91Ó°ÊÓ

The given statement is:

In the last ten years, 35 percent of American adults have had at least one breakup.

Therefore, $p=0.35$.

Nine American adults were chosen at random.

$n=9$

The probability that the chosen number is exactly five is:

$$\begin{gathered} P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ P\left(X=5\right)=\left(\begin{array}{c}9\\ 5\end{array}\right){\left(0.35\right)}^5{\left(1-0.35\right)}^{9-5} \\ =0.1181 \end{gathered}$$

The probability that the chosen number is at most five is:

$$\begin{gathered} P\left(X≤5\right)=P\left(X=0\right)+P\left(X=1\right)+........+P\left(X=5\right) \\ =\left(\begin{array}{c}9\\ 0\end{array}\right){\left(0.35\right)}^0{\left(1-0.35\right)}^{9-0}+\left(\begin{array}{c}9\\ 1\end{array}\right){\left(0.35\right)}^1{\left(1-0.35\right)}^{9-1}+......+\left(\begin{array}{c}9\\ 5\end{array}\right){\left(0.35\right)}^5{\left(1-0.35\right)}^{9-5} \\ =0.0207+0.1004+0.2162+0.2716+0.2194+0.1181 \\ =0.9464 \end{gathered}$$

The probability that the chosen number is at least five is:

$$\begin{gathered} P\left(Xâ‰\yen 5\right)=1-P\left(X<5\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\right] \\ =1-\left[0.0207+0.1004+0.2162+0.2716+0.2194\right] \\ =1-0.8283 \\ =0.1717 \end{gathered}$$

The probability that the chosen number is at least one is:

$$\begin{gathered} P\left(X≤1\right)=P\left(X=0\right)+P\left(X=1\right) \\ =\left(\begin{array}{c}9\\ 0\end{array}\right){\left(0.35\right)}^0{\left(1-0.35\right)}^{9-0}+\left(\begin{array}{c}9\\ 1\end{array}\right){\left(0.35\right)}^1{\left(1-0.35\right)}^{9-1} \\ =0.0207+0.1004 \\ =0.1211 \end{gathered}$$

The probability that the chosen number is at most one is:

$$\begin{gathered} P\left(Xâ‰\yen 1\right)=1-P\left(X=0\right) \\ =1-0.0207 \\ =0.9793 \end{gathered}$$

$$\begin{gathered} P\left(6≤Xâ‰\yen 8\right)=P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right) \\ =0.0424+0.0098+0.0013 \\ =0.0535 \end{gathered}$$

The random variable X has the following probability distribution:

Because the sampling from a finite population is done without replacement and the number of successes follows a hypergeometric distribution.

The hypergeometric distribution is similar to the binomial distribution since the sample is less than 5% of the population.