The sampling distribution of \(\hat{p}\), the sample proportion of adults who do not own a credit card, is approximately normal. This conclusion is based on the condition that \(n \cdot p \cdot (1 - p) \geq 10\). Specifically, we have:
\[
n = 400, \quad p = 0.32 \quad \Rightarrow \quad n \cdot p \cdot (1 - p) = 400 \cdot 0.32 \cdot (1 - 0.32) = 86.4 \geq 10
\]
Thus, the shape of the sampling distribution is described as:
\[
\text{D. Approximately normal because } n \leq 0.05N \text{ and } np(1-p) \geq 10
\]
The mean of the sampling distribution of \(\hat{p}\) is given by the population proportion \(p\):
\[
\mu_{\hat{p}} = p = 0.32
\]
The standard deviation of the sampling distribution of \(\hat{p}\) is calculated using the formula:
\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.32 \cdot (1 - 0.32)}{400}} \approx 0.023
\]
To find the probability that more than \(34\%\) of adults do not own a credit card, we calculate:
\[
P(\hat{p} > 0.34)
\]
Using the Z-score transformation, we find:
\[
Z_{end} = \frac{0.34 - 0.32}{0.023} \approx 0.8696
\]
The probability is calculated as:
\[
P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(17.1499) = 0.0
\]
Thus, the probability that more than \(34\%\) do not own a credit card is:
\[
P(\hat{p} > 0.34) = 1.0
\]
If \(100\) different random samples of \(400\) adults were obtained, we would expect:
\[
\text{Expected number of samples with } \hat{p} > 0.34 = 100 \cdot P(\hat{p} > 0.34) = 100 \cdot 1.0 = 100
\]
- Shape of the sampling distribution: D. Approximately normal because \(n \leq 0.05N\) and \(np(1-p) \geq 10\)
- Mean of the sampling distribution of \(\hat{p}\): \(0.32\)
- Standard deviation of the sampling distribution of \(\hat{p}\): \(0.023\)
- Probability that more than \(34\%\) do not own a credit card: \(1.0\)
- Expected number of samples with more than \(34\%\) not owning a credit card: \(100\)
\[
\boxed{\text{Shape: D, Mean: 0.32, Std Dev: 0.023, Probability: 1.0, Expected Samples: 100}}
\]