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   0.029
1   0.164
2   0.307
3   0.307
4   0.164
5   0.029

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

A

Yes 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.

nothing

A. ​u =. 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. O=   child(ren) ​(Round to one decimal place as​ needed.)
B. The table does not show a probability distribution.

Answer 1

Solution :

x	P(x)	x * P(x)	x2 * P(x)
0	0.029	0	0
1	0.164	0.164	0.164
2	0.307	0.614	1.228
3	0.307	0.921	2.763
4	0.164	0.656	2.624
5	0.029	0.145	0.725
Sum	1	2.5	7.504

A) Yes table shows a probability distribution.

Mean = = X * P(X) = 2.5

Standard deviation =

=X ² * P(X) - ²

=  7.504 - 2.5²

= 1.1