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Seventy-two percent of adults want to live to age 100. You randomly select five adults and ask them whether they want to live to age 100. The random variable represents the number of adults who want to live to age 100. Complete parts (a) through (c) below.

(a) Construct a binomial distribution.

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

(Round to five decimal places as needed.)

Sagot :

GinnyAnswer
To construct a binomial distribution for the given problem, we need to determine the probability of each possible outcome for [tex]\( x \)[/tex], where [tex]\( x \)[/tex] is the number of adults out of five who want to live to age 100.

Given data:
- [tex]\( n = 5 \)[/tex] (number of trials, i.e., number of adults selected)
- [tex]\( p = 0.72 \)[/tex] (probability of success, i.e., the probability that an adult wants to live to age 100)

The binomial probability mass function (pmf) is given by:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \][/tex]

Here, we are given the probabilities for each possible value of [tex]\( x \)[/tex], where [tex]\( x \)[/tex] ranges from 0 to 5. Let's fill in the table with these values.

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $P(x)$ \\ \hline 0 & 0.00172 \\ \hline 1 & 0.02213 \\ \hline 2 & 0.11380 \\ \hline 3 & 0.29263 \\ \hline 4 & 0.37623 \\ \hline 5 & 0.19349 \\ \hline \end{tabular} \][/tex]

Each value of [tex]\( P(x) \)[/tex] needs to be rounded to five decimal places, and the values given already satisfy this requirement.

Hence, the completed binomial distribution table is:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $P(x)$ \\ \hline 0 & 0.00172 \\ \hline 1 & 0.02213 \\ \hline 2 & 0.11380 \\ \hline 3 & 0.29263 \\ \hline 4 & 0.37623 \\ \hline 5 & 0.19349 \\ \hline \end{tabular} \][/tex]
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