For the proprablility distribution table below:
\begin{tabular}{|c|c|c|}
\hline$x$ & $P(x)$ & \\
\hline 0 & 0.06 & \\
\hline 1 & 0.06 & \\
\hline 2 & 0.14 & \\
\hline 3 & 0.11 & \\
\hline 4 & 0.63 & \\
\hline
\end{tabular}
Find the mean. Round to two decimal places.
\[
\mu=
\]

Step 1 ：The probability distribution table is given as follows: \begin{tabular}{|c|c|c|} \hline$x$ & $P(x)$ & \\ \hline 0 & 0.06 & \\ \hline 1 & 0.06 & \\ \hline 2 & 0.14 & \\ \hline 3 & 0.11 & \\ \hline 4 & 0.63 & \\ \hline \end{tabular}

Step 2 ：The mean of a probability distribution is calculated by multiplying each outcome by its probability and then summing these products. In mathematical terms, if x is the outcome and P(x) is the probability of that outcome, the mean (μ) is calculated as: \[\mu = \sum_{i} x_i P(x_i)\]

Step 3 ：So, in this case, we need to multiply each x value by its corresponding P(x) value and then sum these products to find the mean.

Step 4 ：Let's calculate: \[\mu = (0 \times 0.06) + (1 \times 0.06) + (2 \times 0.14) + (3 \times 0.11) + (4 \times 0.63)\]

Step 5 ：After calculating, we find that the mean is approximately 3.19.

Step 6 ：Final Answer: The mean of the probability distribution is \(\boxed{3.19}\).