Question

based on past experience, a bank believes that 8% of the people who receive loans will not make payments on time. the bank has recently approved 200 loans. a) what are the mean and standard deviation of the sampling distribution of the sample proportion? round your answers to three decimal places when needed. mean = standard deviation = b) what is the probability that over 10% of these clients will not make timely payments? round your answer to four decimal places.

Explanation:

Step1: Identify the population proportion and sample size

Let $p = 0.08$ (population proportion of non - timely payers) and $n=200$ (sample size of loans).

Step2: Calculate the mean of the sampling distribution of the sample proportion

The mean of the sampling distribution of the sample proportion $\hat{p}$ is equal to the population proportion $p$. So, $\mu_{\hat{p}}=p = 0.080$.

Step3: Calculate the standard deviation of the sampling distribution of the sample proportion

The formula for the standard deviation of the sampling distribution of the sample proportion is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. Substitute $p = 0.08$ and $n = 200$ into the formula:

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Step4: Standardize for part b

For part b, we want to find $P(\hat{p}>0.10)$. First, calculate the z - score using the formula $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}$. Substitute $\hat{p}=0.10$, $\mu_{\hat{p}} = 0.08$ and $\sigma_{\hat{p}}\approx0.019$ into the formula:

$$ z=\frac{0.10 - 0.08}{0.019}=\frac{0.02}{0.019}\approx1.053 $$

Then, $P(\hat{p}>0.10)=P(Z > 1.053)=1 - P(Z\leqslant1.053)$. From the standard normal table, $P(Z\leqslant1.053)\approx0.8531$. So, $P(\hat{p}>0.10)=1 - 0.8531 = 0.1469$.

Answer:

mean = 0.080
standard deviation = 0.019
b) 0.1469