Question

ive 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	0.1550.155
		2	0.3160.316
		3	0.3160.316
		4	0.1550.155
		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, the random variable x is categorical instead of numerical.

C.

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

D.

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

E.

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

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.

Click to select and enter your answer(s).

Answer 1

A)​Yes, the table shows a probability distribution.

x	f(x)	xP(x)	x2P(x)
0	0.0290	0.0000	0.0000
1	0.1550	0.1550	0.1550
2	0.3160	0.6320	1.2640
3	0.3160	0.9480	2.8440
4	0.1550	0.6200	2.4800
5	0.0290	0.1450	0.7250
	total	2.5000	7.4680
	E(x) =μ=	ΣxP(x) =	2.5000
	E(x2) =	Σx2P(x) =	7.4680
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.2180
	std deviation=	σ= √σ2 =	1.1036

mean of the random variable:μ =2.50

standard deviation of the random variable σ =1.1