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   0.028
1   0.153
2   0.319
3   0.319
4   0.153
5   0.028

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's number values are not associated with probabilities.

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

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

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. muequals 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. sigmaequals nothing ​child(ren) ​(Round to one decimal place as​ needed.)

B. The table does not show a probability distribution.

Answer 1

	x	P(x)
	0	0.028
	1	0.153
	2	0.319
	3	0.319
	4	0.153
	5	0.028
SUM		1

Yes, the table shows a probability distribution

why because

∑P(x) = 1 (Sum of probabilities)

and all probabilities are 0 < P(x) < 1

	x	P(x)	x⋅p(x)	x^2⋅p(x)
	0	0.028	0	0
	1	0.153	0.153	0.153
	2	0.319	0.638	1.276
	3	0.319	0.957	2.871
	4	0.153	0.612	2.448
	5	0.028	0.14	0.7
SUM		1	2.5	7.448
		μ=∑x⋅p(x)	2.5	σ = √(∑x^2⋅p(x)−μ^2)	1.094531863

Mean μ = 2.5

Standard deviation σ = 1.1