Save $K$ A recent study reported that $30 \%$ of the residents of a particular community lived in poverty. Suppose a random sample of 200 residents of this community is taken. We wish to determine the probability that $36 \%$ or more of our sample will be living in poverty. Complete parts (a) and (b) below. a. Before doing any calculations, determine whether this probability is greater than $50 \%$ or less than $50 \%$. Why? A. The answer should be greater than $50 \%$, because the resulting z-score will be positive and the sampling distribution is approximately Normal. B. The answer should be less than $50 \%$, because 0.36 is greater than the population proportion of 0.30 and because the sampling distribution is approximately Normal. C. The answer should be less than $50 \%$, because the resulting z-score will be negative and the sampling distribution is approximately Normal. D. The answer should be greater than $50 \%$, because 0.36 is greater than the population proportion of 0.30 and because the sampling distribution is approximately Normal b. Calculate the probability that $36 \%$ or more of the sample will be living in poverty. Assume the sample is collected in such a way that the conditions for using the CLT are met $P(\hat{p} \geq 0.36)=\square$ (Round to three decimal places as needed )

Step 1 ：The problem is asking for the probability that 36% or more of a sample of 200 residents will be living in poverty, given that the population proportion of residents living in poverty is 30%. This is a problem of sampling distribution of proportions, and we can use the Central Limit Theorem (CLT) to solve it.

Step 2 ：First, we need to calculate the z-score for the sample proportion of 0.36. The z-score is calculated as \((\text{sample proportion} - \text{population proportion}) / \text{standard deviation of the sampling distribution}\). The standard deviation of the sampling distribution (also known as the standard error) can be calculated as \(\sqrt{(p(1-p))/n}\), where p is the population proportion and n is the sample size.

Step 3 ：Let's start by calculating the z-score. The population proportion \(p = 0.3\), the sample size \(n = 200\), and the sample proportion \(p_{\text{sample}} = 0.36\). The standard error \(se = \sqrt{(p(1-p))/n} = 0.0324037034920393\). The z-score \(z = (p_{\text{sample}} - p) / se = 1.851640199545103\).

Step 4 ：The calculated z-score is approximately 1.85. This means that the sample proportion of 0.36 is 1.85 standard deviations above the mean of the sampling distribution.

Step 5 ：Now, we need to find the probability that the z-score is greater than or equal to 1.85. This is equivalent to finding the area to the right of 1.85 in the standard normal distribution.

Step 6 ：The calculated probability is approximately 0.032. This means that there is a 3.2% chance that 36% or more of a sample of 200 residents will be living in poverty, given that the population proportion of residents living in poverty is 30%.

Step 7 ：Final Answer: \(P(\hat{p} \geq 0.36)=\boxed{0.032}\) (rounded to three decimal places)