To construct a binomial distribution with [tex]\( n = 6 \)[/tex] and [tex]\( p = 0.37 \)[/tex], we need to determine the probabilities for each value of [tex]\( x \)[/tex] where [tex]\( x \)[/tex] is the number of successes in [tex]\( n \)[/tex] trials. Here is the completed binomial distribution table with the calculations rounded to the nearest thousandth as needed:
[tex]\[
\begin{tabular}{|c|c|}
\hline \textbf{x} & \textbf{P(x)} \\
\hline 0 & 0.063 \\
\hline 1 & 0.220 \\
\hline 2 & 0.323 \\
\hline 3 & 0.253 \\
\hline 4 & 0.112 \\
\hline 5 & 0.026 \\
\hline 6 & 0.003 \\
\hline
\end{tabular}
\][/tex]
To summarize, the probabilities are as follows:
- [tex]\( P(x = 0) = 0.063 \)[/tex]
- [tex]\( P(x = 1) = 0.220 \)[/tex]
- [tex]\( P(x = 2) = 0.323 \)[/tex]
- [tex]\( P(x = 3) = 0.253 \)[/tex]
- [tex]\( P(x = 4) = 0.112 \)[/tex]
- [tex]\( P(x = 5) = 0.026 \)[/tex]
- [tex]\( P(x = 6) = 0.003 \)[/tex]
