https://file.aiquickdraw.com/mathbot/file/172740899356025k73tsd.jpg

Let's break down the question step by step:

### Given Information:
- 19% of adults in Belize do not suffer from excess weight.
- Sample size \( n = 100 \).
- Proportion of adults who do not suffer from weight-loss woes \( p = 0.19 \).

### (a) Distribution of the Sample Proportion:
To determine if the sample proportion \( \hat{p} \) has an approximate normal distribution, we use the Central Limit Theorem (CLT). The sample proportion \( \hat{p} \) will be approximately normally distributed if:

\[
np \geq 10 \quad \text{and} \quad n(1 - p) \geq 10
\]

Let’s check the conditions:

\[
np = 100 \times 0.19 = 19
\]
\[
n(1 - p) = 100 \times (1 - 0.19) = 100 \times 0.81 = 81
\]

Since both 19 and 81 are greater than 10, the sample proportion can be approximated by a normal distribution.

#### Mean and Standard Deviation of the Sampling Distribution:
The mean \( \mu_{\hat{p}} \) and standard deviation \( \sigma_{\hat{p}} \) of the sample proportion are given by:

\[
\mu_{\hat{p}} = p = 0.19
\]
\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.19 \times 0.81}{100}} = \sqrt{\frac{0.1539}{100}} = \sqrt{0.001539} \approx 0.03922
\]

Thus:
- Mean \( \mu_{\hat{p}} = 0.19 \)
- Standard deviation \( \sigma_{\hat{p}} \approx 0.03922 \)

### (b) Probability that the Sample Proportion Exceeds 0.25:
We are asked to find the probability that \( \hat{p} \) exceeds 0.25. We first need to convert this to a standard normal (z) score:

\[
z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.25 - 0.19}{0.03922} = \frac{0.06}{0.03922} \approx 1.53
\]

Next, we find the probability corresponding to \( z = 1.53 \) using standard normal distribution tables or a calculator.

From the z-table, the area to the left of \( z = 1.53 \) is approximately 0.9370. Therefore, the probability that \( \hat{p} \) exceeds 0.25 is:

\[
P(\hat{p} > 0.25) = 1 - 0.9370 = 0.0630
\]

So, the probability that the sample proportion exceeds 0.25 is approximately 0.0630, or 6.30%.

---

### Summary:
- **(a)** The sample proportion follows an approximate normal distribution with mean 0.19 and standard deviation 0.03922.
- **(b)** The probability that the sample proportion exceeds 0.25 is approximately 0.0630.

---

Would you like further details or have any questions?

Here are 5 related questions for you to explore:
1. How is the Central Limit Theorem used in statistical inference?
2. Why do we check the conditions \( np \geq 10 \) and \( n(1 - p) \geq 10 \) for normal approximation?
3. What is the significance of the z-score in standard normal distributions?
4. How does sample size affect the shape of the sampling distribution of proportions?
5. What are some applications of the normal distribution in real-world statistics?

**Tip**: When solving probability questions involving proportions, always check if the conditions for normality are satisfied before proceeding with calculations.

News reports tell us that the average Belizean is overweight. Many of us have tried to trim down to our weight when we finished high school. And, in fact, only 19% of adults say they do not suffer from weight-loss woes. Suppose that the 19% figure is correct, and that a random sample of 100 adults is selected.

a) Does the distribution of \( \hat{p} \), the sample proportion of adults who do not suffer from excess weight, have an approximate normal distribution? If so, what is its mean and standard deviation?

b) What is the probability that the sample proportion \( \hat{p} \) exceeds 0.25?

Explore the sampling distribution for a sample of Belizean adults regarding weight-loss woes. The problem involves calculating the mean and standard deviation of the sample proportion and finding the probability of the sample proportion exceeding 0.25.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation