two fair dice are each rolled once. Let X denote the absolute value of the difference...

two fair dice are each rolled once. Let X denote the absolute value of the difference between the two numbers that appear.

List all possible values of X

Find the probability distribution of X.

Find the probabilities P(2<X<5) and P(2£X<5).

Find the expected value mand standard deviation of X.

Solutions

Expert Solution

a) Possible values of X are: 0, 1, 2, 3, 4, 5

b) P(X = 0) = P(1,1) + P(2,2) + P(3,3) + P(4,4) + P(5,5) + P(6,6) = 6/36 = 1/6

P(X = 1) = P(1,2) + P(2,1) + P(2,3) + P(3,2) + P(3,4) + P(4,3) + P(4,5) + P(5,4) + P(5,6) + P(6,5) = 10/36 = 5/18

P(X = 2) = P(1,3) + P(3,1) + P(2,4) + P(4,2) + P(3,5) + P(5,3) + P(4,6) +P(6,4) = 8/36 = 2/9

P(X = 3) = P(1,4) + P(4,1) + P(2,5) + P(5,2) + P(3,6) + P(6,3) = 6/36 = 1/6

P(X = 4) = P(1,5) + P(5,1) + P(2,6) + P(6,2) = 4/36 = 1/9

P(X = 5) = P(1,6) + P(6,1) = 2/36 = 1/18

c) P(2 < X < 5) = P(X = 3) + P(X = 4) = 1/6 + 1/9 = 5/18

P(2 < X < 5) = P(X = 2) + P(2 < X < 5) = 2/9 + 5/18 = 1/2

d) Expected value, E(X) = 0 * 6/36 + 1 * 10/36 + 2 * 8/36 + 3 * 6/36 + 4 * 4/36 + 5 * 2/36 = 1.944

E(X²) = 0² * 6/36 + 1² * 10/36 + 2² * 8/36 + 3² * 6/36 + 4² * 4/36 + 5² * 2/36 = 5.833

Var(X) = E(X²) - (E(X))² = 5.833 - 1.944² = 2.0539

Standard deviation = sqrt(2.0539) = 1.43

milcah answered 4 months ago

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