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	0.1610.161
		2	0.3100.310
		3	0.3100.310
		4	0.1610.161
		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=nothing

​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=nothing

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

B.

The table does not show a probability distribution.

Answer 1

Q1) The sum of probabilities here are computed as:
0.029 + 0.161 + 0.310 + 0.31 + 0.161 + 0.029 = 1

Therefore the sum of all probabilities here is 1.
Also each probability is between 0 and 1.

Therefore the given probabilities in the table form a probability distribution here.

Q2) The mean of the random variable here is computed as:

E(X) = 0*0.029 + 1*0.161 + 2*0.31 + 3*0.31 + 4*0.161 + 5*0.029 = 2.5

therefore 2.5 is the required mean value here.

Q3) First we compute the second moment here as:
E(X2) = 02*0.029 + 12*0.161 + 22*0.31 + 32*0.31 + 42*0.161 + 52*0.029 = 7.492

Therefore, the standard deviation here is computed as:

Therefore 1.1145 is the required standard deviation of X here.