In Exercises 1–5, assume that 74% of randomly selected adults have a credit card (based on results from an AARP Bulletin survey). Assume that a group of five adults is randomly selected.

If all five of the adults have credit cards, is five significantly high? Why or

why not?

The number of randomly selected adults are $n=5$.

The probability of randomly selected adults who have a credit card is $p=0.74$.

Referring to exercise 3 of Section Review Exercise.

The mean number of adults in groups of five who have credit cards is $\mu =$3.7 adults.

The standard deviation for the number of adults in groups of five who have credit cards is$\sigma =$1 adult.

Significantly high values are the values which are greater than or equal to$\mu +2\sigma$ .

Using the rule of thumb, the calculations are computed as,

$$\begin{gathered} \mu +2\sigma =3.7+\left(2\times 1\right) \\ =5.7 \end{gathered}$$

Since 5 is less than 5.7, it implies that it is not a significantly high number of adults with credit cards.

Therefore, by using the range rule of thumb, 5 is not a significantly high number of adults with credit cards.

Let x represents the number of adults that have credit card.

In the given scenario, the variable x will follow the binomial distribution.

The probability mass function of the binomial distribution is given as,

$P\left(x\right)={}_n{C}_x{p}^x{q}^{n-x}$

The probability that all five adults have credit cards is computed as,

$$\begin{gathered} P\left(5\right)={}^5{C}_5{\left(0.74\right)}^5{\left(1-0.74\right)}^{5-5} \\ =\frac{5!}{3!\left(5-5\right)!}{\left(0.74\right)}^5{\left(0.26\right)}^0 \\ =0.2219 \\ \approx 0.222 \end{gathered}$$

Thus, the probability that all five adults have credit cards is 0.274.

Also, from the obtained probability it is clear that 0.222 is not less than or equal to 0.05.

Thus, the number of five adults are not said to be significantly high.