Five males with an x-linked genetic disorder have one child each. the random variable x is the number of children among the five who inherit the x-linked genetic disorder. determine whether a probability distribution is given. if a probability distribution is given, find its mean and standard deviation. if a probability distribution is not given, identify the requirements that are not satisfied. x p(x) 0 0.032 1 0.152 2 0.316 3 0.316 4 0.152 5 0.032

Adding up all 6 probabilities gives a total of 1, which means that this is a valid distribution function. Observing that P(0) = P(5), P(1) = P(4), and P(2) = P(3), this is a symmetric distribution.

Its mean will be in the middle, which is at x = 2.5.

The standard deviation is calculated by multiplying each P(x) by the square of deviation (x - average)^2: (0.032)(0 - 2.5)^2 + (0.152)(1 - 2.5)^2 + (0.316)(2 - 2.5)^2 + (0.316)(3 - 2.5)^2 + (0.152)(4 - 2.5)^2 + (0.032)(5 - 2.5)^2 = 1.242, then taking the square root: sqrt(1.242) = 1.1145.