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01.33 points Perous Answers Calculate the expected value E(X) of the given rando

ID: 2909656 • Letter: 0

Question

01.33 points Perous Answers Calculate the expected value E(X) of the given random varlable X. X is the higher number when two dice are rolled, or the common number if doubles are rolled. (So, a roll of 4-3 would be given a value of 4 while a roll of 5-5 would be given a value of S). 4.4722 E(X) Calculate the expected value E(Xx) of the glven random variable X X is the lower number when two dice are rolled, or the common number if doubles are rolled. (So, a roll of 4-3 would be given a value of 3 while a roll of 5-5 would be given a value of S) E(X) 7036

Explanation / Answer

(a)

X can take values: 1,2,3,4,5,6

For X = 1:

Events favourable: (1,1): 1 No.

P(X=1)=1/6 X 1/6 = 1/36 = 0.0278

For X = 2:

Events favouable: (1,2), (2,1),(2,2): 3 Nos.

P(X=2) = 3/36 = 0.0833

For X = 3:
Events favourble: (1,3), (2,3), (3,2), (3,1),(3,3) 5 Nos.

P(X=3) = 5/36 = 0.1389

For X = 4:

Events favourable: (1,4), (2,4), (3,4), (4,1), (4,2), (4,3),(4,4): 7 Nos.

P(X=4) = 7/36 = 0.1944

For X = 5:

Events favourable: (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5): 9 Nos.

P(X=5) = 9/36 = 0.25

For X = 6:

Events favourable: (1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) 11 Nos.

P(X = 6) = 11/36 = 0.3056

x        p                  xp

1         0.0278           0.0278

2        0.0833            0.1667

3        0.1389           0.4167

4        0.1944          0.7778

5       0.25              1.25

6     0.3056            1.8333

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E(X) =                  4.4723

(b)

X can take values: 1,2,3,4,5,6

For X = 1:

Events favourable: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1): 11 Nos.

P(X=1)= 11/36 = 0.3056

For X = 2:

Events favouable: (2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2): 9 Nos.

P(X=2) = 9/36 = 0.25

For X = 3:
Events favourble: (3,3),(3,4),(3,5),(3,6),(4,3),(5,3),(6,3): 7 Nos.

P(X=3) = 7/36 = 0.1944

For X = 4:

Events favourable: (4,4),(4,5),(4,6),(5,4),(6,4): 5 Nos.

P(X=4) = 5/36 = 0.1389

For X = 5:

Events favourable: (5,5),(5,6),(6,5): 3 Nos.

P(X=5) = 3/36 = 0.0833

For X = 6:

Events favourable: (6,6) 1 Nos.

P(X = 6) = 1/36 = 0.0278

x        p                  xp

1         0.3056        0.3056

2        0.25          0.50

3        0.1944      0.5832

4        0.1389        0.5556

5       0.0833          0.4165

6     0.0278        0.1668

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E(X) =                  2.5277