Assume that 12 jurors are randomly selected from a population in which $85 \%$ of the people are MexicanAmericans. Refer to the probability distribution table below and find the indicated probabilities.
\begin{tabular}{|r|r|}
\hline$x$ & \multicolumn{1}{|c|}{$P(x)$} \\
\hline 0 & $0+$ \\
\hline 1 & $0+$ \\
\hline 2 & $0+$ \\
\hline 3 & $0+$ \\
\hline 4 & 0.0001 \\
\hline 5 & 0.0006 \\
\hline 6 & 0.004 \\
\hline 7 & 0.0193 \\
\hline 8 & 0.0683 \\
\hline 9 & 0.172 \\
\hline 10 & 0.2924 \\
\hline 11 & 0.3012 \\
\hline 12 & 0.1422 \\
\hline
\end{tabular}

Find the probability of exactly 5 Mexican-Americans among 12 jurors. Round your answer to four decimal places.
\[
P(x=5)=
\]

Find the probability of 5 or fewer Mexican-Americans among 12 jurors. Round your answer to four decimal places.
\[
P(x \leq 5)=
\]

Step 1 ：The question asks for two probabilities. The first is the probability of exactly 5 Mexican-Americans among 12 jurors. This can be directly read from the probability distribution table provided. The second is the probability of 5 or fewer Mexican-Americans among 12 jurors. This can be calculated by summing up the probabilities of having 0, 1, 2, 3, 4, and 5 Mexican-Americans among the jurors.

Step 2 ：The probability of exactly 5 Mexican-Americans among 12 jurors is \(0.0006\).

Step 3 ：The probability of 5 or fewer Mexican-Americans among 12 jurors is calculated by summing up the probabilities of having 0, 1, 2, 3, 4, and 5 Mexican-Americans among the jurors, which is \(0+0+0+0+0.0001+0.0006=0.0007\).

Step 4 ：Final Answer: The probability of exactly 5 Mexican-Americans among 12 jurors is \(\boxed{0.0006}\). The probability of 5 or fewer Mexican-Americans among 12 jurors is \(\boxed{0.0007}\).