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 Yis given in the accompanying table. Х p(x,y) 0 5 10 15 0 0.03 0.06 0.02 0.10 5 0.04 0.13 0.20 0.10 10 0.01 0.15 0.15 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.) 6:59 PM

Answer 1

a) For each combination, we obtain the values of X + Y here. the expected value of X + Y is computed here as:

$E(X + Y) = \sum (x + y)P(X = x, Y = y)$

The computations here are made as:

p(x,y)	0	5	10	15
0	0.03	0.06	0.02	0.1
5	0.04	0.13	0.2	0.1
10	0.01	0.15	0.15	0.01
x + y	0	5	10	15
0	0	5	10	15
5	5	10	15	20
10	10	15	20	25
E(X + Y)	0	5	10	15
0	0	0.3	0.2	1.5
5	0.2	1.3	3	2
10	0.1	2.25	3	0.25

We add the values in the last table here to get the expected value as:

$E(X + Y) = \sum (x + y)P(X = x, Y = y) = 14.1$

therefore 14.1 is the required expected value here.

b) Similar to above, the expected value of the max of X and Y here is computed as:

$E(Max(X, Y)) = \sum Max(X, Y)P(X = x, Y = y)$

These computations here are made as:

Max(X, Y)	0	5	10	15
0	0	5	10	15
5	5	5	10	15
10	10	10	10	15
E(max(X, Y))	0	5	10	15
0	0	0.3	0.2	1.5
5	0.2	0.65	2	1.5
10	0.1	1.5	1.5	0.15

Adding the values in the last table, we get here:

$E(Max(X, Y)) = \sum Max(X, Y)P(X = x, Y = y) = 9.6$

Therefore 9.60 is the required expected value here.