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Sagot :
To construct a binomial distribution for the situation where [tex]\( n = 6 \)[/tex] and [tex]\( p = 0.34 \)[/tex], we need to determine the probability of each possible outcome (the number of working mothers, out of six, who do not have enough money to cover their health insurance deductibles).
The binomial distribution will give us the probability for each possible value of [tex]\( x \)[/tex], where [tex]\( x \)[/tex] is the number of mothers out of six who do not have enough money.
Here is the completed table with the given probabilities rounded to the nearest thousandth (three decimal places):
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$P(x)$[/tex] \\
\hline
0 & 0.083 \\
\hline
1 & 0.255 \\
\hline
2 & 0.329 \\
\hline
3 & 0.226 \\
\hline
4 & 0.087 \\
\hline
5 & 0.018 \\
\hline
6 & 0.002 \\
\hline
\end{tabular}
Each entry in the table represents the probability of having [tex]\( x \)[/tex] working mothers out of six who do not have enough money to cover their health insurance deductibles. For example, the probability that none of the six mothers ( [tex]\( x = 0 \)[/tex] ) do not have enough money is 0.083, while the probability that exactly three out of the six mothers ( [tex]\( x = 3 \)[/tex] ) do not have enough money is 0.226.
The binomial distribution will give us the probability for each possible value of [tex]\( x \)[/tex], where [tex]\( x \)[/tex] is the number of mothers out of six who do not have enough money.
Here is the completed table with the given probabilities rounded to the nearest thousandth (three decimal places):
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$P(x)$[/tex] \\
\hline
0 & 0.083 \\
\hline
1 & 0.255 \\
\hline
2 & 0.329 \\
\hline
3 & 0.226 \\
\hline
4 & 0.087 \\
\hline
5 & 0.018 \\
\hline
6 & 0.002 \\
\hline
\end{tabular}
Each entry in the table represents the probability of having [tex]\( x \)[/tex] working mothers out of six who do not have enough money to cover their health insurance deductibles. For example, the probability that none of the six mothers ( [tex]\( x = 0 \)[/tex] ) do not have enough money is 0.083, while the probability that exactly three out of the six mothers ( [tex]\( x = 3 \)[/tex] ) do not have enough money is 0.226.
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