The bank has recently approved 200 loans and found that \( 9 \% \) did not make payments on time. Let \( p \) represent the proportion all loans issued by the bank that do not make payments on time. Use the sample to estimate the mean and standard error of the sampling distribution of the sample proportions. Round answers to 4 decimal places as needed. (a) mean \( = \) (b) standard error \( = \)

To find the mean and standard error of the sampling distribution of the sample proportions, we can follow these steps: (a) The mean of the sampling distribution of the sample proportions (also known as the expected proportion) is simply the sample proportion itself. Given that \(9\% = 0.09\), we have: Mean (\(p\)) = \(0.09\) So, rounded to four decimal places: Mean \( = 0.0900\) (b) The standard error (SE) for the sample proportion can be calculated using the formula: \[ SE = \sqrt{\frac{p(1 - p)}{n}} \] where \(p\) is the sample proportion and \(n\) is the sample size. Here, \(p = 0.09\) and \(n = 200\). Now, substituting the values: \[ SE = \sqrt{\frac{0.09 \times (1 - 0.09)}{200}} = \sqrt{\frac{0.09 \times 0.91}{200}} = \sqrt{\frac{0.0819}{200}} = \sqrt{0.0004095} \approx 0.0202 \] So, rounded to four decimal places: Standard error \( = 0.0202\) Final answers: (a) 0.0900 (b) 0.0202