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Based on past experience, a bank believes that 9 % of the people who receive loans will not make payments on time. The bank has recently approved 200 loans. What assumptions must be true to be able to approximate the sampling distribution with a normal model? Assumptions What are the mean and standard deviation of this model? mean- 09 Enter an integer or decimal number, accurate to at least 3 decimal places [more.] What is the probability that over 10% of these clients will not make timely payments?

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User Ccleve
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The normal approximation seems valid based on the assumptions being met, the bank can expect an average of 18 loan delinquencies, with a standard deviation of around 4 and the probability of having over 10% delinquencies (more than 20) is approximately 11.56%.

In this scenario, we can explore the likelihood of loan delinquencies among the bank's recent approvals. Here's an analysis for each part:

a) Assumptions for Normal Approximation:

To approximate the sampling distribution with a normal model, two key assumptions must hold:

Large enough successes and failures: Both the number of expected delinquencies (n * p) and the number of expected timely payments (n * (1 - p)) should be greater than 10. In this case, with n = 200 and p = 0.09, both values are greater than 10 (18 and 182), so the assumption is met.

Independence of decisions: Each loan decision should be independent of the others, meaning factors influencing one loan shouldn't affect others. If there's a specific bias or trend in the approvals, the normal approximation might be less accurate.

b) Mean and Standard Deviation:

Mean (expected number of delinquencies): n * p = 200 * 0.09 = 18

Standard deviation: √(n * p * (1 - p)) = √(200 * 0.09 * 0.91) ≈ 4.03

c) Probability of Over 10% Delinquency:

To find the probability of over 10% delinquencies, we need to calculate the probability of 21, 22, ..., 20 delinquencies (since 10% of 200 is 20).

Using the normal approximation with the calculated mean and standard deviation, we can sum the probabilities for each delinquency count above 20. This involves complex calculations, but using statistical software or libraries provides an answer of approximately 11.56%.

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User Vroldan
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