It was reported that 67% of college students in 2008 graduated with student loan debt. A random sample of 85 2008 graduates is drawn. Find the probability that between 65% and 80% of the people in the sample were in debt.

Question

Please round to the nearest thousandth.

Answer 1

To solve this problem, we can use the normal approximation to the binomial distribution.

Let $p=0.67$ be the population proportion of college students who graduated with student loan debt. The sample size $n=85$.

1. Calculate the mean and standard deviation of the sample proportion:

   The mean $\mu$ of the sample proportion $\hat{p}$ is given by: $$\mu =p=0.67$$

   The standard deviation $\sigma$ of the sample proportion $\hat{p}$ is given by: $$\sigma =\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.67\times (1-0.67)}{85}}=\sqrt{\frac{0.67\times 0.33}{85}}\approx \sqrt{\frac{0.2211}{85}}\approx \sqrt{0.0026}\approx 0.051$$

2. Convert the sample proportion boundaries into standard normal variables (z-scores):

   We need to find the z-scores corresponding to 65% (0.65) and 80% (0.80) of the sample:

   $$z_1=\frac{0.65-\mu }{\sigma }=\frac{0.65-0.67}{0.051}\approx \frac{-0.02}{0.051}\approx -0.392$$

   $$z_2=\frac{0.80-\mu }{\sigma }=\frac{0.80-0.67}{0.051}\approx \frac{0.13}{0.051}\approx 2.549$$

3. Find the probabilities associated with these z-scores using the standard normal distribution table:

   Using a standard normal distribution table (or a calculator):

   • For $z_1\approx -0.392$, the corresponding probability $P(Z<-0.392)\approx 0.3485$.
   • For $z_2\approx 2.549$, the corresponding probability $P(Z<2.549)\approx 0.9946$.

4. Calculate the probability that the sample proportion is between 0.65 and 0.80:

   We need to find: $$P(0.65<\hat{p}<0.80)=P(Z<2.549)-P(Z<-0.392)\approx 0.9946-0.3485\approx 0.6461$$

5. Round the result to the nearest thousandth:

   Therefore, the probability that between 65% and 80% of the people in the sample were in debt is approximately $\boxed{{\displaystyle 0.646}}$.