In debt? According to financial records, $24\%$ of U.S. adults have more debt on their credit cards than they have money in their savings accounts. Suppose that we take a random sample of 100 U.S. adults. Let $D=$ the number of adults in the sample with more debt than savings.

a. Explain why D can be modeled by a binomial distribution even though the sample was selected without replacement.

b. Use a binomial distribution to estimate the probability that 30 or more adults in the sample have more debt than savings.

The number of adults in the sample who have more debt than savings is denoted by the letter D.

Sample size $=100$

Percentage of adults in the United States who have more debt than savings$=24\%$

The following is the concept applied:

$10\%$condition.

$\mathrm{n}<0.10\mathrm{N}$

According to the rule, if the sample represents less than $10\%$ of the population, it is safe to assume that the trials are independent and may be modelled using the binomial distribution, regardless of the without replacement sample.

The sample size $(=100)$ is much less than $10\%$ of all people in the United States.

Furthermore, each adult's study is conducted independently. The financial situation of one adult has no bearing on the financial situation of another adult.

$\text{So,}D~Bin(100,0.24)$

Hence, A binomial distribution can be used to model D.

The number of adults in the sample who have more debt than savings is denoted by the letter D.

Sample size $=100$

Percentage of adults in the United States who have more debt than savings $=24\%$

Concept applied:

$10\%$condition

$\mathrm{n}<0.10\mathrm{N}$

Consider,

Using TI- 83 plus

a) First we have to Click on $2^{nd}$ and then click on Dist.

b) Next Go to binomcdf with $\downarrow$ key.

c) Then Hit Enter

d) Type$100,0.24,29$

e) Then Hit Enter

The probability comes to be $0.88913$.

$$\begin{gathered} P(D\ge 30)=1-0.88913 \\ P(D\ge 30)=0.11087 \end{gathered}$$