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(a) Construct a binomial distribution using [tex]$n = 6$[/tex] and [tex]$p = 0.37$[/tex].

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$P(x)$[/tex] \\
\hline
0 & [tex]$\square$[/tex] \\
\hline
1 & [tex]$\square$[/tex] \\
\hline
2 & [tex]$\square$[/tex] \\
\hline
3 & [tex]$\square$[/tex] \\
\hline
4 & [tex]$\square$[/tex] \\
\hline
5 & [tex]$\square$[/tex] \\
\hline
6 & [tex]$\square$[/tex] \\
\hline
\end{tabular}

(Round to the nearest thousandth as needed.)



Answer :

GinnyAnswer
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]

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