The 2010 census found that 13.9% of adults in the United States identified themselves as of Hispanic origin.  An opinion poll plans to interview 1500 adults at random. What are the mean and standard deviation of the sampling distribution of the proportion of individuals of Hispanic origin for samples of this size? Use a normal approximation to find the probability that such a sample will contain 12% or fewer individuals of hispanic origin.

To solve this problem, we will use the properties of the sampling distribution of a sample proportion. Let's break down the steps to find the mean, standard deviation, and the probability.

### Step 1: Find the mean and standard deviation of the sampling distribution

1. **Mean of the sampling distribution (μ\_p̂):**
   The mean of the sampling distribution of the sample proportion is equal to the population proportion, \( p \).

   \[
   \mu_{\hat{p}} = p = 0.139
   \]

2. **Standard deviation of the sampling distribution (σ\_p̂):**
   The standard deviation of the sampling distribution of the sample proportion is given by:

   \[
   \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}
   \]

   where:
   - \( p \) is the population proportion,
   - \( n \) is the sample size.

   Substituting the given values:

   \[
   \sigma_{\hat{p}} = \sqrt{\frac{0.139 \cdot (1 - 0.139)}{1500}}
   \]

   \[
   \sigma_{\hat{p}} = \sqrt{\frac{0.139 \cdot 0.861}{1500}}
   \]

   \[
   \sigma_{\hat{p}} = \sqrt{\frac{0.119679}{1500}}
   \]

   \[
   \sigma_{\hat{p}} = \sqrt{0.000079786}
   \]

   \[
   \sigma_{\hat{p}} \approx 0.00893
   \]

### Step 2: Use a normal approximation to find the probability

3. **Convert the problem to a standard normal distribution problem:**
   We need to find the probability that the sample proportion \( \hat{p} \) is 12% or fewer.

   \[
   P(\hat{p} \leq 0.12)
   \]

   First, convert \( \hat{p} \) to the corresponding z-score using the formula:

   \[
   z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}
   \]

   Substituting the values:

   \[
   z = \frac{0.12 - 0.139}{0.00893}
   \]

   \[
   z = \frac{-0.019}{0.00893}
   \]

   \[
   z \approx -2.13
   \]

4. **Find the probability corresponding to the z-score:**
   Using the standard normal distribution table, we find the probability that \( z \leq -2.13 \).

   \[
   P(z \leq -2.13) \approx 0.0166
   \]

### Conclusion

- The **mean** of the sampling distribution of the proportion of individuals of Hispanic origin is \( 0.139 \).
- The **standard deviation** of this sampling distribution is approximately \( 0.00893 \).
- The **probability** that a sample of 1500 adults will contain 12% or fewer individuals of Hispanic origin is approximately \( 0.0166 \) or 1.66%.

Would you like more details or have any questions about this calculation? Here are some related questions for further practice:

1. How would the mean of the sampling distribution change if the sample size increased to 2000?
2. What is the probability that the sample proportion will be between 13% and 15%?
3. How would the standard deviation of the sampling distribution change if the population proportion was 0.20?
4. If the sample size was reduced to 1000, what would the new standard deviation be?
5. What is the z-score for a sample proportion of 14%?
6. How do you interpret a z-score in the context of this problem?
7. Can you explain the central limit theorem and its relevance to this problem?
8. How would you verify the normality assumption in this context?

**Tip:** Always check if the sample size is large enough to use the normal approximation by ensuring \( np \) and \( n(1-p) \) are both greater than 10.

Explore the mean and standard deviation of the sampling distribution of the proportion of individuals of Hispanic origin for a sample size of 1500. Learn how to use a normal approximation to find the probability that such a sample will contain 12% or fewer individuals of Hispanic origin.

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