Question

Based on past experience, a bank believes that 13% of the people who receive loans will not make payments on time. The bank has recently approved 500 loans. .Answer the following questions. What are the mean and standard deviation of the proportion of clients in this group who may not make timely payments? mu (p^) = SD(p^) = (Round to three decimal places as needed.) What assumptions underlie your model? Are the conditions met? With reasonable assumptions about the sample, all the conditions are met. The success/failure condition is not met. The 10% and success/failure conditions are not met. The 10% condition is not met. The randomization and 10% conditions are not met. The randomization condition is not met. The randomization and success/failure conditions are not met. Without unreasonable assumptions, none of the conditions are met. What is the probability that over 16% of these clients will not make timely payments? P(p^ > 0.16) = (Round to three decimal places as needed.)

Answer 1

Answer:

The sample proportion mean and standard deviation are

$\mu=p=0.13\\\\\sigma=\sqrt{\frac{p(1-p)}{N} }= \sqrt{\frac{0.13*0.87}{500} }=0.015$

With reasonable assumptions about the sample, all the conditions are met.

The probability that over 16% of these clients will not make timely payments is P=0.023.

Step-by-step explanation:

We have a sample of 500 loans, of which a proportion of p=0.13 is expected to not be paid on time.

The sample proportion mean and standard deviation are

$\mu=p=0.13\\\\\sigma=\sqrt{\frac{p(1-p)}{N} }= \sqrt{\frac{0.13*0.87}{500} }=0.015$

The assumptions underlying this model are:

- The sample is randomly selected from the population.

- The sampling distribution of p-hat is approximately normal. There are at least 10 successes and 10 failures in the sample.

- Individual observations are independent of each other.

With reasonable assumptions about the sample, all the conditions are met.

What is the probability that over 16% of these clients will not make timely payments?

$z=\frac{x-\mu}{\sigma} =\frac{0.16-0.13}{0.015}=\frac{0.03}{0.015}=2\\\\P(p>0.16)=P(z>2)=0.023$

The probability that over 16% of these clients will not make timely payments is P=0.023.