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	0.035
		1	0.157
		2	0.308
		3	0.308
		4	0.157
		5	0.035

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's number values are not associated with probabilities.

D.

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

E.

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

Answer 1

Solution :

The sum of the probability is equal to 1 .

P(X) = 1

A.

​Yes, the table shows a probability distribution.

x	P(x)	x * P(x)	x2 * P(x)
0	0.035	0	0
1	0.157	0.157	0.157
2	0.308	0.616	1.232
3	0.308	0.924	2.772
4	0.157	0.628	2.512
5	0.035	0.175	0.875
Sum	1	2.5	7.548

Mean = = X * P(X) = 2.5

Standard deviation =

=X ² * P(X) - ²

=  7.548 - 2.5²

= 1.1393