Question

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.

Short answer

1. A binomial distribution can be used to model D.
2. More than or equal to 30 persons in the sample have more debts than savings, with a probability of $0.11087.$
   

Step-by-step solution

Part (a) Step 1: Given information

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.

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

Part (a) Step 2:  Calculation

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.

Part (b) Step 1: Given information

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

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

Part (b) Step 2:  Calculation

Consider,

$$\begin{gathered} P(D\ge 30)=1-P(D<30) \\ P(D\ge 30)=1-P(D\le 29) \end{gathered}$$

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 $0.88913$.

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