A discrete random variable X takes values in the set 1, 2, 3, 4, 5 . Its probability mass function is given by P(X=x) = cases kx & for x = 1, 2 k & for x = 3 (x-2)k^2 & for x = 4, 5 cases where k is a constant. The value of P(1 < X 4) is equal to

Step-by-step solution

The sum of the probabilities for all possible values of X must be 1 . P(1) + P(2) + P(3) + P(4) + P(5) = 1 k(1) + k(2) + k + (4-2)k^2 + (5-2)k^2 = 1 k + 2k + k + 2k^2 + 3k^2 = 1 5k^2 + 4k - 1 = 0 (5k - 1)(k + 1) = 0 k = 1 5 or k = -1 . Since probability cannot be negative, we reject k = -1 . Thus, k = 1 5 . We need to find P(1 P(X = 2) + P(X = 3) + P(X = 4) 2k + k + 2k^2 3k + 2k^2 Substituting k = 1 5 : 3 ( 1 5 ) + 2 ( 1 25 ) = 3 5 + 2 25 = 15 + 2 25 = 17 25 = 0.68 . Answer: 0.68