Question

according to a study conducted in one city, 35% of adults in the city have credit card debts of more than $2000. a simple random sample of n = 300 adults is obtained from the city. describe the sampling distribution of p, the sample proportion of adults who have credit card debts of more than $2000. round the standard deviation of the sampling distribution to four decimal places

o a. binomial, μp = 105, σp = 8.20
o b. exactly normal, μp = 0.35, σp = 0.0280
o c. approximately normal, μp = 0.35, σp = 0.0280
o d. approximately normal, μp = 0.35, σp = 0.0008

Explanation:

Step1: Check the conditions for normal approximation

For the sampling distribution of \(\hat{p}\), we check \(np = 300\times0.35=105\) and \(n(1 - p)=300\times(1 - 0.35)=300\times0.65 = 195\). Since \(np\geq10\) and \(n(1 - p)\geq10\), the sampling distribution of \(\hat{p}\) is approximately normal.

Step2: Calculate the mean of the sampling distribution

The mean of the sampling distribution of \(\hat{p}\) is \(\mu_{\hat{p}}=p\). Given \(p = 0.35\), so \(\mu_{\hat{p}}=0.35\).

Step3: Calculate the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution of \(\hat{p}\) is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\). Substitute \(p = 0.35\) and \(n = 300\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

C. Approximately normal, \(\mu_{\hat{p}} = 0.35\), \(\sigma_{\hat{p}}=0.0280\)