Question

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it​ does, find the mean and standard deviation.

x	0	1	2	3	4	5
​P(x)	0.01160.0116	0.08340.0834	0.23990.2399	0.34520.3452	0.24840.2484	0.07150.0715

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

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

B.

The table is not 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

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

B.

The table is not a probability distribution.

Answer 1

The sum of all individual probabilities = 0.0116 + 0.0834 + 0.2399 + 0.3452 + 0.2484 + 0.0715 = 1

Since the sum of all probabilities is equal to 1, it is a probability distribution table.

Mean() = E(X) = 0 * 0.0116 + 1 * 0.0834 + 2 * 0.2399 + 3 * 0.3452 + 4 * 0.2484 + 5 * 0.0715 = 2.9

E(X^2) = 0^2 * 0.0116 + 1^2 * 0.0834 + 2^2 * 0.2399 + 3^2 * 0.3452 + 4^2 * 0.2484 + 5^2 * 0.0715 = 9.9117

Variance = E(X^2) - (E(X))^2

= 9.9117 - (2.9)^2

= 1.5017

Standard deviation () = = 1.2