We are tasked with finding the probability that the sample proportion $\hat{p}$ does not exceed 0.2, given that the true population proportion is $p = 0.1$. We will use the normal approximation for the binomial distribution since we are dealing with a sample of $n = 85$, and then calculate the required probability.

Step 1: Identify the known parameters

• Population proportion $p = 0.1$
• Sample size $n = 85$
• Sample proportion threshold $\hat{p} = 0.2$

Step 2: Calculate the mean and standard deviation of the sampling distribution of $\hat{p}$

The sampling distribution of $\hat{p}$ is approximately normal with:

• Mean $\mu_{\hat{p}} = p = 0.1$
• Standard deviation (standard error) $\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}$

Now, calculate the standard error:

$\sigma_{\hat{p}} = \sqrt{\frac{0.1(1 - 0.1)}{85}} = \sqrt{\frac{0.1 \times 0.9}{85}} = \sqrt{\frac{0.09}{85}} = \sqrt{0.00105882} \approx 0.0325$

Step 3: Convert the problem to a z-score

We want to find the probability that $\hat{p} \leq 0.2$. To do this, we first convert the sample proportion $\hat{p} = 0.2$ into a z-score using the formula:

$z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}$

Substituting the values:

$z = \frac{0.2 - 0.1}{0.0325} = \frac{0.1}{0.0325} \approx 3.0769$

Step 4: Find the probability corresponding to the z-score

Using standard normal distribution tables or a calculator, the cumulative probability for $z = 3.0769$ is approximately 0.9989.

Step 5: Conclusion

The probability that the sample proportion does not exceed 0.2 is approximately 0.9989.

Final Answer:

$\boxed{0.9989}$

Would you like further details or clarifications on any part of this solution?

Related Questions:

1. What is the probability that the sample proportion exceeds 0.2?
2. How would the result change if the sample size was 100 instead of 85?
3. What is the z-score if the sample proportion is 0.15 instead of 0.2?
4. What assumptions are made in applying the normal approximation in this context?
5. How would you interpret a z-score of 3.0769 in terms of standard deviations?

Tip:

When using the normal approximation for proportions, ensure the sample size is large enough by verifying that $np \geq 10$ and $n(1-p) \geq 10$.