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 0 1 2 3 4 5P(x) 0.033 0.149 0.318 0.318 0.149 0.033

a) It is probability distribution

b) mean = 2.5

c) Standard Deviation = 1.11

Step-by-step explanation:

We are given the following in the question:

x          0           1           2         3          4          5

P(x)  0.033   0.149   0.318   0.318   0.149   0.033

Probability distribution:

Since, the summation of all the probabilities is 1 and each value of probability is between 0 and 1, the given is a probability distribution.

Mean of probability distribution:

Thus, the mean of distribution is 2.5

Standard deviation of probability distribution:

Thus, the standard deviation of the probability distribution is 1.11

hope this .

r3t40

207/8 is the right