Question

# Five males with an​ X-linked genetic disorder have one child each. The random variable x is...

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
1   0.153
2   0.319
3   0.319
4   0.153
5   0.028

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

A. ​Yes, the 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.

A. muequals nothing ​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. sigmaequals nothing ​child(ren) ​(Round to one decimal place as​ needed.)

B. The table does not show a probability distribution.

 x P(x) 0 0.028 1 0.153 2 0.319 3 0.319 4 0.153 5 0.028 SUM 1

Yes, the table shows a probability distribution

why because

∑P(x) = 1 (Sum of probabilities)

and all probabilities are 0 < P(x) < 1

 x P(x) x⋅p(x) x^2⋅p(x) 0 0.028 0 0 1 0.153 0.153 0.153 2 0.319 0.638 1.276 3 0.319 0.957 2.871 4 0.153 0.612 2.448 5 0.028 0.14 0.7 SUM 1 2.5 7.448 μ=∑x⋅p(x) 2.5 σ = √(∑x^2⋅p(x)−μ^2) 1.094531863

Mean μ = 2.5

Standard deviation σ = 1.1

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