Based on past experience, a bank believes that \( 11 \% \) of the people who receive loans will not make payments on time. The bank has recently approved 500 loans, which are a random representative sample. Complete parts a through c. \( \mu(\hat{p})=.11 \) SD \( (\hat{p})=.014 \) (Round to three decimal places as needed.) b) What assumptions underlie your model? Are the conditions met? A. With reasonable assumptions about the sample, all the conditions are met. B. The randomization condition is not met. C. The success/failure condition is not met. D. The \( 10 \% \) and success/failure conditions are not met. E. The \( 10 \% \) condition is not met. F. The randomization and success/failure conditions are not met. G. The randomization and \( 10 \% \) conditions are not met. H. Without unreasonable assumptions, none of the conditions are met. c) What is the probability that over \( 13 \% \) of these clients will not make timely payments? P( \( \hat{p}>0.13)=\square \) (Round to three decimal places as needed.)

**b)** To use the normal approximation for the sampling distribution of the sample proportion \( \hat{p} \), we must check the following conditions: 1. **Randomization:** The sample of 500 loans is stated to be a random representative sample. 2. **10% Condition:** The sample size \( n=500 \) must be less than 10% of the population of all loans. Assuming this is true, this condition is met. 3. **Success/Failure Condition:** We need both \( np \) and \( n(1-p) \) to be at least 10. - \( np = 500 \times 0.11 = 55 \) - \( n(1-p) = 500 \times 0.89 = 445 \) Since both numbers are much larger than 10, this condition is met. Thus, all the conditions are met which corresponds with answer choice **A**. --- **c)** We need to find the probability that over \( 13\% \) (i.e., \( \hat{p} > 0.13 \)) of clients will not make timely payments. The sampling distribution of \( \hat{p} \) is approximately normal with: - Mean \( \mu_{\hat{p}} = 0.11 \) - Standard deviation \( \sigma_{\hat{p}} = 0.014 \) Step 1. Compute the \( z \)-score for \( \hat{p} = 0.13 \): \[ z = \frac{0.13 - 0.11}{0.014} = \frac{0.02}{0.014} \approx 1.429 \] Step 2. Find the probability \( P(Z > 1.429) \). Using standard normal tables or a calculator: \[ P(Z \leq 1.429) \approx 0.9236 \] \[ P(Z > 1.429) = 1 - 0.9236 \approx 0.0764 \] Thus, the probability that over \( 13\% \) of the clients will not make timely payments is approximately: \[ P(\hat{p} > 0.13) \approx 0.076 \]