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 \( n = 85 \).
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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: \[ \sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.67 \times (1 - 0.67)}{85}} = \sqrt{\frac{0.67 \times 0.33}{85}} \approx \sqrt{\frac{0.2211}{85}} \approx \sqrt{0.0026} \approx 0.051 \]
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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:
\[ z_1 = \frac{0.65 - \mu}{\sigma} = \frac{0.65 - 0.67}{0.051} \approx \frac{-0.02}{0.051} \approx -0.392 \]
\[ z_2 = \frac{0.80 - \mu}{\sigma} = \frac{0.80 - 0.67}{0.051} \approx \frac{0.13}{0.051} \approx 2.549 \]
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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 \( P(Z < 2.549) \approx 0.9946 \).
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Calculate the probability that the sample proportion is between 0.65 and 0.80:
We need to find: \[ P(0.65 < \hat{p} < 0.80) = P(Z < 2.549) - P(Z < -0.392) \approx 0.9946 - 0.3485 \approx 0.6461 \]
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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 \( \boxed{0.646} \).