Question

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	0.0290.029
		1	0.1490.149
		2	0.3220.322
		3	0.3220.322
		4	0.1490.149
		5	0.0290.029

Does the table show a probability​ distribution? Select all that apply.

A.

​Yes, the table shows a probability distribution.

B.

​No, not every probability is between 0 and 1 inclusive.

C.

​No, the random variable x is categorical instead of numerical.

D.

​No, the random variable​ x's number values are not associated with probabilities.

E.

​No, the sum of all the probabilities is not equal to 1.

Find the mean of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

muμequals=

​child(ren) ​(Round to one decimal place as​ needed.)

B.

The table does not show a probability distribution.

Find the standard deviation of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

sigmaσequals=

​child(ren) ​(Round to one decimal place as​ needed.)

B.

The table does not show a probability distribution.

Click to select and enter your answer(s).

Answer 1

a) The sum of all individual probabilities = 0.029 + 0.149 + 0.322 + 0.322 + 0.149 + 0.029 = 1

Option - A) Yes, the table shows a probability distribution.

b) = E(X) = 0 * 0.029 + 1 * 0.149 + 2 * 0.322 + 3 * 0.322 + 4 * 0.149 + 5 * 0.029 = 2.5

c) E(X^2) =  0^2 * 0.029 + 1^2 * 0.149 + 2^2 * 0.322 + 3^2 * 0.322 + 4^2 * 0.149 + 5^2 * 0.029 = 7.444

= E(X^2) - (E(X))^2

= 7.444 - (2.5)^2 = 1.194

= 1.1