From the image you provided, it appears you are working on a statistics problem related to sampling distributions. Let's break it down step by step.

Problem Overview:

1. Scenario:

   • 37% of adults do not own a credit card.
   • A simple random sample of 600 adults is obtained.

2. Questions:

   • (a) Find the standard deviation of the sampling distribution of the sample proportion $\hat{p}$.
   • (b) Find the probability that more than 41% of the sample do not own a credit card.
   • (c) Interpret the probability from part (b) and estimate the expected number of samples where more than 41% do not own a credit card in 100 random samples.

Let's work through it:

Part (a): Standard Deviation of the Sampling Distribution of $\hat{p}$

The formula for the standard deviation (often called the standard error) of the sampling distribution of the sample proportion $\hat{p}$ is:

$\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}$

Where:

• $p = 0.37$ (the population proportion of adults who do not own a credit card),
• $n = 600$ (the sample size).

Now, let's compute that:

$\sigma_{\hat{p}} = \sqrt{\frac{0.37(1 - 0.37)}{600}} = \sqrt{\frac{0.37 \times 0.63}{600}}$

I'll calculate that next.The standard deviation of the sampling distribution of $\hat{p}$ is approximately 0.020 (rounded to three decimal places).

Part (b): Probability that more than 41% of the sample do not own a credit card

To find this probability, we first convert the proportion to a z-score using the formula:

$z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}$

Where:

• $\hat{p} = 0.41$,
• $p = 0.37$,
• $\sigma_{\hat{p}} \approx 0.020$.

I'll calculate the z-score and then find the probability using the standard normal distribution.The z-score is approximately 2.03. Now, let's find the probability of getting a z-score greater than 2.03 using the standard normal distribution table (or a calculator). This corresponds to the area to the right of $z = 2.03$.

We can use a standard normal distribution to find this value.The probability that more than 41% of the sample do not own a credit card is approximately 0.0212 (or 2.12%).

Part (c): Interpretation and estimation for 100 random samples

Given that the probability is about 2.12%, we expect about 2.12% of the 100 random samples to result in more than 41% of adults not owning a credit card. To estimate this:

$\text{Expected number of samples} = 100 \times 0.0212 = 2.12$

Rounding to the nearest integer, we expect about 2 samples.

Would you like further clarification or additional details on any part? Here are some related questions to explore:

1. How would the standard deviation change if the sample size were increased?
2. What does a z-score of 2.03 imply in terms of the distribution?
3. What would be the probability if we asked about less than 41% of the sample not owning a credit card?
4. How do we interpret the sampling distribution of proportions in real-world contexts?
5. If the population proportion were different, how would that impact the probability calculation?

Tip: Always check the context of proportions in statistics, especially in large samples. Small changes can have a significant impact on the probabilities.