Question

5.1.7 Question Help 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 0.028 1 0.153 0.153 2 0.319 0.319 3 0.319 0.319 4 0.153 0.153 5 0.028 0.028 Does the table show a probability​ distribution? Select all that apply. A. ​Yes, the table shows a probability distribution.

Answer 1

Solution :

The sum of the probability is equal to 1 then the table is probability distribution .

P(X) = 1

We have to check here ,

P(X) = P(X = 0) + P(X = 1)+  P(X = 2)+  P(X = 3)+  P(X = 4)+ P(X = 5)

= 0.028 + 0.153 + 0.319 + 0.319 + 0.153 + 0.028

P(X) = 1

Yes, the table shows a probability distribution.

Mean = = X * P(X)

= 0 * 0.028 + 1 * 0.153 + 2 * 0.319 + 3 * 0.319 + 4 * 0.153 + 5 * 0.028

= 2.5

= 2.5

Standard deviation =

=X ² * P(X) - ²

= 0 * 0.028 + 1² * 0.153 + 2² * 0.319 + 3² * 0.319 + 4² * 0.153 + 5² * 0.028 - 2.5²

= 1.0945

Standard deviation =