According to a 2009 Reader's Digest article, people throw away about 10% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 85 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.2?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer = 
 (Enter your answer as a number accurate to 4 decimal places.)

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 \( n(1-p) \geq 10 \).

According to a 2009 Reader's Digest article, people throw away about 10% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 85 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.2?

This detailed solution explains how to calculate the probability that the sample proportion does not exceed 0.2, given a true proportion of 0.1 and a sample size of 85. It includes key concepts such as the standard error, z-score calculation, and the use of the normal approximation for proportions. Suitable for advanced high school statistics students.

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