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)	0.00020.0002	0.00530.0053	0.04500.0450	0.19190.1919	0.40890.4089	0.34870.3487

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.

Click to select and enter your answer(s).

Answer 1

To be a valid probability distribution, the sum of all probability must be equal to one.

Sum = 0.0002 + 0.0053 + 0.045 + 0.1919 + 0.4089 + 0.3487 = 1

µ = E(X) = 0 * 0.0002 + 1 * 0.0053 + 2 * 0.045 + 3 * 0.1919 + 4 * 0.4089 + 5 * 0.3487 = 4.0501 = 4.1 (ans)

Option-A) µ = 4.1

E(X²) = 0² * 0.0002 + 1² * 0.0053 + 2² * 0.045 + 3² * 0.1919 + 4² * 0.4089 + 5² * 0.3487 = 17.1723

Variance = E(X²) - (E(X))² = 17.1723 - 4.1² = 0.3623

Standard deviation = sqrt(0.3623) = 0.6

Option-A) σ = 0.6