Five males with an X-linked genetic disorder have one child each. The random variable $\mathrm{x}$ 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.
\begin{tabular}{c|c}
\hline $\mathbf{x}$ & $\mathbf{P}(\mathbf{x})$ \\
\hline 0 & 0.031 \\
\hline 1 & 0.157 \\
\hline 2 & 0.312 \\
\hline 3 & 0.312 \\
\hline 4 & 0.157 \\
\hline 5 & 0.031 \\
\hline
\end{tabular}
Find the mean of the random variable $\mathrm{x}$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $\mu=\square$ child(ren) (Round to one decimal place as needed.)
B. The table does not show a probability distribution.
Answer
Final Answer: The mean of the random variable \(x\) is \(\boxed{2.5}\) children.
Step 1 ：Check if the given table represents a probability distribution. For a table to represent a probability distribution, the sum of all probabilities must be equal to 1.
Step 2 ：Calculate the sum of the probabilities: \(0.031 + 0.157 + 0.312 + 0.312 + 0.157 + 0.031 = 1\). So, the table does represent a probability distribution.
Step 3 ：Calculate the mean of the distribution. The mean of a probability distribution is calculated by multiplying each outcome by its probability and summing these products.
Step 4 ：Calculate the mean: \((0*0.031) + (1*0.157) + (2*0.312) + (3*0.312) + (4*0.157) + (5*0.031) = 2.5\).
Step 5 ：Final Answer: The mean of the random variable \(x\) is \(\boxed{2.5}\) children.
