) = 0.21
Therefore, E(M) = 0.03*0 + 0.24*5 + 0.52*10 + 0.21*15 = 9.55
Therefore 9.55 is the expected value required here.
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